MEASURE THEORY

ARIEL YADIN

Course: 201.1.0081 Fall 2014-15 Lecture notes updated: January 22, 2015 (partial solutions)

Contents

Lecture 1. Introduction 4 1.1. Measuring things 4 1.2. Elementary measure 5 This lecture has 6 exercises. 11

Lecture 2. Jordan measure 12 2.1. Jordan measure 12 This lecture has 15 exercises. 24

Lecture 3. Lebesgue outer measure 25 3.1. From finite to countable 25 This lecture has 5 exercises. 29

Lecture 4. Lebesgue measure 30 4.1. Definition of Lebesgue measure 30 4.2. Lebesgue measure as a measure 37 This lecture has 12 exercises. 43

Lecture 5. Abstract measures 44 5.1. σ-algebras 44 5.2. Measures 47 5.3. Fatou’s Lemma and continuity 50 This lecture has 13 exercises. 52

Lecture 6. Outer measures 53 1 2

6.1. Outer measures 53 6.2. Measurability 54 6.3. Pre-measures 57 This lecture has 12 exercises. 62

Lecture 7. Lebesgue-Stieltjes Theory 63 7.1. Lebesgue-Stieltjes measure 63 7.2. Regularity 68 7.3. Non-measureable sets 69 7.4. Cantor set 69 This lecture has 6 exercises. 70

Lecture 8. Functions of measure spaces 71 8.1. Products 71 8.2. Measurable functions 73 8.3. Simple functions 78 This lecture has 20 exercises. 80

Lecture 9. Integration: positive functions 81 9.1. Integration of simple functions 81 9.2. Integration of positive functions 83 This lecture has 7 exercises. 89

Lecture 10. Integration: general functions 90 10.1. Real valued functions 90 10.2. Complex valued functions 91 10.3. Convergence 93 10.4. Riemann vs. Lebesgue integration 97 This lecture has 10 exercises. 101

Lecture 11. Product measures 102 11.1. Sections 105 11.2. Product integrals 107 3

11.3. The Fubini-Tonelli Theorem 114 This lecture has 19 exercises. 120

Lecture 12. Change of variables 121 12.1. Linear transformations of Lebesgue measure 121 12.2. Change of variables formula 124 This lecture has 4 exercises. 128

Lecture 13. Lebesgue-Radon-Nykodim 129 13.1. Signed measures 129 13.2. The Lebesgue-Radon-Nikodym Theorem 136 This lecture has 27 exercises. 146

Lecture 14. Convergence 147 14.1. Modes of convergence 147 14.2. Uniform and almost uniform convergence 153 This lecture has 10 exercises. 157

Lecture 15. Differentiation 158 15.1. Hardy-Littlewood Maximal Theorem 158 15.2. The Lebesgue Differentiation Theorem 161 15.3. Differentiation and Radon-Nykodim derivative 162 This lecture has 6 exercises. 164

Lecture 16. The Riesz Representation Theorem 165 16.1. Compactly supported functions 165 16.2. Linear functionals 168 16.3. The Riesz Representation Theorem 169 This lecture has 5 exercises. 174

Total number of exercises: 177 174 4

Measure Theory

Ariel Yadin

Lecture 1: Introduction

1.1. Measuring things

Already the ancient Greeks developed a theory of how to measure length, area, and volume and area of 1, 2 and 3 dimensional objects. In this setting (i.e. in Rd for d 3) ≤ it stands to reason that the “size” or “measure” of an object must satisfy some basic axioms:

If m(A) is the measure of a set A, it should be the same for any reflection, • translation or rotation of A. That is, m(A) = m(A + x) = m(UA) where U is a rotation matrix. For example, [0, 1]2 should heave the same measure as [ 2, 1]2 which should n o − − have the same measure as (x, y): x + y 1 , the diamond of side-length | | | | ≤ √2 1. If A can be broken into disjoint pieces, the sum of their measures should be the • measure of A. That is, m(A B) = m(A) + m(B). ] Example: [0, 1] [0, 2] should have measure that is the sum of the measures × of [0, 1]2 and [0, 1] (1, 2]. ×

We use to denote disjoint union; that is, A B is not only notation for a set, X ] ] but this notation claims that A B = . The small + sign remind us of the additive ∩ ∅ property above. This is already quite fruitful. If the unit square (0, 1]2 is of measure 1, then:

We can determine the measure of any square of rational side length. • n X m((0, n]2) = m((j 1, j] (k 1, k]) = n2 − × − j,k=1 5

and n 2 X j 1 j k 1 k 2 1 2 m((0, 1] ) = m(( − , ] ( − , ]) = n m((0, ] ). n n × n n n j,k=1 We can measure any rectangle of rational side length: decompose it into squares. • We can measure right-angle triangles: disjoint union of two is a rectangle. • We can then measure any triangle, by bounding it in a rectangle and subtracting • the excess right-angle triangles. With triangles we can measure any polygon. •

X What about measuring a disc?

1.2. Elementary measure

Let us first formally define the above.

Definition 1.1 (Boxes). Consider Rd for some d 1. • ≥ I R is an interval if I is one of [a, b], (a, b), (a, b], [a, b) for some < a b < • ⊂ −∞ ≤ . Note that we allow the singleton a = [a, a] as an interval, and the empty set ∞ { } = (a, a). The length or measure of such I is defined to be I = m(I) = b a. ∅ | | − d B R is a box if B = I1 Id where Ij are intervals. The volume or • ⊂ × · · · × measure of such a box B is defined to be B = m(B) = I I . | | | 1| · · · | d| d E R is an elementary set if E = B1 Bn for some finite number of • ⊂ ∪ · · · ∪ boxes. d d 0 = 0(R ) denotes the set of elementary sets in R . •E E

Exercise 1.1. Show that is closed under finite unions, finite intersections, E0 set-difference, symmetric difference and translations. That is, show that if E,F are elementary sets then so are:

E F and E F , • ∪ ∩ E F := x E : x F = E F c, • \ { ∈ 6∈ } ∩ E F := (E F ) (F E), • 4 \ ∪ \ 6

E + x := y + x : y E . • { ∈ }

Exercise 1.2. Show that if E is an elementary set then there exist B1,...,Bn pairwise disjoint boxes such that E = B B . 1 ]···] n

It would be tempting to define the measure of E = B B as m(B )+ +m(B ). 1]···] n 1 ··· n But we are not guarantied that the decomposition is a unique one.

Proposition 1.2 (Discretisation Formula). Let I be an interval. Then, • 1 m(I) = lim #(I 1 ). n n Z →∞ n ∩ Consequently, if B Rd is a box then, ⊂ 1 m(B) = lim #(B 1 d), n d n Z →∞ n ∩ d and if E = B1 Bn R is an elementary set then ]···] ⊂ 1 1 d m(B1) + + m(Bn) = lim #(E ). n d n Z ··· →∞ n ∩ Proof. Let a < b and I be an interval with endpoints a, b. For n large enough, note that

1  z #(I Z) = # z Z : I # z Z : na z nb ∩ n ∈ n ∈ ≤ { ∈ ≤ ≤ } # z Z : na z nb nb + 1 (na 1) + 1 = n(b a) + 3. ≤ { ∈ b c ≤ ≤ d e} ≤ − − − Similarly,

#(I 1 Z) # z Z : na < z < nb n(b a) 1. ∩ n ≥ { ∈ } ≥ − − Dividing by n and taking n we obtain the formula for intervals. → ∞ Now, if B = I I then 1 × · · · × d 1 d 1 1 B Z = (I1 Z) (Id Z), ∩ n ∩ n × · · · × ∩ n 7 which implies that d 1 d Y 1 #(B Z ) = #(Ij Z). ∩ n ∩ n j=1 Dividing by nd and taking n gives the formula for boxes. → ∞ The final assertion is a consequence of the fact that if A B = then ∩ ∅ d d d (A B) 1 Z = (A 1 Z ) (B 1 Z ). ] ∩ n ∩ n ] ∩ n

ut

Definition 1.3. If E = B B is an elementary set we define the measure of • 1 ]···] n E to be 1 1 m(E) = m(B1) + + m(Bn) = lim #(E ), n d n Z ··· →∞ n ∩ which is well defined by Proposition ??.

Exercise 1.3. Give an example of a set A [0, 1] such that the limit ⊂  1 lim #(A 1 ) n n Z →∞ n ∩ does not exist.

1 1 lim #(A 1 ) and lim #((A + x) 1 ) n n Z n n Z →∞ n ∩ →∞ n ∩ both exist but are not equal.

Solution to ex:1.3. :( ♣ For the first example: Set

 2m 1 n A = n− : Z n, m 1 , 2m 1 2 . 2 3 ≥ − ≤ Since for any z 1 we can write z = (2m 1)2k for some k 0 and m 1, we have ≥ − ≥ ≥ that

n  z n n #(A 2− Z) # n : 1 z 2 = 2 . ∩ ≥ 2 ≤ ≤ 8

So

1 1 n n lim sup n #(A n Z) lim sup 2− #(A 2− Z) = 1. n ∩ ≥ n ∩ →∞ →∞ k 2m 1 z On the other hand, if x A 3− Z then for some m, n 1 and z Z, x = n− = k ∈ ∩ ≥ ∈ 2 3 which implies that 2nz = 3k(2m 1). The left hand side is even while the right hand − k side is odd, which is a contradiction. So A 3− Z = for all k, and we conclude that ∩ ∅ 1 1 n n lim inf #(A ) lim inf 3− #(A 3− ) = 0. n n n Z n Z →∞ ∩ ≤ →∞ ∩ For the second example take A = Q [0, 1] and x = π.:) ∩ X This last exercise shows that we do not want to use the discretization formula to define the measure of general sets, but we can use it for more than just elementary sets.

Exercise 1.4. Show that the measure of elementary sets has the following prop- erties.

m : [0, ). • E0 → ∞ (Additivity) m(E F ) = m(E) + m(F ). • ] m( ) = 0. • ∅ If B is a box, then m(B) = B . • | | (Monotonicity) If E F then m(E) m(F ). • ⊂ ≤ (Subadditivity) m(E F ) m(E) + m(F ). • ∪ ≤ (Translation invariance) For any x Rd, m(E + x) = m(E). • ∈

Solution to ex:1.4. :( ♣ The discretization formula guaranties that m is non-negative and additive, m( ) = 0 ∅ and also that if B is a box then m(B) = B . | | For monotonicity recall that F E is elementary, and F = (F E) (F E). Since \ ∩ ] \ E F we have that F E = E. So m(F ) = m(E) + m(F E) m(E). ⊂ ∩ \ ≥ Subadditivity follows from E F = E (F E) so ∪ ] \ m(E F ) = m(E) + m(F E) m(E) + m(F ). ∪ \ ≤ 9

Translation invariance is another consequence of the discretization formula: For any interval I with endpoints a < b, we have that z x I if and only if z nI +nx. Thus, n − ∈ ∈ if z x I then z [ na + nx , nb + nx ] Z and if z [ na + nx , nb + nx ] Z n − ∈ ∈ b c d e ∩ ∈ d e b c ∩ then z x I. Since n − ∈ #([ na+nx , nb+nx ] Z) n(b a)+3 and #([ na+nx , nb+nx ] Z) n(b a) 1, b c d e ∩ ≤ − d e b c ∩ ≥ − − we have

 z # z Z : x I n(b a) 3. ∈ n − ∈ − − ≤ Dividing by n and taking a limit we get that

m(I + x) = lim 1 # z : z x I = b a = m(I). n n Z n →∞ ∈ − ∈ − d d If B = I1 Id is a box in R , then for any x R , B+x = (I1+x1) (Id+xd), ×· · ·× ∈ ×· · ·× Qd Qd so m(B + x) = j=1 m(Ij + xj) = j=1 m(I) = m(B). Finally, if E = B B is an elementary set, then E +x = (B +x) (B +x). 1 ]···] k 1 ]···] k So m(E + x) = m(E).:) X

Exercise 1.5. Show that if m : [0, ) is additive and translation invariant, 0 E0 → ∞ i.e. m(E F ) = m(E)+m(F ) and m(E +x) = m(E), then there exists a constant c 0 ] ≥ such that m0 = cm.

Solution to ex:1.5. :( ♣ If E = B B is an elementary set then m (E) = m (B ) + + m (B ). So we 1 ]···] k 0 0 1 ··· 0 k only need to show that m (B) = c B for any box B. 0 | | First, note that since n n ] ] [0, 1) = 1 [j 1, j) and [0, n ) = 1 [j 1, j), n − k k − j=1 j=1 additivity and translation invariance guaranty that

1 d 1 d d d m0([0, ) ) = m0([0, 1) ) and m0([0, q) ) = q m0([0, 1)) n nd · 10 for any q Q (make sure to fill in the details here). By translation invariance, for any ∈ aj bj Q we then have ≤ ∈ d Y d m0([a , b ) [a , b )) = (b a ) m0([0, 1) ). 1 1 × · · · × d d j − j · j=1

d Set C = m0([0, 1) ). Let B = I I . Assume that the endpoints of I are a < b . 1 × · · · × d j j j + + Fix ε > 0 such that for all j we have b a > 2ε. Let a−, a , b−, b be rational j − j j j j j + + numbers such that a (a , a + ε), a− (a ε, a ), b (b , b + ε), b− (b ε, b ). j ∈ j j j ∈ j − j j ∈ j j j ∈ j − j + + + + Let I = [a−, b ) and I− = [a , b−). Let B = I± I±, so B B B . j j j j j j ± 1 × · · · × d − ⊂ ⊂ Since m0 is additive and non-negative, it is also monotone. Monotonicity tells us that m (B ) m (B) m (B+). 0 − ≤ 0 ≤ 0 Because a±, b± Q we get that j j ∈ d d + Y + Y m0(B ) = (b a−) C (b a + 2ε) C, j − j · ≤ j − j · j=1 j=1

d d Y + Y m0(B−) = (b− a ) C (b a 2ε) C. j − j · ≥ j − j − · j=1 j=1 Setting M = max (b a ) we have that j j − j d + Y d 1 m0(B ) (b a ) C C 2ε dM − , − j − j · ≤ · · j 1 − d Y d 1 (b a ) C m0(B−) C 2ε dM − . j − j · − ≤ · · j 1 − Taking ε 0 we get that → d Y d m0(B) = (b a ) C = m0([0, 1) ) m(B). j − j · · j=1

:) X

Exercise 1.6. Let E Rd and F Rk be elementary sets. Show that E F is ⊂ ⊂ ×  d+k elementary (in R ) and that m(E F ) = m(E) m(F ). × · 11

Number of exercises in lecture: 6 Total number of exercises until here: 6 12

Measure Theory

Ariel Yadin

Lecture 2: Jordan measure

2.1. Jordan measure

We have seen that the measure of elementary sets is a good way to measure length, area and volume for squares and rectangles, and anything that can be composed of finite unions of such. How about measuring a triangle? Or a circle? We saw that the discretisation formula also has its limitations. However, Jordan measure is essentially the way to use the discretisation formula. Jordan measure is essentially approximating general shapes by elementary ones.

Definition 2.1 (Jordan measure). Let A Rd be a bounded set (i.e. A B(0, r) := • ⊂ ⊂ x : x r ). Define: { | | ≤ } The Jordan inner measure of A • J (A) := sup m(A). ∗ 0 E A E 3 ⊂ The Jordan outer measure of A •

J ∗(A) := inf m(F ). 0 F A E 3 ⊃

If A is such that J (A) = J ∗(A) we say that A is Jordan measurable. We then • ∗ define the Jordan measure of A as this common value, m(A) = J (A) = J ∗(A). ∗

Exercise 2.1. Show that J ∗,J are monotone. Show that J ∗ is subadditive. ∗ 

Solution to ex:2.1. :( ♣ Let A C. Let (E ) , (F ) be sequences of elementary sets such that E A, C F ⊂ n n n n n ⊂ ⊂ n 13 and m(En) J (A), m(Fn) J ∗(C). Note that for every n, → ∗ →

J (C) m(En) J (A) and J ∗(A) m(Fn) J ∗(C). ∗ ≥ → ∗ ≤ → Also, if (G ) is a sequence of elementary sets such that A G and m(G ) J (A), n n ⊂ n n → ∗ then A C G F so ∪ ⊂ n ∪ n

J ∗(A C) m(G F ) m(G ) + m(F ) J ∗(A) + J ∗(C). ∪ ≤ n ∪ n ≤ n n →

:) X

Note that every elementary set E is Jordan measurable and m(E) = J ∗(E) = J (E). ∗ Jordan sets are those which are “almost elementary”.

Exercise 2.2. Show that the following are equivalent for bounded A Rd. ⊂  A is Jordan measurable. • For every ε > 0 there exist elementary sets E A F , E,F , such that • ⊂ ⊂ ∈ E0 m(F E) < ε. \ For every ε > 0 there exists an elementary set E A such that J (A E) < • E0 3 ⊂ ∗ \ ε.

Solution to ex:2.2. :( ♣ For any ε > 0 there exist elementary sets E A F such that m(E) > J (A) ε and ⊂ ⊂ ∗ − m(F ) < J ∗(A) + ε. So if A is Jordan measurable then m(F ) < J ∗(A) + ε = J (A) + ε < ∗ m(E) + 2ε. Since E F we have that m(F ) = m(E) + m(F E), so m(F E) < 2ε. ⊂ \ \ Now assume that there exist E,F , E A F such that m(F E) < ε. Then, ∈ E0 ⊂ ⊂ \ A E F E, so J (A E) m(F E) < ε. \ ⊂ \ ∗ \ ≤ \ Finally assume that for every ε > 0 there exists an elementary set E A such that ε ⊂ J ∗(A Eε) < ε. Since Eε A we have that m(Eε) J (A). Using the subadditivity of \ ⊂ ≤ ∗ J ∗ we get

J ∗(A) J ∗(A Eε) + m(Eε) < ε + J (A). ≤ \ ∗ 14

Taking ε 0 and recalling that J (A) J ∗(A) by definition, we have that J ∗(A) = → ∗ ≤ J (A) and aso A is jordan measurable.:) X ∗

Exercise 2.3. Show that for Jordan measurable sets A, C the following holds.  A C,A C,A C,A C are all Jordan measurable. • ∪ ∩ \ 4 m(A) 0. • ≥ (Additivity) If A C = then m(A C) = m(A) + m(C). • ∩ ∅ ] (Monotonicity) If C A then m(C) m(A). • ⊂ ≤ (Subadditivity) m(A C) m(A) + m(C). • ∪ ≤ (Translation invariance) m(A + x) = m(A). •

Exercise 2.4. Show that a bounded set A is Jordan measurable if and only if for every ε > 0 there exists an elementary set E such that A E and J (E A) < ε. ⊂ ∗ \

Solution to ex:2.4. :( ♣ Let B be a box containing the bounded set A. Suppose that there exists an elementary set F B A such that J ((B A) F ) < ε. ⊂ \ ∗ \ \ Then, with E = B F we have that A E and J (E A) = J ((B A) F ) < ε. \ ⊂ ∗ \ ∗ \ \ Thus we obtain that A is Jordan measurable iff B A is Jordan measurable iff for \ every ε > 0 there exists an elementary set F , F B A, such that J ((B A) F ) < ε ⊂ \ ∗ \ \ iff for every ε > 0 there exists an elementary set E, A E, such that J (E A) < ε.:) ⊂ ∗ \ X

Exercise 2.5. [Tao, ex. 1.1.7] Let B Rd be a closed box, and let f : B R be a ⊂ → continuous function. Show that

The graph (x, f(x)) : x B is Jordan measurable with Jordan measure 0. • { ∈ } 15

The volume under the graph (x, t): x B, 0 t f(x) is Jordan measur- • { ∈ ≤ ≤ } able.

Solution to ex:2.5. :( ♣ f is uniformly continuous in B; that is, for every ε > 0 there exists δ > 0 such that for any x y < δ we have f(x) f(y) < ε. || − || | − | SN Write B = n=1 Bδ(xn) where (Bδ(xn))n are disjoint boxes of diameter less than δ, with x the center of B . Uniform continuity of f gives that for any x B there exists (a n δ ∈ unique) n such that x B (x ), and so (x, f(x)) B (x ) (f(x ) ε, f(x )+ε) =: Q . ∈ δ n ∈ δ n × n − n δ,n Thus, (x, f(x)) : x B S Q . Since m(Q ) = m(B (x )) 2ε, we have that { ∈ } ⊂ δ,n δ,n δ n · X X J ∗( (x, f(x)) : x B ) m(Q ) 2ε m(B (x )) = 2εm(B). { ∈ } ≤ δ,n ≤ δ n n n

Taking ε 0 gives the first assertion. → For the second assertion, let V = (x, t): x B, 0 t f(x) . Fix ε, δ > 0 and { ∈ ≤ ≤ } S B = n Bδ(x) as above. For every n set

mn = inf f(x) Mn = sup f(x). x Bδ(xn) x Bδ(xn) ∈ ∈ Note that M m ε by uniform continuity. Now, set K = mn . For 0 k | n − n| ≤ n b ε c ≤ ≤ K 1 set Q = B (x ) ε[k, k + 1) and Q˜ = B (x ) [εK ,M ]. n − n,k δ n × n δ n × n n For any n, if 0 k < K and (x, t) Q then x B (x ) and t < εK m f(x). ≤ n ∈ n,k ∈ δ n n ≤ n ≤ So

Kn 1 [ [− Q V. n,k ⊂ n k=0 On the other hand, if x B and 0 t f(x) then there exists n such that x B (x ) ∈ ≤ ≤ ∈ δ n and so f(x) [m ,M ]. Thus, either t < εK in which case (x, t) Q for some ∈ n n n ∈ n,k 0 k < K or εK t f(x) M in which case (x, t) Q˜ . Thus, ≤ n n ≤ ≤ ≤ n ∈ n

Kn 1 Kn 1 [ [− [ [− ][ Q V Q Q˜ . n,k ⊂ ⊂ n,k n n k=0 n k=0 n 16

S SKn 1 Now, let Λ := m( n k=0− Qn,k). Then, X X J ∗(V ) Λ + m(Q˜ ) = Λ + m(B (x )) (M εK ) ≤ n δ n · n − n n n X Λ + m(B (x )) (M m + ε) Λ + m(B) 2ε ≤ δ n · n − n ≤ · n J (V ) + m(B) 2ε. ≤ ∗ · Taking ε 0 completes the proof.:) → X

Exercise 2.6. [Tao, p. 31, ex. 1.2.6] Show that it is not true that  J ∗(E) = sup J ∗(U). E U open ⊃

Solution to ex:2.6. :( ♣ Take E = [0, 1] Q. If U E is an open set, then U = , because E does not contain \ ⊂ ∅ S any open intervals. Thus, supU E,U open m∗(U) = 0. However, if E n In where (In)n ⊂ ⊂ are disjoint intervals, then it must be that [0, 1] S I . So P m(I ) m([0, 1]) = 1. ⊂ n n n n ≥ Thus, m∗(E) = 1.:) X

Exercise 2.7. [Tao, p. 12, ex. 1.1.13] Prove the discretisation formula  1  1 d m(A) = lim d # A Z N N ∩ N →∞ for all Jordan measurable subsets A Rd. ⊂

Solution to ex:2.7. :( ♣ We know the discretisation formula holds for all elementary sets. For simplicity of the 1 1 d
presentation, let us denote IN (A) = d # A Z . N ∩ N 17

Let A be Jordan measurable, and let ε > 0. Let E A F be elementary sets such ⊂ ⊂ that m(F ) < m(A) + ε and m(E) > m(A) ε. Note that − d d d E 1 Z A 1 Z F 1 Z , ∩ N ⊂ ∩ N ⊂ ∩ N so I (E) I (A) I (F ). Taking N , N ≤ N ≤ N → ∞

m(A) < m(E) + ε = ε + lim IN (E) ε + lim IN (A) N ≤ N →∞ →∞ ε + lim IN (F ) = ε + m(F ) < 2ε + m(A). ≤ N →∞ Taking ε 0 we get the formula.:) → X

n Exercise 2.8. [Tao, p. 12, ex. 1.1.14] A dyadic cube of scale 2− is a box of the form:

 z1 z1+1
 zd zd+1
D (z , . . . , z ) := n , n n , n , n 1 d 2 2 × · · · × 2 2 d For n N and (z1, . . . , zd) Z . Note that these are half-open half-closed. ∈ ∈ n Let (A, n) be the number of dyadic cubes of scale 2− that are contained in A, and E∗ let (A, n) be the number of dyadic cubes of scale 2 n that intersect A. E∗ − Show that a bounded set A is Jordan measurable if and only if

dn lim 2− ( ∗(A, n) (A, n)) = 0. n ∗ →∞ E − E Show that Jordan measurable sets admit

dn dn m(A) = lim 2− ∗(A, n) = lim 2− (A, n). n n ∗ →∞ E →∞ E

Solution to ex:2.8. :( ♣ dn If limn 2− ( ∗(A, n) (A, n)) = 0: Let S (A, n) be the set of dyadic cubes of →∞ E − E∗ ∗ n scale 2− that are contained in A, and let S∗(A, n) be the set of dyadic cubes of scale n 2− that intersect A. Then, [ [ S (A, n) A S∗(A, n), ∗ ⊂ ⊂ 18 and these are elementary sets. Since dyadic cubes of the same scale are always disjoint, we get that

dn [ [ dn 2− (A, n) = m( S (A, n)) J (A) J ∗(A) m( S∗(A, n)) = 2− ∗(A, n). E∗ ∗ ≤ ∗ ≤ ≤ E

Thus, taking n we get that J ∗(A) = J (A). Moreover, we have that → ∞ ∗ dn dn m(A) = lim 2− ∗(A, n) = lim 2− (A, n). n n ∗ →∞ E →∞ E If A is Jordan measurable: Let B = I I be a box where I is the interval 1 × · · · × d j with endpoints aj < bj, define (B)n := [a 2 n, b + 2 n] [a 2 n, b + 2 n], the enlargement of B • 1 − − 1 − × · · · × d − − d − n by 2− in each coordinate direction. (B) := (a + 2 n, b 2 n) (a + 2 n, b 2 n), shrinking B by 2 n in • n 1 − 1 − − × · · · × d − d − − − each coordinate direction.

n Under these definitions, if a dyadic cube Dn(z) of scale 2− intersects B then it is n contained in (B) ; if Dn(z) is not contained in B then Dn(z) does not intersect (B)n. Finally note that

d d n Y n Y n m((B) ) m((B) ) = (b a + 2 2− ) (b a 2 2− ) − n j − j · − j − j − · j=1 j=1

X Y  n d S n d S  = (b a ) (2 2− ) −| | ( 2 2− ) −| | j − j · · − − · S 1,...,d j S ⊂{ } ∈ d n 2 4 2− m(B). ≤ · · · Fix ε > 0 and let E A F be elementary sets such that m(F ) m(A) + ε and ⊂ ⊂ ≤ m(E) m(A) ε. Suppose that E = Un E and F = Um F where E ,F are ≥ − k=1 k j=1 j n j boxes. n Let Dn(z) be a dyadic cube of scale 2− such that Dn(z) intersects A but is not contained in A. Then, there exists j such that Dn(z) intersects some Fj, so Dn(z) is n contained in (Fj) . Also, Dn(z) is not contained in Ek for all k, so for all k we get that

Dn(z) does not intersect (Ek)n. Thus, if we define n m ] n [ n (E)n := (Ek)n and (F ) := (Fj) k=1 j=1 19 then n n X X  d 2 n  d 2 n m((E) ) = m((E ) ) m(E ) 1 2 2 − = m(E) 1 2 2 − , n k n ≥ k · − · · − · k=1 k=1 m m n X n X  d 2 n  d 2 n m((F ) ) m((F ) ) m(F ) 1 + 2 2 − = m(F ) 1 + 2 2 − , ≤ j ≤ j · · · · j=1 j=1 dn dn and since for any Jordan measurable set J we have that 2− (J) m(J) 2− ∗(J), E∗ ≤ ≤ E dn dn n n 2− ( ∗(A, n) (A, n)) 2− ( ((F ) , n) ∗((E)n, n)) m((F ) ) m((E)n) E − E∗ ≤ E∗ − E ≤ − d 2 n m(F ) m(E) + 2 2 − (m(F ) + m(E)) ≤ − · · d 2 n ε + 2 2 − (2m(A) + ε). ≤ · · Taking n , we have that for all ε > 0, → ∞ dn 0 lim 2− ( ∗(A, n) (A, n)) ε, n ∗ ≤ →∞ E − E ≤ so this limit must be 0.:) X

Exercise 2.9. [Tao, p. 13, ex. 1.1.18] Show that for any bounded set A:  J (A¯) = J (A). • ∗ ∗ J (A◦) = J (A). • ∗ ∗ A is Jordan measurable if and only if J (∂A) = 0. • ∗

Solution to ex:2.9. :( ♣ First, A A¯ so J (A) J (A¯), and we only need to prove J (A¯) J (A). Let ε > 0. ⊂ ∗ ≤ ∗ ∗ ≤ ∗ Then choose disjoint boxes (B )N such that A S B and P m(B ) J (A) + ε. n n=1 ⊂ n n n n ≤ ∗ Note that [ [ A¯ B B¯ , ⊂ n ⊂ n n n and since B¯n are also boxes, X X J ∗(A¯) m(B¯ ) = m(B ) J ∗(A) + ε. ≤ n n ≤ n n 20

Taking ε 0 completes the proof of the first item. → For the second item, since A◦ A we only need to show that J (A) J (A◦). Fix ⊂ ∗ ≤ ∗ ε > 0 and let (B )N be a finite number of disjoint boxes such that S B A and n n=1 n n ⊂ P n m(Bn) J (A) ε. Then, since Bn◦ are disjoint boxes, and since m(Bn◦) = m(Bn), ≥ ∗ − from S B A we deduce that n n◦ ⊂ ◦

X X J (A◦) m(Bn◦) = m(Bn) J (A) ε. ∗ ≥ ≥ ∗ − n n

Taking ε 0 completes the second item. → For the final item: First, note that (∂A) = . If D (z) ∂A for some dyadic cube ◦ ∅ n ⊂ D (z), then (∂A) (D (z)) = , a contradiction. So ∂A contains no dyadic cubes, n ◦ ⊃ n ◦ 6 ∅ which is to say that (∂A, n) = 0 for all n. Thus, ∂A is Jordan measurable with E∗ J(∂A) = 0 if and only if 2 dn (∂A, n) 0. − E∗ → Note that if Dn(z) is a dyadic cube, then it intersects A and is not contained in A, if and only if it intersects ∂A. That is, ∗(∂A, n) = ∗(A, n) (A, n). E E − E∗ Now, by a previous exercise, A is Jordan measurable, if and only if 2 dn( (A, n) − E∗ − dn (A, n)) 0 if and only if 2− ∗(∂A, n) 0, if and only if ∂A is Jordan measurable E∗ → E → with J(∂A) = 0, if and only if J ∗(∂A) = 0.

The last step follows from the fact that if J ∗(∂A) = 0 then J (∂A) J ∗(∂A) = 0 ∗ ≤ and so ∂A is Jordan measurable with J(∂A) = 0.:) X

Exercise 2.10. Show that any triangle ABC in R2 is Jordan measurable.  2 Show that any compact convex polygon in R is Jordan measurable.

 Exercise 2.11. Show that the ball B(x, r) = x Rd : x < r is Jordan mea- ∈ | |  d surable with measure m(B(x, r)) = cdr where cd > 0 is a constant depending only on the dimension d. 21

Exercise 2.12. Show that if A Rd,C Rk are Jordan measurable, then A C ⊂ ⊂ × is Jordan measurable and m(A C) = m(A) m(C). × ·

Solution to ex:2.12. :( ♣ Fix ε > 0. Let E A F,E C F such that E,E ,F,F and m(F E) < ε, ⊂ ⊂ 0 ⊂ ⊂ 0 0 0 ∈ E0 \ m(F E ) < ε. Note that E E A C F F and E E ,F F are elementary 0 \ 0 × 0 ⊂ × ⊂ × 0 × 0 × 0 sets. Thus, since we assumed that A, C are Jordan measurable,

J ∗(A C) m(F F 0) = m(F ) m(F 0) < (m(E) + ε) (m(E0) + ε) (m(A) + ε) (m(C) + ε). × ≤ × · · ≤ ·

Also, since m(A) = m(A E) + m(E) m(F E) + m(E) and m(C) = m(C E ) + \ ≤ \ \ 0 m(E ) m(F E ) + m(E ), 0 ≤ 0 \ 0 0

J (A C) m(E E0) (m(A) ε) (m(C) ε). ∗ × ≥ × ≥ − · −

Taking ε 0 we have that J ∗(A C) = J (A C) = m(A) m(C).:) X → × ∗ × ·

Exercise 2.13. Show that if J ∗(A) = 0 then A is Jordan measurable. Show that if m(A) = 0 for Jordan measurable A, then any C A is Jordan measur- ⊂ able.

Exercise 2.14. give an example of a sequence (An)n of Jordan measurable sets  S such that A := n An is bounded but not Jordan measurable. 22

Theorem 2.2. Let L : Rd Rd be an invertible linear transformation. If A is ••• → Jordan measurable, then L(A) is also, and m(L(A)) = det L m(A). | |

Proof. If E is an elementary set, then L(E) is Jordan measurable. Indeed, recall that any invertible matrix L can be written as a product of elementary operation matrices: L = T T where each T is either multiplication of a row by a scalar c, addition of 1 ··· n j one row to another row, or swapping of two rows. That is, Tj is in one of the following families of matrices:

 1 0 0   1 0 0   1 0 0  ··· ··· . . ··· ......  . .. .   . .. .   . .  M = c R = 1 0 1 S =  0 1  c,j  ··· ···  i,j  ··· ··· ···  i,j  ··· 1 0 ···   . .. .   . .. .   . ··· .··· .  ...... 0 1 0 1 0 1 ··· ··· ··· That is Mc,j is obtain by multiplying the j-th row of the identity matrix by c; Ri,j is obtained by adding the i-th row in the identity matrix to the j-th row; Si,j is obtained by swapping the i-th and j-th rows of the identity matrix. It is immediate to compute that det M = c, det R = det S = 1. Also, if c,j | i,j| | i,j| B = I I we have that M (B) = I cI I and for i < j, 1 × · · · × d c,j 1 × · · · × j × · · · × d S (B) = I I I I . Thus, we have that for any box B, and any i,j 1 × · · · j × · · · × i × · · · × d linear map L M ,S , m(L(B)) = det L m(B). If E = B B is an ∈ { c,j i,j} | | · 1 ]···] k elementary set, where (B )k are boxes, then L(E) = L(B ) L(B ), so L(E) is j j=1 1 ]···] k an elementary set and

k k X X m(L(E)) = m(L(B )) = det L m(B ) = det L m(E). j | | · j | | · j=1 j=1 Finally, if A is Jordan measurable, the for any ε > 0 choose elementary sets E A F ⊂ ⊂ such that m(F ) < m(E) + ε. So, L(E) L(A) L(F ) and ⊂ ⊂

J ∗(L(A)) m(L(F )) = det L m(F ) det L m(E) + det L ε ≤ | | · ≤ | | · | | · det L J(A) + det L ε det L m(F ) + det L ε ≤ | | · | | · ≤ | | · | | · det L m(E) + det L 2ε = m(L(E)) + det L 2ε J (L(A)) + det L 2ε. ≤ | | · | | · | | · ≤ ∗ | | ·

Taking ε 0 we have that J (L(A)) = J ∗(L(A)) = det L J(A). This proves the → ∗ | | · theorem for the case that L M ,S . ∈ { c,j i,j} 23

Now for a somewhat more cumbersome computation:

n d o Ri,j(B) = (x1, . . . , xi, . . . , xi + xj, . . . , xd) R : k , xk Ik , ∈ ∀ ∈ which is the Cartesian product of the parallelepiped (x, x + y): x I , y I with a { ∈ i ∈ j} d 2 box in R − . Indeed, if we apply S = Si,1Sj,2 to this set,

n d o Ri,j(B) = (x1, . . . , xi, . . . , xi + xj, . . . , xd) R : k , xk Ik ∈ ∀ ∈ n d o = S (xi, xi + xj, x3, . . . , xi 1, x1, xi+1, . . . , xj 1, x2, xj+1, . . . , xd) R : k , xk Ik − − ∈ ∀ ∈

= S( (x, x + y): x Ii, y Ij I3 Ii 1 I1 Ii+1 Ij 1 Ij Ij+1 Id). { ∈ ∈ } × × · · · − × × × · · · × − × × × · · · ×

Since S preserves Jordan measure (and measurability), it suffices to show the Jordan measurability of P := (x, x + y): x I , y I and compute its Jordan measure. { ∈ i ∈ j} By translating P we may assume without loss of generality that the endpoints of Ii are 0 < a and the endpoints of I are 0, b. We may also write P = T B T where T ,T j 1 ] ] 2 1 2 are right-angle triangles, with orthogonal sides parallel to the axes, and B is a box, as in Figure 1. In a previous exercise it was shown that triangles are Jordan measurable. A translation of T by the vector (0, b) shows that T T has the Jordan measure of 2 − 1 ] 2 the box I I . Thus, P is Jordan measurable and has Jordan measure J(P ) = I I , i × i | i| · | j| which consequently is J(P ) = m(I I ). i × j

P

T2

B

T1 T1 T (0, b) 2 −

Figure 1. Computation of the Jordan measure of the parallelepiped P . 24

Exercise 2.15. Let P := (x, x + y): x I, y J where I,J are intervals. Show { ∈ ∈ } that P is Jordan measurable and that J(P ) = I J . | | · | |

We conclude that Ri,j(B) is Jordan measurable, and that Y J(Ri,j(B)) = J(S(P B)) = J(P B) = J2(P ) Jd 2(B) = Ii Ij Ik = m(B), × × · − | | · | | · | | k=i,j 6 where P = (x, x + y): x Ii, y Ij and B = I3 Ii 1 I1 Ii+1 Ij 1 { ∈ ∈ } × · · · − × × × · · · × − × d 2 Ij Ij+1 Id R − . Since det Ri,j = 1 we have just proved that for any box × × · · · × ⊂ | | B, the set R (B) is Jordan measurable and J(R (B)) = det R m(B). i,j i,j | i,j| · Just as for M and S we now use the fact that R (A B) = R (A) R (B) to c,j i,j i,j ] i,j ] i,j prove the theorem for any L M ,S ,R . ∈ { c,j i,j i,j} Finally, if L is a general invertible linear map, then L = L L where L 1 ··· n k ∈ M ,S ,R for all k. So for any A that is Jordan measurable, also L (A) is Jordan { c,j i,j i,j} n measurable, and thus Ln 1Ln(A) is as well, and so on to get that L(A) = L1 Ln(A) − ··· is Jordan measurable. We also compute the measure by

J(L(A)) = J(L (L L (A))) = det L J(L L (A)) = 1 2 ··· n | 1| · 2 ··· n = = det L det L J(A) = det L J(A). ··· | 1| · · · | n| · | | ·

ut Number of exercises in lecture: 15 Total number of exercises until here: 21 25

Measure Theory

Ariel Yadin

Lecture 3: Lebesgue outer measure

3.1. From finite to countable

Recall that in order to measure sets from outside we used the outer measure by approximating with finitely many elementary sets.

Exercise 3.1. Show that  J ∗(A) = inf m(B1) + + m(Bn), A B1 Bn ··· ⊂ ∪···∪ where B1,...,Bn are always boxes.

Lebesgue’s idea is that there is no reason to stop with finite, and not countable collections. (Mathematicians are not afraid of infinity anymore...)

Definition 3.1. Let A Rd be a set. The Lebesgue outer measure of A is defined • ⊂ to be

 X [ m∗(A) := inf B :(B ) is a sequence of boxes ,A B . | n| n n ⊂ n n n

Note that we may have m (A) = . X ∗ ∞

Exercise 3.2. Show that in general m (A) J (A) for bounded sets A. ∗ ≤ ∗  Show that for Q := Q [0, 1] we have that ∩

J ∗(Q) = 1 and m∗(Q) = 0. 26

Solution to ex:3.2. :( ♣ If A is bounded then the infimum for Lebesgue outer measure is on a larger set than the infimum for Jordan outer measure.

For Q, we have that Q = [0, 1], so J ∗(Q) = J ∗([0, 1]) = m([0, 1]) = 1 and if Q = q , q ,... is an enumeration of Q then m (Q) P q = 0.:) { 1 2 } ∗ ≤ n | { n} | X

Exercise 3.3. Give an example of an unbounded set with Lebesgue outer measure 0.

Proposition 3.2. Properties of Lebesgue outer measure: • m ( ) = 0. • ∗ ∅ (Monotonicity) If A C then m (A) m (C). • ⊂ ∗ ≤ ∗ (Countable subadditivity) If (A ) is a sequence of sets then m (S A ) • n n ∗ n n ≤ P n m∗(An).

Proof. First, is a box of volume 0; another proof is by noting that [0, 1 ]d for all ∅ ∅ ⊂ n n, so m ( ) inf m([0, 1 ]d) = 0. ∗ ∅ ≤ n n For A C we have that the infimum used to obtain m (A) is over a larger set that ⊂ ∗ the one used to obtain the infimum for m∗(C).

Let (An)n be a sequence of sets. Fix ε > 0. For every n let (Bn,k)k be a sequence of boxes such that A S B and n ⊂ k n,k

X n B m∗(A ) + ε 2− . | n,k| ≤ n · k

Consider the sequence of boxes (Bn,k)n,k. We have that

[ [ A B , n ⊂ n,k n n,k 27 and so

[ X X X n X m∗( A ) B m∗(A ) + ε 2− = m∗(A ) + 2ε. n ≤ | n,k| ≤ n · n n n,k n n n Taking ε 0 we have countable subadditivity. → ut It is now natural to ask for the additivity property: if A C = is it true that ∩ ∅ m (A C) = m (A) + m (C)? As it turns out, this is not always the case, a counter ∗ ] ∗ ∗ example will be given in the future. However, in some cases, when the sets are separated enough, we have additivity.

Proposition 3.3. If A, C are such that dist(A, C) > 0 then m (A C) = m (A) + • ∗ ] ∗ m∗(C). Specifically, if A, C are closed disjoint sets and A is compact then dist(A, C) > 0.

Proof. By sub-additivity it suffices to prove m (A) + m (C) m (A C). We may also ∗ ∗ ≤ ∗ ] assume w.l.o.g. that m (A C) < and by monotonicity that m (A) < , m (C) < . ∗ ] ∞ ∗ ∞ ∗ ∞ Fix ε > 0. Let (B ) be a sequence of boxes such that A C S B and P B n n ] ⊂ n n n | n| ≤ m (A C) + ε. ∗ ] Let r = dist(A, C) > 0. Note that for any n, we may replace the box Bn by a finite number of disjoint boxes Bn,1,...,Bn,k such that the diameter of any Bn,j is less than r. So we obtain a sequence (B ) such that A C S B and P B m (A C) + ε, n0 n ] ⊂ n n0 n | n0 | ≤ ∗ ] and such that any box Bn0 cannot intersect both A and C.

Thus, we have that N = NA NC where NA = n : B0 A = and NC = ] { n ∩ 6 ∅} n : B C = . { n0 ∩ 6 ∅} We now have that

[ [ A B0 and C B0 , ⊂ n ⊂ n n NA n NC ∈ ∈ and so

X X X m∗(A) + m∗(C) B0 + B0 B0 m∗(A C) + ε. ≤ | n| | n| ≤ | n| ≤ ] n NA n NC n ∈ ∈ Taking ε 0 completes the proof. → 28

As for the case where A, C are closed and A is compact, note that dist(x, C) is a continuous function of x (because C is closed) and so achieves a minimum on A when A is compact. ut

Exercise 3.4. Give an example of two closed sets A, C such that A C = but ∩ ∅ dist( A, C) = 0.

The next proposition shows that the notation “m” is an appropriate one, as an ex- tension of elementary measure.

Proposition 3.4. For any elementary set E we have that m (E) = m(E). • ∗

Proof. If suffices to prove that J (E) m (E). ∗ ≤ ∗ Case I. E is closed. Since E is bounded, it is compact, and the Heine-Borel Theorem tells us that every open cover of E has a finite sub-cover. Fix any ε > 0. Let (B ) be a sequence of boxes such that E S B and P B n n ⊂ n n n | n| ≤ m∗(E) + ε.

X The boxes (Bn)n do not form an open cover – they need not be open.

For every n let B be an open box containing B B such that B B + ε2 n. n0 n ⊂ n0 | n0 | ≤ | n| − (Exercise: show this is possible.) Then, P B m (E) + 2ε and (B ) are an open n | n0 | ≤ ∗ n0 n cover of E. Thus, there is a finite sub-cover

N [ E B0 , ⊂ n n=1 for some large enough N. Thus,

N X X J ∗(E) B0 B0 m∗(E) + 2ε. ≤ | n| ≤ | n| ≤ n=1 n Taking ε 0 completes the proof for closed E. → Case II. E is a general elementary set. Write E = B B where (B )n are 1 ]···] n j j=1 disjoint boxes. These need not be closed boxes. 29

Fix ε > 0. For every 1 j n let B B be a closed box such that B B ε . ≤ ≤ j0 ⊂ j | j0 | ≥ | j| − n (Exercise: Show this is always possible.) Then, E = B B is a finite union of 0 10 ]···] n0 disjoint closed boxes, and thus a closed elementary set. Also, n n X X m(E0) = B0 B ε = m(E) ε. | j| ≥ | j| − − j=1 j=1 So by Case I,

m(E) m(E0) + ε = m∗(E0) + ε m∗(E) + ε. ≤ ≤ Taking ε 0 completes the proof. → ut

Exercise 3.5. Show that for any bounded set A we have  J (A) m∗(A) J ∗(A). ∗ ≤ ≤

Show that if A is Jordan measurable then m(A) = m∗(A).

Solution to ex:3.5. :( ♣ We have already seen m (A) J (A). If E is an elementary set such that E A, then ∗ ≤ ∗ ⊂ m(E) = m (E) m (A) by monotonicity. Taking supremum over all such elementary ∗ ≤ ∗ sets contained in A, we get that J (A) m∗(A). This proves the first assertion. ∗ ≤ If A is Jordan measurable, then m(A) = J (A) m∗(A) J ∗(A) = m(A).:) X ∗ ≤ ≤ Number of exercises in lecture: 5 Total number of exercises until here: 26 30

Measure Theory

Ariel Yadin

Lecture 4: Lebesgue measure

4.1. Definition of Lebesgue measure

Recall that A is Jordan measurable if and only if for every ε > 0 there exists an elementary set E such that A E and J (E A) < ε. This motivates the following. ⊂ ∗ \ Later we will see that there is another way to approach the issue of Lebesgue measure, and outer measures in general.

Definition 4.1 (Lebesgue measure). A set A Rd is said to be Lebesgue measur- • ⊂ able if for every ε > 0 there exists an open set U such that A U and m (U A) < ε. ⊂ ∗ \ for Lebesgue measurable A we denote m(A) := m∗(A) and refer to this as the Lebesgue measure of A.

X Lebesgue measurable sets are sets that are “almost open”.

? Why do open sets enter the picture?

Definition 4.2. We say that boxes (B ) are almost disjoint if any two boxes can • n n only intersect at their boundary; that is, for every n = m, B B = . 6 n◦ ∩ m◦ ∅

Lemma 4.3. Any open set U Rd can be written as a countable union of almost • ⊂ disjoint closed boxes.

Proof. For any z Zd define the dyadic cube at scale n 0: ∈ ≥

 z1 z1+1   zd zd+1  D (z) = n , n n , n . n 2 2 × · · · × 2 2 n This is the d-dimensional cube of side-lengths 2− and corner z. The following is the crucial property to verify: If D (z) D (z ) = then one is n ◦ ∩ m 0 ◦ 6 ∅ contained in the other. 31

Specifically, if Dn(z) ( Dm(z0) then n > m. Now, set n d o Γ = (z, n): z Z , n 0, Dn(z) U . ∈ ≥ ⊂ These are all dyadic cubes whose closure is contained in U. Also, set

 Λ = (z, n) Γ: (z0, n0) Γ ,D (z) D 0 (z0) . ∈ 6 ∃ ∈ n ⊂ n These are all dyadic cubes contained in U that are maximal with respect to inclusion. We claim that [ U = Dn(z). (z,n) Λ ∈ One inclusion is obvious, since all closures of dyadic cubes indexed by Λ are contained in U. For the other inclusion, let x U. Then, since U is open (this is the only place ∈ we use that U is an open set!), there is a small ball B(x, ε) U. However, for large ⊂ enough n (so that √d2 n < ε), if x D (z) then D (z) B(x, ε) U (because D (z) − ∈ n n ⊂ ⊂ n n has diameter √d2− ). Since for every n there exists some dyadic cube of scale n that d S contains x (because R = z Zd Dn(z) for every fixed n), we get that for some large ∈ d enough n there exists z Z with (z, n) Γ and x Dn(z). Thus, there must be ∈ ∈ ∈ (z, n) Λ such that x D (z) (by just taking the cube of minimal scale containing x). ∈ ∈ n Since this holds for all x U we get that ∈ [ U D (z). ⊂ n (z,n) Λ ∈ This is of course a countable union. Also, since Λ is the indices of maximal dyadic cubes, any two cannot intersect except for the boundary; that is, they are almost disjoint.

ut This construction has a remarkable consequence.

S Exercise 4.1. Show that if U = n Bn where (Bn)n are almost disjoint boxes  P then m∗(U) = n Bn = J (U). | | ∗ 32

Figure 2. Part of the dyadic decomposition of a set. If the set is open, one can continue to capture all points in the set by taking smaller and smaller squares.

Solution to ex:4.1. :( ♣ P P Pm m∗(U) n Bn so it suffices to show that n Bn J (U). Note that n=1 Bn ≤ | | | | ≤ ∗ | | ≤ P J (U) for all m because B1 Bm U. Taking m we have n Bn J (U). ∗ ∪ · · · ∪ ⊂ → ∞ | | ≤ ∗ :) X

Exercise 4.2. Show that if (Bn)n are almost disjoint and (Bn0 )n are almost disjoint,  S S and if n Bn = n Bn0 then X X B = B0 . | n| | n| n n

Proposition 4.4 (Outer regularity, Lebesgue measure). For any set A we have •

m∗(A) = inf m∗(U). A U open ⊂ Proof. One direction m (A) m (U) for any open U A, is just monotonicity. ∗ ≤ ∗ ⊃ 33

For the other direction, if m (A) = there is nothing to prove. Assume m (A) < . ∗ ∞ ∗ ∞ Fix ε > 0. Let (B ) be boxes such that A S B and P B m (A) + ε. For n n ⊂ n n n | n| ≤ ∗ every n let B be an open box such that B B and B B + ε2 n. Thus, the n0 n ⊂ n0 | n0 | ≤ | n| − S set U = n Bn0 is an open set containing A and X m∗(U) B0 m∗(A) + 2ε. ≤ | n| ≤ n Taking infimum over the left hand side and ε 0 completes the proof. → ut

Exercise 4.3. Show that it is false that  m∗(A) = sup m∗(U). A U open ⊃

Proposition 4.5 (Lebesgue measurable sets). Examples of Lebesgue measurable sets: • If U is open it is Lebesgue measurable. • If m (A) = 0 then A is Lebesgue measurable. • ∗ is Lebesgue measurable. •∅ If (A ) is a sequence of Lebesgue measurable sets, then S A is Lebesgue • n n n n measurable. Any closed set F is Lebesgue measurable. • If A is Lebesgue measurable then so is Ac = Rd A. • \ If (A ) is a sequence of Lebesgue measurable sets, then T A is Lebesgue • n n n n measurable.

Proof. For open sets this is obvious by definition. If m (A) = 0 then by outer regularity for any ε > 0 there exists an open set U A ∗ ⊃ such that m (U A) m (U) < m (A) + ε = ε. So A is Lebesgue measurable. ∗ \ ≤ ∗ ∗ has m ( ) J ( ) = 0. ∅ ∗ ∅ ≤ ∗ ∅ 34

Countable unions. Now, if (An)n are all Lebesgue measurable, then: Fix ε > 0. For every n there exists an open set U A such that m (U A ) < ε2 n. Thus, n ⊃ n ∗ n \ n − S S U := n Un is an open set containing n An such that by subadditivity, [ X m∗(U A ) µ∗(U A ) ε. \ n ≤ n \ n ≤ n n S Since this holds for all ε > 0 we get that n An is Lebesgue measurable. Closed sets. Let F be a closed set. Note that F = S (F B[0, n]) where B[0, n] n ∩ is the closed ball of radius n. So it suffices to prove that F is Lebesgue measurable for closed and bounded F (because then F B[0, n] is Lebesgue measurable for all n). ∩ Heine-Borel guaranties then that F is compact. Fix ε > 0. Let U be an open set such that F U and m (U) m (F ) + ε. The ⊂ ∗ ≤ ∗ set U F is open, so we can write U F = S B where (B ) are almost disjoint and \ \ n n n n closed boxes. For any m > 0, the set C := B B is closed and disjoint from m 1 ∪ · · · ∪ m F . Also, F is compact, so we have additivity:

m∗(F ) + m∗(C ) = m∗(F C ) m∗(F (U F )) = m∗(U) m∗(F ) + ε. m ∪ m ≤ ∪ \ ≤ Since F is bounded, we have that m (F ) < , so we conclude that for all m > 0, ∗ ∞ m X B = m∗(B B ) ε, | n| 1 ∪ · · · ∪ m ≤ n=1 where the equality is because (B ) are almost disjoint. Thus, taking m , n n → ∞ X m∗(U F ) = B ε. \ | n| ≤ n This holds for all ε > 0, so F is Lebesgue measurable. Complements. Now, if A is Lebesgue measurable: For every n let U A be an n ⊃ open set such that m (U A) < 2 n. Let F = U c and let F = S F . Since F are ∗ n \ − n n n n n closed they are Lebesgue measurable, and thus F is Lebesgue measurable as a countable union. Note that m (Ac F ) = m (U A) < 2 n for all n. Since Ac F Ac F , ∗ \ n ∗ n \ − \ ⊂ \ n c c m∗(A F ) inf m∗(A Fn) = 0. \ ≤ n \ Thus, Ac = F (Ac F ) which is the union of two Lebesgue measurable sets, and thus ∪ \ Lebesgue measurable itself. 35

c Countable intersections. If (An)n are all Lebesgue measurable, then so are (An)n and thus also !c \ [ c An = An . n n

ut

Exercise 4.4. Show that the following are equivalent.  (1) A is Lebesgue measurable. (2) For every ε > 0 there exists an open set U A such that m (U A) < ε. ⊃ ∗ \ (3) For every ε > 0 there exists an open set U such that m (U A) < ε. ∗ 4 (4) For every ε > 0 there exists a closed set F such that m (A F ) < ε. ∗ 4 (5) For every ε > 0 there exists a closed set F A such that m (A F ) < ε. ⊂ ∗ \ (6) For every ε > 0 there exists a Lebesgue measurable set C such that m (A C) < ∗ 4 ε.

Solution to ex:4.4. :( ♣ (1) (2) is the definition. ⇐⇒ (2) (3) follows by taking the same open set U A. ⇒ ⊃ (3) (2): Fix ε > 0 and let U be such that m (U A) < ε. Let V U A be an ⇒ ∗ 4 ⊃ 4 open set such that m (V ) < 2ε, by outer regularity. Note that A U (A U) U V , ∗ ⊂ ∪ \ ⊂ ∪ which is an open set, and

m∗((U V ) A) m∗(U A) + m∗(V A) m∗(U A) + m∗(V ) < 3ε. ∪ \ ≤ \ \ ≤ 4 So (1) (2) (3). ⇐⇒ ⇐⇒ (1) (4): A is Lebesgue measurable, so also Ac is. Fix ε > 0 and let U be an open ⇒ set such that m (U Ac) < ε. Let F = U c. So m (F A) = m (U Ac) < ε, and F is ∗ 4 ∗ 4 ∗ 4 closed. (4) (5): Fix ε > 0 and let F be a closed set such that m (A F ) < ε. By outer ⇒ ∗ 4 regularity let U A F be an open set such that m (U) < 2ε. Then F U c is a closed ⊃ 4 ∗ ∩ 36 set and F U c F A A, and ∩ ⊂ ∩ ⊂

c c m∗(A F U ) m∗(A F ) + m∗(A U ) m∗(A F ) + m∗(U) < 3ε. \ ∩ ≤ \ \ ≤ 4 (5) (6): Fix ε > 0 and let F A be a closed set such that m (A F ) < ε. The ⇒ ⊂ ∗ \ C = F is Lebesgue measurable, and m (A C) = m (A F ) < ε. ∗ 4 ∗ \ (6) (3): Fix ε > 0 and let C be Lebesgue measurable such that m (A C) < ε. ⇒ ∗ 4 Let U C be an open set such that m (U C) = m (U C) < ε. Note that A U ⊃ ∗ 4 ∗ \ 4 ⊂ A C S C U, so 4 4 m∗(A U) m∗(A C) + m∗(C U) < 2ε. 4 ≤ 4 4 :) X

Exercise 4.5. Show that if A is Jordan measurable then it is also Lebesgue mea- surable.

Definition 4.6. A family of subsets is a σ-algebra if • F ; •∅∈F is closed under complements, i.e. A implies Ac ; •F ∈ F ∈ F is closed under countable unions, i.e. (A ) implies S A . •F n n ⊂ F n n ∈ F

Exercise 4.6. Show that if is a σ-algebra it is closed under countable intersec- F tions, set difference and symmetric difference.

Exercise 4.7. Show that the family of Lebesgue measurable sets form a σ-algebra.  37

4.2. Lebesgue measure as a measure

Proposition 4.7 (Measure axioms). Let be the family of all Lebesgue measurable • L sets. Recall that for A we define m(A) = m (A). We the have that m : [0, ] ∈ L ∗ F → ∞ with the following properties:

m( ) = 0; • ∅ (Countable additivity) If (A ) are pairwise disjoint Lebesgue measurable • n n ⊂ L sets then ] X m( An) = m(An). n n Proof. We only need to prove the assertion regarding additivity.

Case I. (An)n are all compact. Since m∗ is additive on disjoint compact sets, for any N, N N X ] ] X m(A ) = m( A ) m( A ) m(A ), n n ≤ n ≤ n n=1 n=1 n n where the last two inequalities are monotonicity and countable subadditivity. Taking N we get the compact case. → ∞ Case II. (An)n are all bounded. Fix ε > 0. For every n let Fn be a closed set with F A and m (A F ) < ε2 n. Since F A is bounded, F is compact. By n ⊂ n ∗ n \ n − n ⊂ n n subadditivity, m(A ) m(F ) + ε2 n, so n ≤ n − X X ] ] m(A ) m(F ) + ε = m( F ) + ε m( A ) + ε. n ≤ n n ≤ n n n n n Taking ε 0 completes the bounded case. → Case III. Now for the general case. Let C = B(0, k) B(0, k 1) for all n, and k \ − set D = A C . So (D ) is a sequence of pairwise disjoint bounded Lebesgue n,k n ∩ k n,k n,k measurable sets. Thus, for every n, ] X m(An) = m( Dn,k) = m(Dn,k), k k and ] ] X X m( An) = m( Dn,k) = m(Dn,k) = m(An). n n,k n,k n

ut 38

Exercise 4.8. Show that if A is Lebesgue measurable then  m(A) = sup m(K). A K compact ⊃

Solution to ex:4.8. :( ♣ Monotonicity implies that it suffices to prove that m(A) supA K compact m(K). ≤ ⊃ First suppose that A is bounded, so m(A) < . ∞ Fix ε > 0. Since A is Lebesgue measurable there exists a closed set K A such that ⊂ µ (A K) < ε. Since K A and A is bounded, we have that K is bounded as well, so ∗ \ ⊂ K is compact because it is bounded and closed. So K is a compact set such that K A ⊂ and µ(K) = µ(A) µ(A K) µ(A) ε. This holds for all ε > 0 establishing the claim − \ ≥ − in the case that A is bounded. For the case where A is unbounded, consider A = A (B(0, n) B(0, n 1)) for all n. n ∩ \ − These are all bounded, so for every ε > 0 and every n there exists a compact set K A n ⊂ n such that m(K ) m(A ) ε2 n. Also, (A ) are disjoint so m(A) = P m(A ). Thus, n ≥ n − − n n n n if K = Un K , then K is compact, K A and n0 j=1 j n0 n0 ⊂ n n X X m(K0 ) = m(K ) m(A ) ε m(A) ε. n j ≥ j − → − j=1 j=1 This implies that there exists n large enough so that m(K ) m(A) 2ε. n0 ≥ − Conclusion, for any ε > 0 there exists a compact K A such that m(K) m(A) ε. ⊂ ≥ − This proves the unbounded case.:) X

Finally, let us end with the following criterion for Lebesgue measurability.

Proposition 4.8 (Charath´eodory criterion). The following are equivalent. • A is Lebesgue measurable. • For every box B we have B = m (B A) + m (B A). • | | ∗ ∩ ∗ \ For every elementary set E we have m(E) = m (E A) + m (E A). • ∗ ∩ ∗ \ 39

Proof. If A is Lebesgue measurable and E is elementary, then E = (E A) (E A) which ∩ ] \ are both Lebesgue measurable sets, so by additivity, m(E) = m (E A) + m (E A). ∗ ∩ ∗ \ Un Suppose the second bullet holds. Let E be any elementary set, and write E = j=1 Bj n for disjoint boxes (Bj)j=1. Then by subadditivity, n n X X m∗(E A) + m∗(E A) m∗(B A) + m∗(B A) = B = m(E). ∩ \ ≤ j ∩ j \ | j| j=1 j=1 Since m(E) m (E A) + m (E A) for any set E just by subadditivity, we conclude ≤ ∗ ∩ ∗ \ that equality holds for all elementary sets, proving the third bullet. Now assume the third bullet. We want to show that A is Lebesgue measurable. Assume first that m (A) < . Fix ε > 0. Let (B ) be disjoint boxes such that ∗ ∞ n n P B m (A) + ε and A S B (this exists by the definition of m ). For every n n | n| ≤ ∗ ⊂ n n ∗ let B B be an open box such that B B + ε2 n. Note that U = S B is an n0 ⊃ n | n0 | ≤ | n| − n n0 open set and U A. For every n we have that ⊃ n m∗(B0 A) + m∗(B0 A) = B0 B + ε2− . n ∩ n \ | n| ≤ | n| By subadditivity, X X m∗(A) + m∗(U A) m∗(B0 A) + m∗(B0 A) B + ε m∗(A) + 2ε. \ ≤ n ∩ n \ ≤ | n| ≤ n n Thus, m (U A) 2ε. Since this holds for all ε > 0 this completes the proof for A with ∗ \ ≤ m (A) < . ∗ ∞ Now assume that m (A) = . Let E be any elementary set, and let B be a box. ∗ ∞ Since E B is elementary, and since E (A B) (E B) ((E B) A), we have that ∩ \ ∩ ⊆ \ ∪ ∩ \

m∗(E A B) + m∗(E (A B)) m∗(E B A) + m∗((E B) A) + m(E B) ∩ ∩ \ ∩ ≤ ∩ ∩ ∩ \ \ = m(E B) + m(E B) = m(E). ∩ \ Thus, for any elementary E we have the Carath´eodory criterion for A B, ∩

m(E) = m∗(E (A B)) + m∗(E (A B)). ∩ ∩ \ ∩ Since B is bounded, we have m (A B) < and by the previous part A B is Lebesgue ∗ ∩ ∞ ∩ measurable. Since A = S (A B ) where B are boxes of side length n around 0, we n ∩ n n 40 have that A is Lebesgue measurable as a countable union of Lebesgue measurable sets.

ut

Exercise 4.9. For a bounded set A define the Lebesgue inner measure by  m (A) = m(E) m∗(E A), ∗ − \ for a Lebesgue mesurable set E such that A E and m(E) < . ⊂ ∞ Show that this definition does not depend on the choice of the set E.

Show that m (A) m∗(A) and equality holds if and only if A is Lebesgue measur- ∗ ≤ able.

Solution to ex:4.9. :( ♣ Let E,F be Lebesgue measurable sets such that A F E and m(E) < . ⊂ ⊂ ∞ Fix ε > 0. Let (B ) be a sequence of boxes such that E A S A and P B n n \ ⊂ n n n | n| ≤ m (E A) + ε. Note that ∗ \ [ [ F A (E A) (E F ) (B (E F )) and E F (B (E F )). \ ⊂ \ \ \ ⊂ n \ \ \ ⊂ n ∩ \ n n So,

X c m(E F ) + m∗(F A) m∗(B (E F ) ) + m∗(B (E F )) \ \ ≤ n ∩ \ n ∩ \ n X = B m∗(E A) + ε. | n| ≤ \ n Taking ε 0 and rearranging we have that →

m(E) m∗(E A) m(F ) m∗(F A). − \ ≤ − \ Since E A (E F ) (F A) we have \ ⊂ \ ∪ \

m(E) m∗(E A) m(E) m(E) + m(F ) m∗(F A). − \ ≥ − − \ So we have show that

m(E) m∗(E A) = m(F ) m(F A) − \ − \ 41 whenever A F E. ⊂ ⊂ For general Lebesgue measurable sets E,F such that A F and A E and m(E) < ⊂ ⊂ , m(F ) < , we have that A E F , so ∞ ∞ ⊂ ∩

m(E) m∗(E A) = m(E F ) m∗((E F ) A) = m(F ) m∗(F A). − \ ∩ − ∩ \ − \

This shows that m is well defined. ∗ Now, if A is Lebesgue measurable then for any Lebesgue measurable set E A with ⊃ m(E) < we have that m(E A) = m(E) m(A). So m (A) = m(E) m(E A) = ∞ \ − ∗ − \ m(A) = m∗(A).

On the other hand, assume that m∗(A) = m (A). Fix ε > 0 and let U A be an ∗ ⊃ open set such that A U and m(U) m (A) + ε. Note that ⊂ ≤ ∗

m(U) ε m∗(A) = m (A) = m(U) m∗(U A), − ≤ ∗ − \ so m (U A) ε.:) ∗ \ ≤ X

T Exercise 4.10. Reminder: A Gδ set is a set A = n Un where (Un)n are all open  S (i.e. a countable intersection of open sets). A Fσ set is a set A = n Fn where (Fn)n are all closed (i.e. a countable union of closed sets). A null set is a set with Lebesgue (outer) measure 0. Show that the following are equivalent.

A is Lebesgue measurable. • A = G N is where G is G and N is null. • \ δ A = F N where F is F and N is null. • ∪ σ

Exercise 4.11. Show that if A is Lebesgue measurable then A+x is also Lebesgue measurable and m(A + x) = m(A). 42

Exercise 4.12. Let be the family of Lebesgue measurable sets. Suppose that L m : [0, ] admits the following properties: 0 L → ∞ m ( ) = 0; • 0 ∅ If (A ) are pairwise disjoint Lebesgue measurable sets then m (U A ) = • n n ⊂ L 0 n n P n m0(An); d m0(A + x) = m0(A) for all A , x R ; • ∈ L ∈ m ([0, 1]d) = 1. • 0 Show that m0 is Lebesgue measure.

Solution to ex:4.12. :( ♣ We already know that m0(E) = m(E) for any Jordan measurable set E, and specifically for boxes.

It is simple to prove that m0 is sub-additive and monotone. Step I. We show that for A , if m(A) = 0 then m (A) = 0: Let A be a Lebesgue ∈ L 0 measurable set of 0 measure, m(A) = 0. For any ε > 0, let (Bn)n be a sequence of boxes such that A S B and such that P B ε. Then, m (A) P m (B ) = ⊂ n n n | n| ≤ 0 ≤ n 0 n P B ε, and taking ε 0, we have that m (A) = 0. n | n| ≤ → 0 Step II. We show that if U is open of finite Lebesgue measure, then m0(U) = m(U): Let U be an open set such that m(U) < . Write U = S B where (B ) are ∞ n n n n almost disjoint closed boxes. Let Bn0 = (Bn)◦, so (Bn0 )n are pairwise disjoint. Let F = S B U B . Since n n \ n n0 ] X X m(F ) = m(U) m( B0 ) = B B0 = 0, − n | n| − | n| n n n we have that m (F ) = 0. Note that U = F U B , so 0 ] n n0 X X m0(U) = m0(F ) + m0(B0 ) = B0 = m(U). n | n| n n 43

Step III. We show that for any A , the inequality m (A) m(A) holds: Let A ∈ L 0 ≤ be a Lebesgue measurable set with m(A) < . Fix ε > 0 and let U be an open set such ∞ that A U and m(U) m(A) = m(U A) < ε. Then, m (A) m (U) m(A) + ε. ⊂ − \ 0 ≤ 0 ≤ Taking ε 0 we have that for any A , the inequality m (A) m(A) holds. (The → ∈ L 0 ≤ case where m(A) = is immediate.) ∞ Step IV. We show that for any bounded set A we have m (A) = m(A): Let A ∈ L 0 ∈ L be a bounded set. Let B be a box bounding A B. We have m (B A) = m (B) m (A) ⊂ 0 \ 0 − 0 by additivity of m0. Also,

m0(A) m(A) = m(B) m(B A) m0(B) m0(B A) = m0(A). ≤ − \ ≤ − \ Step V. We show that for any A we have m (A) = m(A): Let A . Write ∈ L 0 ∈ L U d A = An where An = Bn Bn 1, and Bn = [ n, n] . Then by additivity of both m, m0 \ − − and since An are all bounded, X X m0(A) = m0(An) = m(An) = m(A). n n :) X

Number of exercises in lecture: 12 Total number of exercises until here: 38 44

Measure Theory

Ariel Yadin

Lecture 5: Abstract measures

We now review the construction of Lebesgue measure, isolating the main abstract properties, in order to generalize it to other settings.

5.1. σ-algebras

We saw in the discussion of Lebesgue measure that the family of Lebesgue measurable sets form a special structure called a σ-algebra.

Definition 5.1 (σ-algebra). Let X be any set. We denote by 2X = (X) = • P A : A X the set of all subsets of X. { ⊂ } A family 2X is called a σ-algebra (on X) if: F ⊂ ; •∅∈F is closed under complements, i.e. A implies X A ; •F ∈ F \ ∈ F is closed under countable unions, i.e. if (A ) is a sequence in then S A •F n n F n n ∈ . F

Exercise 5.1. Show that if is a σ-algebra on X then: F  is closed under countable intersections, i.e. if (A ) is a sequence in then •F n n F T A . n n ∈ F X . • ∈ F is closed under finite unions and finite intersections. •F is closed under set differences. •F is closed under symmetric differences. •F 45

Exercise 5.2. Suppose 2X is a family of subsets satisfying the following: F ⊂  ; •∅∈F is closed under complements; •F is closed under countable intersections. •F Show that is a σ-algebra. F

T Exercise 5.3. Show that if ( α)α I is a collection of σ-algebras on X, then α Fα F ∈ is also a σ-algebra on X.

Proposition 5.2 (σ-algebra generated by subsets). Let be a collection of subsets • K of X. There exists a σ-algebra, denoted σ( ) such that σ( ) and for every other σ- K K ⊂ K algebra such that we have that σ( ) . F K ⊂ F K ⊂ F That is, σ( ) is the smallest σ-algebra containing . K K We call σ( ) the σ-algebra generated by . K K

Proof. Define

\ σ( ) := : is a σ-algebra on X, . K {F F K ⊂ F}

This is a σ-algebra with the required properties. ut

Exercise 5.4. Show that if then σ( ) σ( ). Also, if and is a K ⊂ L K ⊂ L K ⊂ F F σ-algebra, then σ( ) . K ⊂ F 46

Definition 5.3 (Borel σ-algebra). Given a topological space X, the Borel σ-algebra • is the σ-algebra generated by the open sets. It is denoted (X). B Specifically in the case X = Rd we have that

d = d = (R ) = σ(U : U is an open set ). B B B

A Borel-measurable set, i.e. a set in (X), is called a Borel set. X B

Exercise 5.5. Prove that 

d = d = (R ) = σ(B : B is an open box ) = σ(B0 : B0 is a closed box ). B B B

Solution to ex:5.5. :( ♣ If B is an open box then B so σ(B : B is an opend box ) . ∈ B ⊂ B If U is an open set then it is a countable union of closed boxes, so U σ(B : ∈ B is a closed box ), which shows that σ(B : B is a closed box ). B ⊂ 0 0 If B = [a , b ] [a , b ] is a closed box, then for 0 1 1 × · · · × d d

B = (a 1 , b + 1 ) (a 1 , b + 1 ), n 1 − n 1 n × · · · × d − n d n we have that B = T B and B are all open boxes. Thus, B σ(B : B is an open box ), 0 n n n 0 ∈ which implies that σ(B : B is a closed box ) σ(B : B is an open box ).:) 0 0 ⊂ X 47

Exercise 5.6. Show that  (R) = σ((a, b): < a < b < ) = σ([a, b]: < a < b < ) B −∞ ∞ −∞ ∞ = σ((a, b]: < a < b < ) = σ([a, b): < a < b < ) −∞ ∞ −∞ ∞ = σ((a, ): < a < ) = σ([a, ): < a < ) ∞ −∞ ∞ ∞ −∞ ∞ = σ(( , a): < a < ) = σ(( , a]: < a < ). −∞ −∞ ∞ −∞ −∞ ∞

5.2. Measures

Definition 5.4. A pair (X, ) where is a σ-algebra on X is call a measurable • F F space. Elements of are called measurable sets. F Given a measurable space (X, ), a function µ : [0, ] is called a measure (on F F → ∞ (X, )) if F µ( ) = 0; • ∅ (Additivity) For all sequences (A ) of pairwise disjoint sets in , we have • n n ⊂ F F that ] X µ( An) = µ(An). n n (X, , µ) is called a measure space. F

A measure space (X, , µ) is called finite if µ(X) < . It is called σ-finite if X F ∞ X = S A where A and µ(A ) < for all n. n n n ∈ F n ∞

Exercise 5.7. Show that any measure is finitely additive; that is, if (X, , µ) is a F measure space then for any disjoint sets A, B we have µ(A B) = µ(A) + µ(B). ∈ F ] 48

Example 5.5. The counting measure: Take = 2X and µ(A) = A . F | | If f : X [0, ) is a function then → ∞ X µ(A) := sup f(an) (an)n A ⊂ n is a measure on (X, 2X ). If X is uncountable and

= A X : A is countable, or Ac is countable , F { ⊂ } then (X, ) is a measurable space and F  1 Ac is countable µ(A) = 0 A is countable is a measure on (X, ). F 4 5 4

Exercise 5.8. Give an example of a set X such that    A = , µ(A) := ∞ | | ∞ 0 A < , | | ∞ is not a measure on (X, 2X ).

Proposition 5.6 (Basic properties of measures). Let (X, , µ) be a measure space. • F Then:

(Monotonicity) For A B we have µ(A) µ(B). • ⊂ ∈ F ≤ (Subadditivity) If (A ) then µ(S A ) P µ(A ). • n n ⊂ F n n ≤ n n

Proof. Monotonicity follows from B = A (B A) and finite additivity, so µ(B) = ] \ µ(A) + µ(B A) µ(A). \ ≥ 49

For subadditivity, let (A ) and set n n ⊂ F n 1 [− B = A ( A ). n n \ j j=1 So ] [ B = A and B A . n n n ⊂ n n n Thus,

[ ] X X µ( A ) = µ( B ) = µ(B ) µ(A ). n n n ≤ n n n n n

ut

Definition 5.7. For a measure space (X, , µ) a set N is called a null set (or • F ∈ F µ-null set) if µ(N) = 0. A measure space (X, , µ) such that for all null sets N we have that any A N F ∈ F ⊂ is also measurable (i.e. in ) is called complete. F

Exercise 5.9. Let (X, , µ) be a measure space. Let be the set of all µ-null F N sets. Define ¯ := A F : A ,F N . F { ∪ ∈ F ⊂ ∈ N } Show that ¯ is a σ-algebra. F Show that if we defineµ ¯ : ¯ [0, ] by F → ∞ µ¯(A F ) = µ(A) A ,F N , ∪ ∀ ∈ F ⊂ ∈ N thenµ ¯ is a well defined complete measure on ¯; moreover,µ ¯ = µ and if ν is a complete F F measure on ¯ such that ν = µ then ν =µ ¯. F F

Exercise 5.10. Show that if µ , . . . , µ are measures on (X, ) the for any non- 1 n F negative numbers a , . . . , a the function µ := P a µ is also a measure on (X, ). 1 n j j j F 50

5.3. Fatou’s Lemma and continuity

Proposition 5.8 (Monotone convergence). Let (X, , µ) be a measure space. • F If A A is an increasing sequence in then 1 ⊂ 2 ⊂ · · · F [ lim µ(An) = µ( An). n →∞ n If A A is an decreasing sequence in then under the condition that there 1 ⊃ 2 ⊃ · · · F exists n such that µ(A ) < we have that n ∞ \ lim µ(An) = µ( An). n →∞ n

Proof. For the increasing sequence case define Bn = An An 1. Then, (Bn)n are pairwise \ − disjoint, so n n ] X lim µ(An) = lim µ( Bj) = lim µ(Bj) n n n →∞ →∞ j=1 →∞ j=1 X ] [ = µ(Bn) = µ( Bn) = µ( An). n n n For the decreasing sequence case, since µ(A ) < we define B = A A (so B = k ∞ n k \ n n ∅ if n k). Note that µ(B ) + µ(A ) = µ(A ) for n k and since µ(A ) µ(A ) < ≤ n n k ≥ n ≤ k ∞ we have that µ(Bn) = µ(Ak) µ(An). Note that (Bn)n k is an increasing sequence such − ≥ S T T that n k Bn = Ak n k An. Since n k An Ak we have ≥ \ ≥ ≥ ⊂ \ [ \ µ(Ak) = µ( An) + µ( Bn) = µ( An) + lim µ(Bn) n n k n k n →∞ ≥ ≥ \ = µ( An) + lim (µ(Ak) µ(An)). n − n →∞ Since µ(A ) < we may subtract it from both sides to get the proposition. k ∞ ut

Let (An)n be a sequence of subsets of X. Define \ [ [ \ lim sup An = Ak and lim inf An = Ak. n k n n k n ≥ ≥ 51

lim sup A is the set of all x X that appear in infinitely many of the subsets A . n ∈ n Similarly, lim inf A is the set of all x X that appear in all but finitely many of the n ∈ subsets An.

We say that the sequence (An)n converges if lim inf An = lim sup An, and denote this common set by lim An = lim sup An = lim inf An in this case.

Exercise 5.11. Show that lim inf A lim sup A . n ⊆ n 

S Exercise 5.12. Show that if (An)n is an increasing sequence then lim An = n An  T and if (An)n is a decreasing sequence then lim An = n An.

Lemma 5.9 (Fatou’s Lemma). Let (X, , µ) be a measure space. • F If (A ) is a sequence in then n n F

µ(lim inf An) lim inf µ(An). n ≤ →∞ If in addition µ(S A ) < then n n ∞

µ(lim sup An) lim sup µ(An). ≥ n →∞ T Proof. For every n we have that k n Ak An, so ≥ ⊂ \ µ( Ak) inf µ(Ak). ≤ k n k n ≥ ≥ T Note that Bn := k n Ak for an increasing sequence so ≥ [ lim inf µ(An) = lim inf µ(Ak) lim µ(Bn) = µ( Bn) = µ(lim inf An). n n k n ≥ n →∞ →∞ ≥ →∞ n S For A := n An \ [ [ \ A lim sup A = A A = (A A ) = lim inf(A A ). \ n \ k \ k \ n n k n n k n ≥ ≥ 52

Thus, if we have µ(A) < then µ(A lim sup A ) = µ(A) µ(lim sup A ) and µ(A ∞ \ n − n \ A ) = µ(A) µ(A ), so n − n

µ(A) µ(lim sup An) = µ(A lim sup An) lim inf(µ(A) µ(An)) = µ(A) lim sup µ(An), − \ ≤ n − − n →∞ →∞ which completes the proof by subtracting µ(A) from both sides. ut

Exercise 5.13. Show that if lim A exists and µ(S A ) < then n n n ∞  µ(lim An) = lim µ(An). n →∞

Number of exercises in lecture: 13 Total number of exercises until here: 51 53

Measure Theory

Ariel Yadin

Lecture 6: Outer measures

6.1. Outer measures

Definition 6.1. Let X be any set. An outer measure on X is a function µ : 2X • ∗ → [0, ] such that ∞ µ ( ) = 0; • ∗ ∅ µ (A) µ (B) for all A B; • ∗ ≤ ∗ ⊂ µ (S A ) P µ (A ). • ∗ n n ≤ n ∗ n

Exercise 6.1. Show that Lebesgue outer measure m∗ is an outer measure (on d R ).

Analogously to the way we defined Lebesgue outer measure, we have:

Exercise 6.2. Let 2X such that , and let ρ : [0, ] such that E ⊂ ∅ ∈ E E → ∞ ρ( ) = 0. Define ∅ ( ) X [ µ∗(A) := inf ρ(E ): A E , n , E , n ⊂ n ∀ n ∈ E n n where inf = . Show that µ is an outer measure. ∅ ∞ ∗

Solution to ex:6.2. :( ♣ Since S and we have that µ ( ) = 0. ∅ ⊂ n ∅ ∅ ∈ E ∗ ∅ 54

If A B then any sequence (E ) participating in the infimum for µ (B) also par- ⊂ n n ∗ ticipates in the infimum for A. So µ (A) µ (B). ∗ ≤ ∗ For a sequence (A ) : If µ (A ) = for some n then there is nothing to prove. So n n ∗ n ∞ assume that µ (A ) < for all n. ∗ n ∞ Fix ε > 0 and for every n let (E ) be such that P ρ(E ) µ (A ) + ε2 n n,k k ⊂ E k n,k ≤ ∗ n − and A S E . Then, S A S E and n ⊂ k n,k n n ⊂ n,k n,k [ X X µ∗( A ) ρ(E ) µ∗(A ) + ε. n ≤ n,k ≤ n n n,k n Taking ε 0 completes the proof.:) → X Compare this to the case that is the collection of boxes in Rd. E 6.2. Measurability

Recall the Carath´eodory’s criterion for Lebesgue measurability: A Rd is Lebesgue ⊂ measurable if and only if for every elementary set E we have m(E) = m (E A) + ∗ ∩ m (E A). This motivates the following definition. ∗ \ Definition 6.2 (Measurable sets). Let µ be an outer measure on X. A subset A X • ∗ ⊂ is called µ -measurable (or simply measurable) if for every subset E X we have ∗ ⊂

µ∗(E) = µ∗(E A) + µ∗(E A). ∩ \

Exercise 6.3. Show that A is µ -measurable if and only if for every E X such ∗ ⊂ that µ (E) < we have ∗ ∞ c µ∗(E A) + µ∗(E A ) µ∗(E). ∩ ∩ ≤

d Exercise 6.4. Let m be Lebesgue measure in R . We have seen that m∗ is an  outer measure. Show that A is Lebesgue measurable if and only if it is m∗-measurable. 55

Solution to ex:6.4. :( ♣ If A is m∗-measurable, then we have already seen that A is Lebesgue measurable, since it is enough to require that m(E) = m (E A) + m (E Ac) for every elementary set ∗ ∩ ∗ ∩ E. Assume that A is Lebesgue measurable. Then B = m (B A) + m (B Ac) for | | ∗ ∩ ∗ ∩ every box B. Let E X be any subset such that m (E) < . Fix ε > 0. Let (B ) be pairwise ⊂ ∗ ∞ n n disjoint boxes such that E U B and P B m (E) + ε. E A U (B A) ⊂ n n n | n| ≤ ∗ ∩ ⊂ n n ∩ and E Ac U (B Ac), so ∩ ⊂ n n ∩

c X c X m∗(E A) + m∗(E A ) m(B A) + m(B A ) = B m∗(E) + ε. ∩ ∩ ≤ n ∩ n ∩ | n| ≤ n n Taking ε 0 shows that A is m -measurable.:) → ∗ X

Theorem 6.3 (Charath´eodory’s Theorem). Let µ be an outer measure on X and ••• ∗ let be the collection of all µ -measurable subsets. Then, is a σ-algebra, and µ F ∗ F ∗ restricted to is a complete measure on (X, ). F F

Proof. First, since µ (E) = µ (E X) + µ (E ) for all E X. ∅ ∈ F ∗ ∗ ∩ ∗ ∩ ∅ ⊂ Also, is closed under complements, because µ (E) = µ (E A) + µ (E Ac) is a F ∗ ∗ ∩ ∗ ∩ symmetric equation in A, Ac. Now, if A, B then A B = (A B) (A Bc) (Ac B), so for any E with ∈ F ∪ ∩ ] ∩ ] ∩ µ (E) < , ∗ ∞

c µ∗(E) = µ∗(E A) + µ∗(E A ) ∩ ∩ c c c c = µ∗(E A B) + µ∗(E A B ) + µ∗(E A B) + µ∗(E A B ) ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ c µ∗(E (A B)) + µ∗(E (A B) ). ≥ ∩ ∪ ∩ ∪ Thus A B as well. ∪ ∈ F This shows that is closed under finite unions. F 56

Note that if A, B ,A B = then ∈ F ∩ ∅ c µ∗(A B) = µ∗((A B) A) + µ∗((A B) A ) = µ∗(A) + µ∗(B). ] ] ∩ ] ∩ So µ is finitely additive on . ∗ F Now, if (A ) are pairwise disjoint sets in then for all n, n n F n n n ] ] ] c µ∗(E A ) = µ∗(E A A ) + µ∗(E A A ) ∩ j ∩ j ∩ n ∩ j ∩ n j=1 j=1 j=1 n 1 n ]− X = µ∗(E A ) + µ∗(E A ) = = µ∗(E A ). ∩ n ∩ j ··· ∩ j j=1 j=1 Thus, for any n, n n n ] ] X ] µ∗(E) = µ∗(E A ) + µ∗(E A ) µ∗(E A ) + µ∗(E A ). ∩ n \ n ≥ ∩ j \ n j=1 j=1 j=1 n Taking n we have that for A = U A , → ∞ n n X c c µ∗(E) µ∗(E A ) + µ∗(E A ) µ∗(E A) + µ∗(E A ). ≥ ∩ n ∩ ≥ ∩ ∩ n This implies that A and that the above inequalities are all equalities, so ∈ F X µ∗(E A) = µ∗(E A ). ∩ ∩ n n Thus, taking E = A we get that µ is countable additive on and that is closed ∗ F F under countable disjoint unions. So we are only left with showing that is closed under countable unions. F Now if (A ) is any sequence in (not necessarily disjoint), then set B = A n n F n n \ Sn 1 − A . So (B ) are pairwise disjoint. Because is closed under complements and j=1 j n n F finite unions we have that B for all n. Thus, S A = U B , which proves n ∈ F n n n n ∈ F that is closed under countable unions. F

Finally, to show that (X, , µ∗ ) is indeed complete: If A admits µ∗(A) = 0 and F F then by monotonicity, for any E with µ (E) < , ∗ ∞ c c µ∗(E A) + µ∗(E A ) µ∗(E A ) µ∗(E). ∩ ∩ ≤ ∩ ≤ So A . ∈ F ut 57

Exercise 6.5. [Folland p.32 ex.17] Let µ∗ be an outer measure on X and let (An)n be a sequence of pairwise disjoint µ -measurable subsets. Show that for any A X, ∗ ⊂ ] X µ∗(A ( A )) = µ∗(A A ). ∩ n ∩ n n n

6.3. Pre-measures

It is also interesting to what degree this resulting measure is unique.

Definition 6.4. An algebra is a collection of subsets 2X that is closed under • A ⊂ finite unions and complements.

Exercise 6.6. Let B be a box in Rd. Show that  = E B : E is elementary A { ⊂ } is an algebra over B.

Exercise 6.7. Show that an algebra is closed under finite intersections, set A differences, symmetric differences. Show that . ∅ ∈ A

Exercise 6.8. Let be an algebra such that for all sequences (A ) that are A n n ⊂ A pairwise disjoint we have that U A . n n ∈ A Show that is a σ-algebra. A 58

Definition 6.5 (Pre-measure). Given an algebra on X, a pre-measure is a func- • A tion µ : [0, ] such that µ ( ) = 0 and for any sequence (A ) that are 0 A → ∞ 0 ∅ n n ⊂ A pairwise disjoint and admit U A we have that µ (U A ) = P µ (A ). n n ∈ A 0 n n n 0 n Lemma 6.6. Let µ be a pre-measure on over X. Define • 0 A ( ) X [ µ∗(A) := inf µ (E ): A E , n , E . 0 n ⊂ n ∀ n ∈ A n n

Then µ∗ is an outer measure such that µ∗ = µ0 and every set in is µ∗-measurable. A A Proof. We have already seen that such a definition gives rise to an outer measure. For any A let (E ) be a sequence in such that A S E . Let B = ∈ A n n A ⊂ n n n Sn 1 A (E − E ). Then, B for all n and E B for all n. Thus, ∩ n \ j=1 j n ∈ A n \ n ∈ A µ (E ) = µ (B ) + µ (E B ) µ (B ) for all n, and also because U B = A , 0 n 0 n 0 n \ n ≥ 0 n n n ∈ A X X µ (A) = µ (B ) µ (E ). 0 0 n ≤ 0 n n n Taking infimum over all such sequences (E ) we obtain that µ (A) µ (A). n n ⊂ A 0 ≤ ∗ Since µ (A) µ (A) by definition, we get that µ (A) = µ (A). This completes the first ∗ ≤ 0 0 ∗ assertion. For the second assertion, let A and E X such that µ (E) < . Fix ε > 0 and ∈ A ⊂ ∗ ∞ let (E ) be such that E S E and P µ (E ) µ (E) + ε. For all n we have n n ⊂ A ⊂ n n n 0 n ≤ ∗ µ (E A) + µ (E Ac) = µ (E ). Thus, 0 n ∩ 0 n ∩ 0 n X X c X X c µ∗(E) ε + µ (E A) + µ (E A ) = ε + µ∗(E A) + µ∗(E A ) ≥ − 0 n ∩ 0 n ∩ − n ∩ n ∩ n n n n c ε + µ∗(E A) + µ∗(E A ). ≥ − ∩ ∩ So A is µ -measurable. ∗ ut Theorem 6.7 (Charath´eodory’s Extension Theorem). Let µ be a pre-measure on ••• 0 an algebra over X. Let = σ( ). Then, there exists a measure µ on (X, ) such A F A F that µ = µ0. A 59

Also, if ν is a measure on (X, ) such that ν = µ0 then ν(A) µ(A) for all A , F A ≤ ∈ F and ν(A) = µ(A) for all A with µ(A) < . ∈ F ∞ Moreover, if X = S A for A , µ (A ) < (i.e. µ is σ-finite), then ν = µ. n n n ∈ A 0 n ∞ 0

Proof. Lemma 6.6 tells us that µ0 extends to a measure µ on the µ∗-measurable sets, which form a σ-algebra that contains , and thus contains = σ( ). Here A F A ( ) X [ µ∗(A) := inf µ (E ): A E , n , E . 0 n ⊂ n ∀ n ∈ A n n Now, if A , and (E ) is such that A S E , then ∈ F n n ⊂ A ⊂ n n X X ν(A) ν(E ) = µ (E ). ≤ n 0 n n n Taking infimum over all such sequences we get that ν(A) µ (A) = µ(A) for all A . ≤ ∗ ∈ F If µ(A) = µ (A) < then for any ε > 0 we may choose (E ) such that A ∗ ∞ n n ⊂ S E and P µ (E ) µ (A) + ε. Thus, µ(S E ) µ(A) + ε which implies that n n n 0 n ≤ ∗ n n ≤ µ(S E A) ε. Hence, n n \ ≤ n [ [ [ µ(A) µ( En) = lim µ0( Ej) = ν( En) ≤ n n →∞ j=1 n [ [ ν(A) + ν( E A) ν(A) + µ( E A) ν(A) + ε. ≤ n \ ≤ n \ ≤ n n Thus, taking ε 0 implies that ν(A) = µ(A). → Finally, if µ is σ-finite, then we may write X = U X for X pairwise disjoint 0 n n ∈ A and µ (X ) < . So for any A we have A = U (A X ). Since µ(A X ) < 0 n ∞ ∈ F n ∩ n ∩ n ∞ for all n we get that

X X µ(A) = µ(A X ) = ν(A X ) = ν(A). ∩ n ∩ n n n

ut

Exercise 6.9. [Folland, p.32, ex.18] Let 2X be an algebra. Let be all count- A ⊂ Aσ able unions of sets in ; let be all countable intersections of sets in . Let µ be A Aσδ Aσ 0 a pre-measure on and let µ be the induced outer measure. A ∗ 60

(1) Show that for any E X and any ε > 0 there exists A such that ⊂ ∈ Aσ µ (A) µ (E) + ε and E A. ∗ ≤ ∗ ⊂ (2) Show that if µ (E) < , then E is µ -measurable if and only if there exists ∗ ∞ ∗ B with E B and µ (B E) = 0. ∈ Aσδ ⊂ ∗ \ (3) Show that if µ is σ-finite then the assumption in (b) above that µ (E) < is 0 ∗ ∞ superfluous.

Solution to ex:6.9. :( ♣

Fix ε > 0. Let (A ) be a sequence in such that E A := S A and • n n A ⊂ n n P µ (A ) µ (E) + ε. Then since µ (A) P µ (A ) = P µ (A ) we are n 0 n ≤ ∗ ∗ ≤ n ∗ n n 0 n done. ( ) Let (A ) be sets in such that B = T S A and such that • ⇐ n,k n,k A k n n,k ∈ Aσδ E B and µ (B E) = 0. ⊂ ∗ \ Let F X. Then, since A are all µ -measurable, then also B is µ - ⊂ n,k ∗ ∗ measurable. So,

c c c c µ∗(F E) + µ∗(F E ) µ∗(F B) + µ∗(F E B ) + µ∗(F E B) ∩ ∩ ≤ ∩ ∩ ∩ ∩ ∩ c µ∗(F B) + µ∗(F B ) + µ∗(B E) = µ∗(F ). ≤ ∩ ∩ \

This holds for any F so E is µ∗-measurable. (*) Note: this direction did not require µ (E) < . ∗ ∞ ( ) If E is µ -measurable, then for any n > 0 there exists A such that ⇒ ∗ n ∈ Aσ E A and µ (A ) µ (E) + 1 . Thus, for B = T A , since E B, ⊂ n ∗ n ≤ ∗ n n n ∈ Aσδ ⊂ c µ∗(B) = µ∗(B E) + µ∗(B E ) = µ∗(E) + µ∗(B E). ∩ ∩ \ Also, since B A for all n, ⊂ n

µ∗(B) inf µ∗(An) µ∗(E). ≤ n ≤ Thus, µ (B E) 0. ∗ \ ≤ 61

The only direction we need to show is ( ) because the other direction holds • ⇒ without the finite measure assumption. If X = U F such that µ (F ) < for all n, then for any A X, since F n n 0 n ∞ ⊂ n are all µ∗-measurable, X µ∗(A) = µ∗(A F ). ∩ n n (This was shown in a previous exercise.) For every k we have that E F is measurable and has finite outer measure. ∩ k So let B F be a set such that E F B and µ (B (E F )) = 0. k ⊂ k Aσδ ∩ k ⊂ k ∗ k \ ∩ k Let B = U B . We have that E B and B F Ec = B (E F ). So, k k ⊂ ∩ k ∩ k \ ∩ k c X c X µ∗(B E ) = µ∗(B F E ) = µ∗(B (E F )) = 0. ∩ ∩ k ∩ k \ ∩ k k k Thus we are left with showing that B . ∈ Aσδ For every k write

\ [ k Bk = An,j, n j where Ak and Ak F for all n, j, k. Writing C := U S Ak we have n,j ∈ A n,j ⊂ k n k j n,j that C . We have that B = T (C F ) = F T C , so n ∈ Aσ k n n ∩ k k ∩ n n ] ] \ \ B = B = F C = C . k k ∩ n n ∈ Aσδ k k n n

:) X

Exercise 6.10. [Folland p.32, ex.19] Let µ∗ be an outer measure induced from a  c pre-measure µ0 on X such that µ0(X) < . Define µ (A) = µ0(X) µ∗(A ). Show ∞ ∗ − that µ (A) µ∗(A) and that A is measurable if and only if µ (A) = µ∗(A). ∗ ≤ ∗

Exercise 6.11. [Folland p.32, ex.23] Let be the collection of finite unions of sets A  of the form (a, b] Q, for all a < b . Show that is an algebra over Q. ∩ −∞ ≤ ≤ ∞ A 62

What is the σ-algebra generated by ? A Define µ ( ) = 0 and µ (A) = for = A . Show that µ is a pre-measure, and 0 ∅ 0 ∞ ∅ 6 ∈ A 0 Q that µ0 can be extended to a measure on 2 in more than one way.

Exercise 6.12. [Folland p.33] Let (X, , µ) be a finite measure space. Let µ be F ∗ the outer measure induced by µ. Let Y be such that µ (Y ) = µ (X). 6∈ F ∗ ∗ Show that if A, B are such that A Y = B Y , then µ(A) = µ(B). ∈ F ∩ ∩ Define := A Y : A . FY { ∩ ∈ F} Show that is a σ-algebra on Y . FY Define ν on (Y, ) by ν(A Y ) = µ(A), which is well defined by the above. Prove FY ∩ that ν is a measure.

Number of exercises in lecture: 12 Total number of exercises until here: 63 63

Measure Theory

Ariel Yadin

Lecture 7: Lebesgue-Stieltjes Theory

7.1. Lebesgue-Stieltjes measure

Define 1 = (R) to be the collection of finite disjoint unions of intervals of the form A A (a, b], (a, ), ( , b], for < a < b < . ∞ −∞ ∅ −∞ ∞

Exercise 7.1. Show that 1 is an algebra on R and that σ( 1) = 1 (the Borel A A B σ-algebra).

Lemma 7.1. Let F : R R be a non-decreasing function that is right-continuous • → (continue ´adroite). Define µ ( ) = 0 and for a < b < , µ ((a, b]) = F (b) 0 ∅ −∞ ≤ ∞ 0 − F (a). Also, define µ((a, )) = F ( ) F (a). (Here by F ( ) and F ( ) we mean ∞ ∞ − −∞ ∞ the limits at and respectively.) For any A that is a finite disjoint union of −∞ ∞ ∈ A1 such intervals, A = I I , define µ (A) = µ (I ) + + µ (I ). 1 ]···] n 0 0 1 ··· 0 n Then, µ is a well defined pre-measure on . 0 A1 Proof. Let denote the collection of all intervals of the form (a, b], (a, ), ( , b], for I ∞ −∞ ∅ < a < b < . −∞ ∞ Well defined. To show that µ is well defined: Note that if I,J , then also 0 ∈ I I J .Now, if (a, b] = (a , b ] (a , b ], then without loss of generality a = a < ∩ ∈ I 1 1 ]···] n n 1 b = a < b = = a < b = b. So, 1 2 2 ··· n n n X F (b) F (a) = F (b ) F (a ). − j − j j=1 Also, a similar identity holds when writing ( , b] or (a, ) as finite disjoint unions. −∞ ∞ Thus, if A = I I = J J then for all i we have that I = (I J ) 1 ]···] n 1 ]···] k i i ∩ 1 ]···] 64

(Ii Jk), so ∩ k X µ (I ) = µ (I J ), 0 i 0 i ∩ j j=1 and we get that n k X X X µ (I ) = µ (I J ) = µ (J ). 0 i 0 i ∩ j 0 j i=1 i,j j=1

This shows that µ0 is well defined.

Finite additivity. This also proves that µ0 is finitely additive. Indeed, if A1,...,An are pairwise disjoint sets in , then we may write A = Unj I for I , and then A1 j k=1 j,k j,k ∈ I n n nj n nj n ] ] ] X X X µ0( Aj) = µ0( Ij,k) = µ0(Ij,k) = µ0(Aj). j=1 j=1 k=1 j=1 k=1 j=1

We will use the finite additivity of µ0 in what follows. Pre-measure. To show that it is a pre-measure we need to consider (A ) n n ⊂ A1 that are pairwise disjoint and U A = A . Since every A is a finite disjoint n n ∈ A1 n union of intervals, we may assume without loss of generality that An is an interval for all n. Since A = I I is also a finite disjoint union of intervals, we have that 1 ]···] k U (A I ) = I which is a countable disjoint union of intervals whose union is an n n ∩ j j interval. So using the additivity of µ0, we may assume without loss of generality that

A is an interval. Also, we may re-order (An)n so that An = (an, bn] and an < bn = an+1 Un for all n. Set Bn = j=1 Aj. A B , so µ (A) = µ (B ) + µ (A B ). Thus, \ n ∈ A1 0 0 n 0 \ n n X µ (A) = µ (B ) + µ (A B ) µ (A ). 0 0 n 0 \ n ≥ 0 j j=1 Taking n we get that µ (A) P µ (A ). So we only need to prove the other → ∞ 0 ≥ n 0 n inequality. Start with the case A = (a, b] with < a < b < . Fix ε > 0. F was assumed to −∞ ∞ be right-continuous. So we may choose δ > 0 and δn > 0 such that for all n,

n F (a + δ) < F (a) + ε and F (bn + δn) < F (bn) + ε2− .

We have the open cover [a + δ, b] S (a , b + δ ), and since [a + δ, b] is compact, there ⊂ n n n n is a finite sub-cover. That is, there exists m > 0 such that 65

[a + δ, b] Sm (a , b + δ ), • ⊂ k=1 k k k and for all k, we have b + δ (a , b + δ ). • k k ∈ k+1 k+1 k+1 In this case we get that

µ (A) = F (b) F (a) F (b) F (a + δ) + ε F (b + δ ) F (a ) + ε 0 − ≤ − ≤ m m − 1 m 1 X− = F (b + δ ) F (a ) + F (a ) F (a ) + ε m m − m j+1 − j j=1 m 1 X− F (b + δ ) F (a ) + F (b + δ ) F (a ) + ε ≤ m m − m j j − j j=1 m 1 X− = F (b + δ ) F (b ) + F (b ) F (a ) + ε j j − j j − j j=1 m m X X j µ (A ) + ε2− + ε ≤ 0 j j=1 j=1

X X n X µ (A ) + ε2− + ε µ (A ) + 2ε. ≤ 0 n ≤ 0 n n n n Taking ε 0 completes the case where A = (a, b]. → If A = ( , b] then for any large enough r > 0 we have that ( r, b] = U (A ( r, b]). −∞ − n n∩ − Since µ (A ( r, b]) µ (A ( r, b]) + µ (A ( r, b]c) = µ (A ), 0 n ∩ − ≤ 0 n ∩ − 0 n ∩ − 0 n X X F (b) F ( r) = µ (( r, b]) = µ (A ( r, b]) = µ (A ), − − 0 − 0 n ∩ − 0 n n n and taking r we get that → ∞ X µ (A) = F (b) F ( ) µ (A ). 0 − −∞ ≤ 0 n n If A = (a, ), then similarly, for any large enough r > 0 we have that (a, r] = ∞ U (A (a, r]), so n n ∩ X X F (r) F (a) = µ (A (a, r]) µ (A ), − 0 n ∩ ≤ 0 n n n and taking r completes the proof. → ∞ ut

??? THEOREM 7.2 (Lebesgue-Stieltjes measure). If F : R R is a right-continuous → non-decreasing function, then there exists a unique measure µF on (R, ) such that for B 66 all a < b,

µ ((a, b]) = F (b) F (a). F − In fact, µ = µ if and only if F G is constant. F G − Moreover, one may recover F from µF : If µ is a measure on (R, ) that is finite on B all bounded sets in , if we define B  µ((0, x]) x > 0   F (x) = 0 x = 0    µ((x, 0]) x < 0 − then F is non-decreasing and right-continuous, and µF = µ.

Proof. Lemma 7.1 and Charath´odory’sExtension Theorem tell us that we may extend the pre-measure µ0((a, b]) = F (b) F (a) uniquely to (R, ). Also, if µF = µG then − B F (b) F (a) = G(b) G(a) for all a < b which implies that F = G + F (0) G(0). − − − If µ is a measure on (R, ) that is finite on all bounded sets in , one may check B B that by the definition of F in the theorem, µ((a, b]) = F (b) F (a) for all a < b. So F − is non-decreasing (since µ is non-negative). F is right continuous from the continuity properties of measures: For any x and ε 0 n & \ F (x + ε ) F (x) = µ((x, x + ε ]) µ( (x, x + ε ]) = µ( ) = 0. n − n → n ∅ n

ut

X The measure µF is called the Lebesgue-Stieltjes measure associated to F .

Exercise 7.2. Show that µF satisfies the following properties.  µ ( a ) = F (a) F (a ). • F { } − − µ ((a, b)) = F (b ) F (a). • F − − µ ([a, b]) = F (b) F (a ). • F − − µ ([a, b)) = F (b ) F (a ). • F − − − 67

Exercise 7.3. Show that for F (x) = x the associated measure µF is Lebesgue  measure on R.

Solution to ex:7.3. :( ♣ Note that µ is a measure with µ((0, 1]) = 1 and µ((a, b] + x) = F (b + x) F (a + x) = F − b a = µ((a, b]). So µ is the extension of Jordan measure on elementary sets, and must − be Lebesgue measure.:) X

Exercise 7.4. Let A R be Lebesgue measurable. Suppose that m(A) > 0. Show ⊂ that for any 0 < ε < 1 there exists an open interval I such that m(A I) (1 ε)m(I). ∩ ≥ −

Solution to ex:7.4. :( ♣ First suppose the 0 < m(A) < . Define ∞

n m(A I) o β = sup m(I∩) : I is an open interval .

Of course β 1. We need to show that β = 1. ≤ Fix ε > 0 and let (In)n be disjoint intervals (not necessarily open or closed) such that A U I and P λ(I ) m(A) + ε. ⊂ n n n n ≤ Recall that Lebesgue measure of all points is 0. So m(A I) is the same for open, ∩ closed, half-open-closed intervals, and similarly for m(I). Thus, m(A I ) βm(I ) ∩ n ≤ n for all n, and we obtain

X X m(A) = m(A I ) β m(I ) β(m(A) + ε). ∩ n ≤ n ≤ n n

Taking ε 0 we get that β = 1. → 68

In the case that m(A) = , let N be large enough so that m(A ( N,N)) > 0. ∞ ∩ − Since m(A ( N,N)) 2N < , by the previous part, for any 0 < ε < 1 there exists ∩ − ≤ ∞ an open interval I such that

m(A I) m(A ( N,N) I) > (1 ε)m(I). ∩ ≥ ∩ − ∩ −

:) X

7.2. Regularity

We have already seen outer and inner regularity for Lebesgue measure. This is a more general fact, for Lebesgue-Stieltjes measures.

Proposition 7.3 (Outer and inner regularity). For any Lebesgue measurable set A • we have

µF (A) = inf µF (U) = sup µF (K). A U open A K compact ⊂ ⊃ Proof. If µ(A) = then any U A also admits µ(U) = so outer regularity is ∞ ⊃ ∞ simple in this case. So assume that µ(A) < . By the definition of the outer measure ∞ from the pre-measure µ0, for every ε > 0 there exists intervals ((an, bn])n such that A S (a , b ] and P µ((a , b ]) µ(A) + ε. (Note that even if A is unbounded ⊂ n n n n n n ≤ this is possible with only bounded intervals.) For every n there exists δn > 0 such that n S F (bn + δn) < F (bn) + ε2− . Thus, taking U = n(an, bn + δn), which is an open set containing A, we get that X X µ(U) µ((a , b + δ ]) = F (b + δ ) F (a ) ≤ n n n n n − n n n X X F (b ) F (a ) + ε = µ((a , b ]) + ε µ(A) + 2ε. ≤ n − n n n ≤ n n Taking infimum of the left hand side, and ε 0 we have outer regularity. → For inner regularity, first assume that A is bounded. Consider B = A A. Fix ε > 0. \ There exists an open set U such that B U and µ(U) < µ(B) + ε. Let K = A U. ⊂ \ This is a closed bounded set, so it is compact. Also, K A and ⊂ µ(B) + µ(A) = µ(A) µ(K) + µ(U) µ(K) + µ(B) + ε, ≤ ≤ 69 so µ(K) µ(A) ε. Taking supremum of such K A and ε 0 we have inner ≥ − ⊂ → regularity for bounded A. If A is unbounded: Let ε > 0. For every r > 0 there exists a compact K A ( r, r) r ⊂ ∩ − such that µ(K ) µ(A ( r, r)) ε. Note that µ(A ( r, r)) µ(A) by monotone r ≥ ∩ − − ∩ − → convergence. So

sup µ(K) lim µ(A ( r, r)) ε = µ(A) ε. A K compact ≥ r ∩ − − − ⊃ →∞ Taking ε 0 completes the proof. → ut

7.3. Non-measureable sets

In this section we will show that the Lebesgue measurable sets are not all of 2R; L that is, there exist non-Lebesgue-measurable sets (and specifically non-Borel sets).

Exercise 7.5. Let Q = Q [0, 1). Define an equivalence relation on R by r r0 if ∩ ∼  r r0 Q. Let R be a set of representatives for the equivalence classes of this relation − ∈ (chosen by the axiom of choice!). For every q Q let R = r + q (mod 1) : r R [0, 1) . ∈ q { ∈ ∩ } Show that if R then also R for all q Q and λ(R ) = λ(R) where λ is ∈ L q ∈ L ∈ q Lebesgue measure. Show that this leads to a contradiction, so R . 6∈ L

7.4. Cantor set

The Cantor set C [0, 1] is defined as follows. Start with C = [0, 1]. Given C ⊂ 0 n define Cn+1 by “removing the middle third of each interval composing Cn”; formally:

1 [ 2 1
Cn+1 = 3 Cn 3 + 3 Cn .

Then take \ C = Cn. n 70

Exercise 7.6. Show that C is Lebesgue measurable and has Lebesgue measure 0. (*) Show that C has cardinality of [0, 1].

Solution to ex:7.6. :( ♣ Note that for every n the set Cn is the union of two measurable sets so is measurable by induction. Thus, C is measurable as an intersection of measurable sets. Let us calculate the measure of C . Since C [0, 1], we have that 1 C [0, 1 ] and n n ⊂ 3 n ⊂ 3 2 + 1 C [ 2 , 1]. This implies that these sets are disjoint, and 3 3 n ⊂ 3 1 2 1 1 1 2 λ(Cn+1) = λ( 3 Cn) + λ( 3 + 3 Cn) = 3 λ(Cn) + 3 λ(Cn) = 3 λ(Cn).

n Thus, λ(Cn) = (2/3) and

λ(C) inf λ(Cn) = 0. ≤ n :) X

Number of exercises in lecture: 6 Total number of exercises until here: 69 71

Measure Theory

Ariel Yadin

Lecture 8: Functions of measure spaces

8.1. Products

Let (X, ), (Y, ) be two measurable spaces. What is the natural measurable struc- F G ture on X Y ? Naturally, any set A B for A ,B should be measurable in the × × ∈ F ∈ G product.

Exercise 8.1. Let (X, ), (Y, ) be two measurable spaces. Let π : X Y X F G X × → and π : X Y Y be the natural projections. Show that Y × → 1 1 σ(A B : A ,B ) = σ(π− (A), π− (B): A ,B ). × ∈ F ∈ G X Y ∈ F ∈ G

Solution to ex:8.1. :( ♣ We only need to show that every generator in one σ-algebra is in the other. If A ,B then ∈ F ∈ G 1 1 π− (A) = A Y and π− (B) = X Y, X × Y × which are both in the left σ-algebra. 1 1 On the other hand, A B = π− (A) π− (B) which is in the right σ-algebra.:) × X ∩ Y X

Exercise 8.2. Generalize the previous exercise: If ((X , )) is a sequence of n Fn n measurable spaces, then

1 σ(A A : A n) = σ(π− (A ): A ), 1 × · · · × n × · · · n ∈ Fn ∀ n n n ∈ Fn 72 where π : X X X is the natural projection. n 1 × · · · × n × · · · → n

X We use the notations

n O O X = X X and X X n 1 × · · · × n × · · · 1 × · · · × n n j=1 and n O O = σ( A : A n) and = σ( n A : A j). Fn ⊗n n n ∈ Fn ∀ Fj ⊗j=1 j j ∈ Fj ∀ n j=1 These are called product σ-algebras. Also, O (X , ) = ( X , ) n Fn ⊗n n ⊗nFn n and similarly for finite products. These latter spaces are called product (measure) spaces.

Exercise 8.3. Show that (Rd) = d (R). B ⊗j=1B 

Exercise 8.4. Show that if = σ( ) for some sets , then Fn En En  O 1 = σ(π− (E ): E ). Fn n n n ∈ En n

Solution to ex:8.4. :( ♣ Let O 1 σ = = σ(π− (A ): A ) 1 Fn n n n ∈ Fn n by a previous exercise. Let σ = σ(π 1(E ): E ). 2 n− n n ∈ En 73

It is immediate that σ σ . 2 ⊂ 1 So we only need to show that σ σ . For this it suffices to prove that for any n and 1 ⊂ 2 any A we have that π 1(A ) σ . n ∈ Fn n− n ∈ 2 Fix n and set = A X : π 1(A) σ . Then: π 1( ) = σ so . Gn ⊂ n n− ∈ 2 n− ∅ ∅ ∈ 2 ∅ ∈ Gn Also, if A then π 1(Ac) = π 1(A)c σ , so Ac as well; i.e. , is closed under ∈ Gn n− n− ∈ 2 ∈ Gn G complements. If (A ) is a sequence in then π 1(S A ) = S π 1(A ) σ ; k k ⊂ Gn Gn n− k k k n− k ∈ 2 so is closed under countable unions as well. In conclusion is a σ-algebra. By Gn Gn definition of σ , we have that , so = σ( ) . 2 En ⊂ Gn Fn En ⊂ Gn Since this holds for all n, we conclude that for any n and any A we have n ∈ Fn π 1(A ) σ .:) n− n ∈ 2 X

Of course now one would like to construct a product measure space from two (or more) measure spaces. For this, it is convenient to first go through the theory of measurable functions.

8.2. Measurable functions

Definition 8.1. Let (X, ), (Y, ) be measurable spaces. A function f : X Y is • F G → called ( , )-measurable, or just measurable, if f 1(A) for all A . F G − ∈ F ∈ G That is, f pulls back measurable sets to measurable sets. Sometimes we denote this by f :(X, ) (Y, ). F → G

Proposition 8.2. Let (X, ), (Y, ) be measurable spaces. Let f : X Y . Suppose • F G → that = σ( ). G K f is measurable if and only if for every K , f 1(K) . ∈ K − ∈ F

Proof. One direction is trivial. For the other direction, verify that A Y : f 1(A) is a σ-algebra containing ⊂ − ∈ F , and so it contains as well. K G ut

Exercise 8.5. Show that if X,Y are topological spaces and f : X Y is a con- → tinuous function, then f is Borel measurable (i.e. ( (X), (Y ))-measurable). B B 74

d For a function f : R C we say that f is Borel if f is ( d, (C))-measurable X → B B and f is Lebesgue if f is ( , (C))-measurable. Note that Borel implies Lebesgue, but L B the converse is not necessarily true.

Exercise 8.6. Show that for measurable spaces (X, ), (Y, ), (Z, ) and measur- F G H able functions f :(X, ) (Y, ) and g :(Y, ) (Z, ), we have that g f : X Z F → G G → H ◦ → is ( , )-measurable. F H

Exercise 8.7. Let (X, ), (Y, ), (Z, ) be measurable spaces, and π : X Y F G H X × → X, π : X Y Y the natural projections. Show that f : Z X Y is measurable if Y × → → × and only if π f and π f are measurable. X ◦ Y ◦

Exercise 8.8. Let (X, ), (Y, ), (Z, ) be measurable spaces. Let f : X F G H → Y, g : X Z be measurable functions. Show that F : X Y Z defined by → → × F (x) = (f(x), g(x)) is a measurable function.

Solution to ex:8.8. :( ♣ If π : Y Z Y, π : Y Z Z are the natural projections, then π F = f, π F = g Y × → Z × → Y ◦ Z ◦ which are measurable.:) X

Exercise 8.9. Show that f :(X, ) C is measurable if and only if Ref, Imf F → are measurable. 75

Exercise 8.10. Show that if f, g :(X, ) C are measurable then f + g and fg F → are also.

Solution to ex:8.10. :( ♣ Let F (x) = (f(x), g(x)) and ψ(x, y) = x + y, φ(x, y) = xy. Then ψ, φ are continuous, and F is measurable. So f + g = ψ F and fg = φ F are measurable.:) ◦ ◦ X

Exercise 8.11. [Folland p.48] Show that for f :(X, ) R, if for all q Q we F → ∈ have that f 1( , q] then f is measurable. − −∞ ∈ F

Exercise 8.12. Show that any monotone function f : R R is Borel. → 

We will also like to have functions with values . One must be careful when X ±∞ adding and when multiplying 0 , but as a topological space we can consider ∞ − ∞ · ∞ [ , ]. The Borel σ-algebra is defined in the same way as for R: it is the σ-algebra −∞ ∞ generated by intervals.

Proposition 8.3. Let (f ) be a sequence of measurable functions f :(X, ) • n n n F → [ , ]. Then, −∞ ∞

sup fn inf fn lim inf fn lim sup fn n n n n are all measurable.

If the limit limn fn(x) = f(x) exists for all x then f is measurable. →∞ 76

Proof. For any a , ≤ ∞

1 \ 1 (sup f )− ( , a] = f − ( , a] . n −∞ n −∞ ∈ F n n Since ( , a] generate ([ , ]) we get that sup f is measurable. −∞ B −∞ ∞ n n Similarly,

1 [ 1 (inf fn)− ( , a] = f − ( , a] . n −∞ n −∞ ∈ F n A bit more cumbersome to check, but not too difficult is

1 \ [ 1 (lim inf f )− ( , a] = f − ( , a], n −∞ k −∞ n k n ≥ and

1 [ \ 1 (lim sup f )− ( , a] = f − ( , a]. n −∞ k −∞ n k n ≥ Which gives the measurability of lim inf fn and lim sup fn. Finally, if f = lim f exists then f = lim sup f = lim inf f and so is measurable. n n n ut

Exercise 8.13. Show that for f :(X, ) [ , ], F → −∞ ∞  + f = max 0, f and f − = max 0, f { } { − } are measurable if f is.

Exercise 8.14. Show that for measurable f :(X, ) C the functions f : X R F → | | →  and sgnf : X R defined by f (x) = f(x) and → | | | |   f(x)  f(x) f(x) = 0, sgnf(x) = | | 6 0 f(x) = 0, are measurable. Show that arg f is measurable. 77

Exercise 8.15. [Folland p.48] Let (X, ) be a measurable space. Let f : X F →  1 1 [ , ] and let Y = f − (R). Show that f is measurable if and only if f − ( ) −∞ ∞ ∞ ∈ 1 , f − ( ) and f is measurable as a function f :(Y, Y ) R, where F −∞ ∈ F Y Y F → = A Y : A . FY { ∩ ∈ F}

Exercise 8.16. [Folland p.48] Let (X, ) be a measurable space. Let f, g : X F → [ , ]. −∞ ∞ Define fg(x) = f(x)g(x) where 0 = 0 ( ) = 0 and x ( ) = x = ±∞ · · ±∞ · ±∞ ±∞ · ±∞ for x > 0 and x ( ) = x = if x < 0. · ±∞ ±∞ · ∓∞ Show that if f, g are measurable then fg is measurable. For a [ , ] define ∈ −∞ ∞  f(x) + g(x) if f(x) = g(x) = or f(x) g(x) < , (f + g)a(x) = 6 − ±∞ | | ∧ | | ∞ a if f(x) = g(x) = − ±∞ (that is, replacing the problematic with a). ∞ − ∞ Show that if f, g are measurable then (f + g)a is measurable.

Exercise 8.17. Show that if (f ) is a sequence of measurable functions on (X, ) n n F  then A = x : limn fn(x) exists is a measurable set. Show that for any a R, { →∞ } ∈  limn fn(x) if the limit exists, f(x) = →∞ a otherwise is measurable. 78

8.3. Simple functions

Definition 8.4 (Simple functions). Let (X, ) be a measurable space. • F The indicator function of a set A X, denoted 1 is the function ⊂ A  1 x A 1A : X R 1A(x) := ∈ → 0 x A. 6∈ A simple function is a measurable function f : X [0, ) such that f(X) is a → ∞ finite set.

Exercise 8.18. Let (X, ) be a measurable space, and let A X. Show that 1 F ⊂ A is measurable if and only if A . ∈ F

Exercise 8.19. Let (X, ) be a measurable space. Show that for any simple func- F tion f we can write f = Pn a 1 where 0 < a < a < < a and A for all j=1 j Aj 1 2 ··· n j ∈ F j. Show that in fact, we have the standard representation X f = a1f −1(a). 0

Solution to ex:8.19. :( ♣ Since f(X) is finite, and since f 0, we may write f(X) = a < a < < a for ≥ { 1 2 ··· n} a > 0 (if 0 f(X)) or f(X) = 0 = a < a < < a (in the case that 0 f(X)). 1 6∈ { 0 1 ··· n} ∈ Since f is measurable, A := f 1(a ) for all j. Finally, since X = f 1(f(X)) = j − j ∈ F − U f 1(a ) = U A , for every x X there exists a unique j(x) such that x A . j − j j j ∈ ∈ j(x) 79

[0, ) ∞

n 2− (k + 1)

n 2− k

n ϕn 1 2− − ϕn

X

An,k

Figure 3. Approximating functions with simple functions

Note that f(x) = a = a 1 (x). Since (A ) are disjoint, if 1 (x) = 0 then j(x) j(x) Aj(x) j j Aj 6 1 (x) = 0 for all i = j. Altogether this gives Ai 6 X X f(x) = a1 f(a)=x = a1f −1(a)(x). { } a f(X) 0

The following is a very useful building block in the theory of measurable functions.

Proposition 8.5. Let (X, ) be a measurable space. Let f : X C be a Borel • F → function.

If f 0 then there exists a monotone sequence of simple functions 0 ϕ • ≥ ≤ 1 ≤ ϕ ϕ such that ϕ f. In fact, the convergence is uniform on 2 ≤ · · · ≤ n ≤ · · · n % any set for which f is bounded. If f is real-valued then f = f + f and f +, f are non-negative Borel functions. • − − − In general, f = Ref + iImf where Ref, Imf are real valued and Borel. So • f = f f + i(f f ) where all f are non-negative Borel functions. 1 − 2 3 − 4 j 80

Proof. We only prove the first bullet. The other bullets have actually been proven in previous exercises. If f 0 then for every n define ϕ := min n, 2 n 2nf . ≥ n { − b c} n n Note that ϕn(X) 2− k : k N , k 2 n which is finite, so ϕn is simple (why ⊂ { ∈ ≤ } 1 n n is it measurable?). Also, one may verify that for the sets An,k = f − [2− k, 2− (k + 1)) and E = f 1[n + 2 n, ) we have that the standard representation of ϕ is n − − ∞ n 2nn X n ϕn = 2− k1An,k + n1En . k=0 To show that ϕ ϕ , note that 2 r 2r for any r 0, so n ≤ n+1 b c ≤ b c ≥ n n n n n 1 n ϕ = n 2− 2 f (n + 1) 2− 2 f (n + 1) 2− 2 2 f = ϕ . n ∧ b c ≤ ∧ b c ≤ ∧ 2 b · c n+1

Now, if A is a set on which f is bounded, say supx A f(x) = M < , then for all ∈ ∞ n n n n n > M we have that supx A 2− 2 f(x) < n so ϕn(x) = 2− 2 f(x) for all x A. ∈ b c b c ∈ n Thus, supx A ϕn(x) f(x) 2− for all n > M. So ϕn f uniformly on A. ∈ | − | ≤ → ut

Exercise 8.20. Let f :(X, ) [0, ] be a measurable function. Show that F → ∞ there exists a sequence of simple functions 0 ϕ ϕ ϕ such that ≤ 1 ≤ 2 ≤ · · · ≤ n ≤ · · · ϕ f. n %

Solution to ex:8.20. :( ♣ c 1 Write f = f1 + 1 c where 0 = 0 = 0 and A = f ( ). A ∞ A · ∞ ∞ · − ∞ f1A is measurable and non-negative, so there is a monotone sequence a simple func- tions ψ f1 . Define ϕ = ψ 1 + n1 c .:) n % A n n A A X

Number of exercises in lecture: 20 Total number of exercises until here: 89 81

Measure Theory

Ariel Yadin

Lecture 9: Integration: positive functions

9.1. Integration of simple functions

Let (X, , µ) be a measure space. Let ϕ be a simple function, with standard repre- F Pn sentation ϕ = j=1 aj1Aj . Define the integral of ϕ with respect to µ as

n Z X ϕdµ := ajµ(Aj). j=1 For A define ∈ F Z Z ϕdµ := ϕ1Adµ. A Proposition 9.1. Let ϕ, ψ be simple functions on a measure space (X, , µ). Then: • F For all r > 0 we have R rϕdµ = r R ϕdµ. • If ϕ = Pn a 1 for pairwise disjoint (A )n , then R ϕdµ = Pn a µ(A ). • j=1 j Aj j j=1 j=1 j j R (ϕ + ψ)dµ = R ϕdµ + R ψdµ. • If ϕ ψ then R ϕdµ R ψdµ. • ≤ ≤

Proof. For r > 0, rϕ is a simple function such that rϕ(X) = ra : a ϕ(X) . So, { ∈ } X rϕ = ra1ϕ−1(a) 0

X X ϕ = a1Ea,b and ψ = b1Ea,b . a ϕ(X),b ψ(X) a ϕ(X),b ψ(X) ∈ ∈ ∈ ∈ So by the above,

Z X Z Z (ϕ + ψ)dµ = (a + b)µ(Ea,b) = ϕdµ + ψdµ. a ϕ(X),b ψ(X) ∈ ∈ Finally, if ϕ ψ we claim that for all a ϕ(X), b ψ(X) we have that aµ(E ) ≤ ∈ ∈ a,b ≤ bµ(E ), because either E = or if x E then a = ϕ(x) ψ(x) = b. Thus, a,b a,b ∅ ∈ a,b ≤ Z X X Z ϕdµ = aµ(E ) bµ(E ) = ψdµ. a,b ≤ a,b a ϕ(X),b ψ(X) a ϕ(X),b ψ(X) ∈ ∈ ∈ ∈

ut

Proposition 9.2. Let ϕ be a simple functions on a measure space (X, , µ). The • F map A R ϕdµ is a measure on (X, ). 7→ A F

Proof. The function ϕ1 is just the zero function, which has 0 integral by definition. ∅ For any E , ∈ F X ϕ1E = a1ϕ−1(a) E. ∩ a ϕ(X) ∈ 1 The sets (ϕ− (a) E)a ϕ(X) are pairwise disjoint so ∩ ∈ Z X 1 ϕdµ = aµ(ϕ− (a) E). ∩ E a ϕ(X) ∈ 83 U Now if A = n An then Z Z X 1 X X 1 X ϕdµ = aµ(ϕ− (a) A) = a µ(ϕ− (a) A ) = ϕdµ. ∩ ∩ n A a ϕ(X) a ϕ(X) n n An ∈ ∈

ut

9.2. Integration of positive functions

We use L+(X, ) to denote the set of measurable functions f :(X, ) [0, ]. F F → ∞

Definition 9.3. Let (X, , µ) be a measure space. For f L+(X, ) define • F ∈ F Z Z fdµ := sup  ϕdµ : 0 ϕ f , ϕ is simple . ≤ ≤

Exercise 9.1. Show that if f g for f, g L+(X, ) then R fdµ R gdµ. ≤ ∈ F ≤ Show that for any c > 0, R cfdµ = c R fdµ.

Solution to ex:9.1. :( ♣ This simple functions participating in the supremum for f also participate in the supre- mum for g. If c > 0 then for any simple function ϕ, ϕ f if and only if cϕ cf. Taking ≤ ≤ R R supremums over cϕdµ = c ϕdµ yields the result.:) X

It is usually difficult to compute supremums over such big sets (all simple functions dominated by some f L+). The next fundamental result will make life much simpler ∈ when wishing to compute integrals of positive functions.

??? THEOREM 9.4 (Monotone Convergence). Let (X, , µ) be a measure space. If F (f ) is a sequence in L+(X, ) such that 0 f f for all n (i.e. a monotone n n F ≤ n ≤ n+1 sequence), then for the limit f(x) = limn fn(x) (which always exists as f = supn fn), →∞ Z fdµ = lim fndµ. n →∞ 84

Proof. Because f f for all n we have that R f R f and so lim R f R f. We are n ≤ n ≤ n ≤ left with proving the other inequality. Let 0 < ε < 1 and let ϕ be a simple function such that 0 ϕ f. Define ≤ ≤ A = x X : f (x) (1 ε)ϕ(x) = f (1 ε)ϕ . n { ∈ n ≥ − } { n ≥ − } Note that A A because f f . Thus (A ) is an increasing sequence. Since n ⊂ n+1 n ≤ n+1 n n A R ϕdµ is a measure on (X, ), we get that 7→ A F Z Z ϕdµ ϕdµ An → because

[   A = x : n , f (x) (1 ε)ϕ(x) = x : sup f (x) (1 ε)ϕ(x) n { ∃ n ≥ − } n ≥ − n n = x : f(x) (1 ε)ϕ(x) = X. { ≥ − } Since f f 1 (1 ε)ϕ1 , we get that n ≥ n An ≥ − An Z Z Z fndµ (1 ε) ϕdµ (1 ε) ϕdµ. ≥ − · An → − · Taking supremum over ϕ and ε 0 we get that → Z Z lim fndµ fdµ, n →∞ ≥ which completes the proof. ut

Proposition 9.5 (Poor man’s Fubini). If (f ) is a sequence of functions in L+(X, ) • n n F then Z X X Z fndµ = fndµ. n n

Proof. If f, g L+(X, ), then let (ϕ ) , (ψ ) be monotone sequences of simple func- ∈ F n n n n tions such that ϕ f and ψ g. Then, ϕ + ψ f + g. So, using the Monotone n % n % n n % Convergence Theorem, Z Z Z Z (f + g)dµ = lim (ϕn + ψn)dµ = fdµ + gdµ. n →∞ 85

Now, for a sequence (f ) in L+(X, ), for any n, n n F n n Z X X Z fkdµ = fkdµ. k=1 k=1 Since (f ) are all non-negative, Pn f P f , so by Monotone Convergence again, n n k=1 k % n n n X Z Z X Z X fndµ = lim fkdµ = fndµ. n n →∞ k=1 n

ut

X For a measure µ we say that a measurable set A occurs almost everywhere, or a.e., if µ(Ac) = 0. sometimes we stress the dependence on the measure by saying µ-a.e.

For a function f we may sometimes shorthand f a = x : f(x) a , and X { ≤ } { ≤ } similarly for other measurable sets.

Proposition 9.6. For f L+(X, ), f = 0 a.e. (that is, µ( x : f(x) = 0 ) = 0) if • ∈ F { 6 } and only if R fdµ = 0.

R P 1 Proof. If f is a simple function then fdµ = 0 0 ) = ∈ − { } 0. Now, if f L+, then let ϕ f be an approximating monotone sequence of simple ∈ n % functions. Since 0 ϕ f we have that f = 0 implies that ϕ = 0 for all n and so ≤ n ≤ n R R f = lim ϕn = 0. In the other direction, if since f > 0 = S f > 1 we have that if µ(f > 0) > 0 { } n n 1  1 then there exists n such that µ(f > n ) > 0. For A = f > n , Z Z 1 1 fdµ n 1Adµ = n µ(A) > 0. A ≥

ut

Exercise 9.2. Show that if (f ) is a monotone sequence in L+ such that f f n n n % a.e., then R f dµ R fdµ. n → 86

Solution to ex:9.2. :( ♣ Let A = f f . So g = f 1 is a monotone sequence on L+ that increases to { n % } n n A + c g g := f1 . Also, f (1 1 ) = f 1 c , f(1 1 ) = f1 c L and µ(A ) = 0, so n % A n − A n A − A A ∈ Z Z Z Z fndµ = gndµ and fdµ = gdµ.

Thus, Z Z Z Z fdµ = gdµ = lim gndµ = lim fndµ. n n →∞ →∞ :) X

Lemma 9.7 (Fatou’s Lemma). Let (X, , µ) be a measure space. Let (f ) be a • F n n sequence on L+(X, ). Then, F Z Z lim inf fndµ lim inf fndµ. n ≤ →∞ R R Proof. For every j k we have that infn k fn fj and so infn k fndµ infj k fjdµ. ≥ ≥ ≤ ≥ ≤ ≥ + The sequence gk := infn k fn is a monotone sequence in L converging to gk ≥ % lim inf fn. By Monotone Convergence, Z Z Z Z lim inf fndµ = lim gkdµ lim inf fjdµ = lim inf fndµ. k ≤ k j k n →∞ →∞ ≥ →∞

ut

Exercise 9.3. Assume that (f ) is a sequence in L+ such that R sup f dµ < . n n n ∞ Show that Z Z lim sup fndµ lim sup fndµ. n ≤ →∞

Solution to ex:9.3. :( ♣ Define g = (sup f ) f . Then (g ) L+. Since g , f are non-negative functions, n n − n n n ⊂ n n 87

R f R g + R f = R sup f < , and we can subtract R f from both sides to get n ≤ n n n ∞ n that R g = R sup f R f for all n. Also, since lim inf g , lim sup f are non-negative n n − n n n functions, and since lim inf gn + lim sup fn = sup fn, we have that Z Z Z Z lim sup f dµ lim inf g dµ + lim sup f dµ = sup f dµ < , n ≤ n n n ∞ R and subtracting lim sup fndµ from both sides we have that Z Z Z lim inf g dµ = sup f dµ lim sup f dµ. n n − n However, by Fatou’s Lemma, Z Z Z Z lim inf g dµ lim inf g dµ = sup f dµ lim sup f dµ. n ≤ n n − n Since R sup f dµ < we may subtract it from both sides.:) n ∞ X

Exercise 9.4. Let f L+(X, ) for a measure space (X, , µ). Show that ∈ F F ν(A) := R fdµ is a measure on (X, ). Show that for any g L+(X, ), A F ∈ F Z Z gdν = gfdµ.

Solution to ex:9.4. :( ♣ ν( ) = 0 is simple. ∅ U P If A = n An then f1A = n f1An and so Z X Z X ν(A) = fdµ = fdµ = ν(An) A n An n (by poor man’s Fubini). This shows that ν is a measure. Pn Now, if g is a simple function with standard representation g = j=1 aj1Aj , then n n Z X X Z Z gdν = ajν(Aj) = aj fdµ = gfdµ. j=1 j=1 Aj 88

Now for general g L+, let ϕ g be a monotone sequence of simple functions ∈ n % approximating g. So ϕ f gf. Monotone Convergence guaranties that n % Z Z Z Z gdν = lim ϕndν = lim ϕnfdµ = gfdµ. n n →∞ →∞ :) X

Exercise 9.5. [Folland, p.52, ex.13] Suppose (fn)n is a sequence of non-negative functions such that f f a.e. and R f R f < . n → n → ∞ Show that for any measurable A, R f R f. A n → A Show that this is not necessarily true if R f = . ∞

Solution to ex:9.5. :( ♣ Since f 1 f1 , by Fatou’s Lemma, for any A, n A → A Z Z f lim inf fn. A ≤ A Since R f < , ∞ Z Z Z Z Z Z Z Z f = f f f lim inf fn = lim sup( fn fn) = lim sup fn. A − Ac ≥ − Ac − Ac A Hence, Z Z Z Z f lim inf fn lim sup fn f. A ≤ A ≤ A ≤ A Now for a counter-example when R f = . Set ∞ 2 fn = 1[0,n−1]n + 1[n−1,n).

So fn 1[0, ) =: f. Also, → ∞ Z Z 1 f = n + n n− = f. n − → ∞ However, for A = [0, 1], Z Z 1 fn = n + 1 n− = 1 = f. A − → ∞ 6 A :) X 89

Exercise 9.6. Let (X, , µ) be a measure space. Assume that if (f ) is a sequence F n n in L+(X, ) such that f f for all n. Assume that R f dµ < and that f f F n+1 ≤ n 1 ∞ n & a.e. R R Show that limn fndµ = fdµ. →∞

Exercise 9.7. Let (X, , µ) be a measure space and f L+(X, ) such that F ∈ F Rfdµ < . ∞ Show that f = = f 1( ) is a null set. { ∞} − ∞ Show that f > 0 is σ-finite (i.e. can be written as a countable union of finite- { } measure sets).

Solution to ex:9.7. :( ♣ If µ(f = ) > 0 then ∞ Z Z = µ(f = ) fdµ fdµ < , ∞ ∞ · ∞ ≤ f= ≤ ∞ { ∞} a contradiction. U Write f > 0 = f = ∞ n 1 < f n . We have already seen that { } { ∞} ] n=1 { − ≤ } f = is a null set (and specifically has finite measure). For any n 1, if µ(n 1 < { ∞} ≥ − f n) = then ≤ ∞ Z Z = (n 1) µ(n 1 < f n) fdµ fdµ < , ∞ − · − ≤ ≤ n 1

Measure Theory

Ariel Yadin

Lecture 10: Integration: general functions

We have already seen that any measurable function can be decomposed into real and imaginary parts, and those into positive and negative parts each. So essentially we can split the task of integrating a functions into integrating four positive functions. The only thing we have to be careful about is . ∞ − ∞

10.1. Real valued functions

Definition 10.1. Let (X, , µ) be a measure space. Let f :(X, ) R be a • F F → measurable function. Consider the integrals I+ = R f +dµ, I = R f dµ (f + = 0 f, f = 0 f). − − ∨ − ∨ − + If at least one integral I ,I− is finite then we define Z Z Z + fdµ := f dµ f −dµ − and we say the integral R fdµ exists. If I+ = I = then R fdµ does not exist (or is − ∞ undefined). If I+ < and I < then we say that f is integrable. ∞ − ∞ R R As usual we define A fdµ = f1Adµ.

Exercise 10.1. Let (X, , µ) be a measure space. Let f :(X, ) R be a mea- F F → surable function. Show that f is integrable if and only if R f dµ < . | | ∞

Proposition 10.2. For a measure space (X, , µ) the set of integrable real-valued • F functions is a vector space, and the integral is a linear functional on this space. 91

Proof. If f, g are integrable then Z Z Z Z f + αg dµ ( f + α g )dµ f dµ + α g dµ < . | | ≤ | | | || | ≤ | | | | | | ∞ So f + αg is also integrable. Now, write h = f + g. So h+ h = f + f + g+ g , which implies that − − − − − − + + + h + f − + g− = h− + f + g . Since these are all positive functions, Z Z Z Z Z Z + + + h + f − + g− = h− + f + g , which after rearranging gives Z Z Z Z Z Z Z Z Z + + + h = h h− = f f − + g g− = f + g. − − − If α > 0 then (αf)+ = αf +, (αf) = αf and ( αf)+ = αf , ( αf) = αf +. So − − − − − − Z Z Z Z Z Z Z Z + + αf = αf αf − = α f and ( αf) = αf − αf = α f. − − − −

ut 10.2. Complex valued functions

Definition 10.3. Let (X, , µ) be a measure space and let f :(X, ) C be a • F F → measurable function. We say that f is integrable if R f dµ < (this makes sense because f L+(X, )). | | ∞ | | ∈ F We define Z Z Z fdµ = Refdµ + i Imfdµ. R R As usual we define A fdµ = f1Adµ. The set of all complex valued integrable functions is denoted L1(X, , µ) = L1(X) = F L1(µ).

Exercise 10.2. f L1(X, , µ) if and only if Ref, Imf are both integrable. ∈ F Thus, the integral of a complex function is well defined. It is a (complex) linear functional the vector space L1(X, , µ). F 92

Solution to ex:10.2. :( ♣ The first assertion follows since f Ref + Imf 2 f .:) | | ≤ | | | | ≤ | | X

Exercise 10.3. Show that for f L1, ∈  Z Z

fdµ f dµ. ≤ | |

Solution to ex:10.3. :( ♣ If R f = 0 this is immediate. Otherwise, if f is real valued then f = f + + f , so | | − Z Z Z Z Z Z + + f = f f − f + f − = f . − ≤ | |

R R f R R If f is complex valued and f = 0, then set α = |R | C. So α f = αf R and 6 f ∈ ∈ Z Z Z Z Z Z Z

f = αf = Re αf = Re(αf) Re(αf) α f = f . ≤ | | ≤ | || | | |

:) X

Exercise 10.4. Show that for f L1 the set f = 0 is σ-finite. ∈ { 6 } 

Proposition 10.4. For f, g L1(X, , µ) we have that the following are equivalent. • ∈ F f = g a.e. • R f g dµ = 0. • | − | For every A , R fdµ = R gdµ. • ∈ F A A

Proof. If f = g a.e. then f g = 0 a.e. so R f g dµ = 0. | − | | − | 93

If R f g dµ = 0 then for any A , | − | ∈ F Z Z Z Z

fdµ gdµ 1A f g dµ f g dµ = 0. A − A ≤ | − | ≤ | − | If R fdµ = R gdµ for all A then also R Re(f g)dµ = R Im(f g)dµ = 0 for A A ∈ F A − A − all A . So without loss of generality we may assume that f, g are real valued. Set ∈ F A = f > g and B = f < g . We then have { } { } Z Z Z f g dµ = (f g)dµ + (g f)dµ = 0. | − | A − B − So f g = 0 a.e., and f = g a.e. | − | ut

10.3. Convergence

Exercise 10.5. Let (f ) be a sequence in L1(X, , µ) such that f f uniformly n n F n → (i.e. for every ε > 0 there exists n such that for all n > n , sup f (x) f(x) < ε). 0 0 x | n − | Show that if µ(X) < then R f dµ R fdµ. ∞ n → Show that if µ(X) = this is not necessarily true. ∞

??? THEOREM 10.5 (Dominated Convergence Theorem). Let (fn)n be a sequence in L1(X, , µ) such that f f a.e. and there exists g L1(X, , µ) such that f g F n → ∈ F | n| ≤ for all n. Then f L1(X, , µ) and R f dµ R fdµ. ∈ F n →

Proof. Let E = f f . So µ(Ec) = 0 and R h = R h for all h L1. Since { n → } E ∈ f 1 lim f 1 g, we have that f L1. | | E ≤ | n| E ≤ ∈ Suppose first that (f ) , f are real valued. Then, g f 0 and g + f 0 so by n n − n ≥ n ≥ Fatou’s Lemma, Z Z Z Z Z Z g + f = lim inf(g + f ) lim inf (g + f ) = g + lim inf f , n ≤ n n Z Z Z Z Z Z g f = lim inf(g f ) lim inf (g f ) = g lim sup f . − − n ≤ − n − n 94

Subtracting the finite quantity R g from both sides of both equations we get that

Z Z Z lim sup f f lim inf f , n ≤ ≤ n which implies these are equal. ut

Proposition 10.6. If (f ) is a sequence in L1(X, , µ) such that P R f dµ < , • n n F n | n| ∞ P R P R then f = n fn is a.e. well defined (the sum converges a.e.), and fdµ = n fndµ.

Proof. Poor man’s Fubini tells us that for g = P f we have R g = P R f < , n | n| n | n| ∞ so g L1. So N = g = is a null set, and off this set the sum f = P f converges ∈ { ∞} n n absolutely. Also, Pn f g for every n. This sequence converges a.e. to f so by Dominated | j=1 j| ≤ Convergence,

n n X Z X Z Z X Z fndµ = lim fjdµ = lim fjdµ = fdµ. n n n →∞ j=1 →∞ j=1

ut

Exercise 10.6. [Folland p.59] Let (f ) , (g ) , f, g all be functions in L1(X, , µ) n n n n F such that f f a.e., g g a.e., f g a.e., and R g dµ R gdµ. n → n → | n| ≤ n n → Show that R f dµ R fdµ. n →

Exercise 10.7. Let (X, , µ) be a measure space and let (X, ¯, µ¯) be its com- F F  pletion. Show that if f : X C is ¯-measurable, then there exists g : X C → F → such that g is -measurable and there exists a set N such that µ(N) = 0 and F ∈ F x : g(x) = f(x) N, so g1 c = f1 c (and so also g = f µ¯-a.e.) { 6 } ⊂ N N 95

Solution to ex:10.7. :( ♣ Recall that

¯ = A F : A ,F N for some N with µ(N) = 0 . F { ∪ ∈ F ⊂ ∈ F }

Suppose that f = 1A F for A F F¯, A , F N and µ(N) = 0. Then if we ∪ ∪ ∈ ∈ F ⊂ ∈ F set g = 1A N then g is measurable, and if g(x) = f(x) then x (A N) (A F ) = ∪ F 6 ∈ ∪ \ ∪ N F N. \ ⊂ P U Now suppose f = k ak1Ak Fk is a simple function for X = k(Ak Fk), Ak , ∪ ∪ ∈ F P Fk Nk with µ(Nk) = 0. Then, defining g = k ak1Ak Nk we have that if ⊂ ∈ F ∪ g(x) = f(x) then x S N which has µ-measure 0. 6 ∈ k k If f 0 and ¯-measurable, then let f f be a sequence of ¯-measurable simple ≥ F n % F functions converging monotony to f. Then for every n there is a set N and a n ∈ F simple -measurable function g such that µ(N ) = 0 and g 1 c = f 1 c . Taking F n n n Nn n Nn S N = N we have that µ(N) = 0 and g 1 c = f 1 c f1 c . Thus, g := f1 c is n n n N n N % N N -measurable. F Finally if f is any ¯-measurable function, write f = f f + i(f f ) for non- F 1 − 2 3 − 4 negative ¯-measurable functions f , . . . , f . For any j, let g be a non-negative - F 1 4 j F measurable function and N be such that g 1 c = f 1 c and µ(N ) = 0. Then j ∈ F j Nj j Nj j for g := g g + i(g g ) and N := S N we have that g is -measurable and 1 − 2 3 − 4 j j F g1N c = f1N c with µ(N) = 0. :) X

Exercise 10.8. [Folland, p.59, ex.24] Let (X, , µ) be a measure space of finite F  measure. Let (X, ¯, µ¯) be its completion. Let f : X R be a bounded non-negative F → function. Show that f is ¯-measurable if and only if there exist sequences (g ) , (h ) of - F n n n n F measurable simple functions such that g f h and R (h g )dµ < 1 for all n ≤ ≤ n n − n n n. R R R Show that in this case, fdµ¯ = limn gndµ = limn hndµ. →∞ →∞ 96

Solution to ex:10.8. :( ♣ If f is ¯-measurable, then we can find a set N such that µ(N) = 0 and f1 c is F ∈ F N -measurable. Let M > 0 be such that 0 f M (as f is bounded). Taking F ≤ ≤ 1 1 g := µ(X)nf1 c and h := µ(X)nf1 c + M1 , n µ(X)n b N c n µ(X)n d N e N we have that

g f1 c f = f1 c + f1 h . n ≤ N ≤ N N ≤ n n 1 o Also, since gn(X), hn(X) k : 0 k µ(X)nM, k N we have that gn, hn ⊂ µ(X)n ≤ ≤ ∈ are simple functions. Finally, for x N,(h g )(x) 1 so R (h g )dµ = 6∈ n − n ≤ µ(X)n n − n R 1 c (h g )dµ n , as required. N n − n ≤ − For the other direction, assume that gn, hn are as in the statement of the exercise. Let A = h g > n 1/2 . n n − n − Note that if a f b then for every n, either a n 1/2 g h b + n 1/2 or ≤ ≤ − − ≤ n ≤ n ≤ − a f b and h g > n 1/2. Also, if for every n, a n 1/2 g h b + n 1/2, ≤ ≤ n − n − − − ≤ n ≤ n ≤ − then 1/2 1/2 a = lim(a n− ) f lim(b + n− ) = b. n − ≤ ≤ n Thus,

\ n 1/2 1/2o [  a f b = a n− g h b + n− ( a f b A ) . { ≤ ≤ } − ≤ n ≤ n ≤ { ≤ ≤ } ∩ n n Since a n 1/2 g h b + n 1/2 we only need to show that the set T A − − ≤ n ≤ n ≤ − ∈ F n n has µ-measure 0. So it suffices to show that infn µ(An) = 0. To this end, Z 1 µ(An) √n (hn gn)dµ √n n− 0. ≤ An · − ≤ · → Thus, a f b is the union of a -measurable set and a subset of a µ-null set, { ≤ ≤ } F and so f is ¯-measurable. F Now if g f h as in the statement of the exercise, since f is ¯-measurable, we n ≤ ≤ n F may find a set N such that µ(N) = 0 and f1 c is -measurable. By replacing ∈ F N F gn with supk n gk we may assume without loss of generality that gn1N c f1N c , so by ≤ % R R R R monotone convergence, g dµ f1 c dµ and g dµ¯ fdµ¯. Also, n → N n → Z Z Z 1 g dµ h dµ g dµ + n− . n ≤ n ≤ n 97 R R So h dµ f1 c dµ as well. n → N R R Hence we are left with showing that f1N c dµ = fdµ,¯ and for this it suffices to prove that for all n, R g dµ = R g dµ.¯ Because g are simple functions and -measurable for n n n F all n, by linearity it suffices that for any set A , we have R 1 dµ = µ(A) =µ ¯(A) = ∈ F A R 1Adµ¯.:) X

10.4. Riemann vs. Lebesgue integration

Recall the Riemann integral of a bounded function f :[a, b] R: → For every partition P = (x )n , a = x < x < < x = b, define k k=0 0 1 ··· n n n X X SP f = sup f(x) (xk xk 1) and sP f = inf f(x) (xk xk 1). − − x [xk−1,xk] · − x [xk−1,xk] · − k=1 ∈ k=1 ∈ b b Define Iaf = inf SP f and iaf = sup sP f where the infimum and supremum are over all partitions P of [a, b]. b b R b If Iaf = iaf we say that f is Riemann integrable and define a fdx to be the common value. n Given a partition of [a, b], P = (xk)k=0, define the functions n n X X ΨP f = sup f(x) 1[xk−1,xk] and ψP f = inf f(x) 1[xk−1,xk]. x [xk−1,xk] · x [xk−1,xk] · k=1 ∈ k=1 ∈ Note that Z Z SP f = ΨP fdλ and sP f = ψP fdλ.

If f is Riemann integrable, we may choose a sequence of partitions (Pk)k such that R b lim SPk f = lim sPk f = a fdx. We may also choose the partitions as refinements of the previous ones in the sequence. In this case the functions Ψk := ΨPk f form a decreasing sequence and ψk := ψPk f form an increasing sequence. So there exist limiting functions Ψ = lim Ψ and ψ = lim ψ . Note that ψ f Ψ. k k ≤ ≤ Since f is bounded, and since λ([a, b]) < , we get that (ψ ) , (Ψ ) are dominated ∞ k k k k sequences, and so by Dominated Convergence, Z Z Z b ψdλ = Ψdλ = fdx. a So R Ψ ψ dλ = R (Ψ ψ)dλ = 0 and so Ψ = ψ a.e., which implies that Ψ = f = ψ a.e. | − | − 98

c Thus, f1A = Ψ1A for some A such that λ(A ) = 0. Since λ is a complete measure on (R, ), we have that f is Lebesgue measurable. Also, L Z Z Z b fdλ = Ψdλ = fdx. [a,b] [a,b] a Conclusion: If f is Riemann integrable on [a, b] then it is also Lebesgue measurable R R b on [a, b] and [a,b] fdλ = a fdx.

Exercise 10.9. Given a bounded function f :[a, b] R define →  Θ(x) = lim sup f(y) and θ(x) = lim inf f(y). −1 n y x n−1 n y x n →∞ | − |≤ →∞ | − |≤ Show that f is continuous at x if and only if Θ(x) = θ(x). Show that a.e. Z Z b b Θdλ = Iaf and θdλ = iaf. [a,b] [a,b] Conclude that f is Reimann integrable if and only if

λ( x : f is not continuous at x ) = 0. { }

Solution to ex:10.9. :( ♣ Write

Θn(x) = sup f(y) and θn(x) = inf f(y). −1 y x n−1 y x n | − |≤ | − |≤ Assume that Θ(x) = θ(x). Let ε > 0. Choose n large enough so that Θ (x) Θ(x) < | n − | ε/2 and θ (x) θ(x) < ε/2. So if y x < 1 we have that | n − | | − | n θ(x) ε/2 < θ (x) f(y) Θ (x) < Θ(x) + ε/2. − n ≤ ≤ n Since Θ(x) θ(x), for all y x < 1 we have that f(x) f(y) < ε. − | − | n | − | That is, for all ε > 0 there exists n large enough so that for all y x < 1 we have | − | n f(y) f(x) < ε. So f is continuous at x. | − | 99

For the other direction, if f is continuous at x, then for any ε > 0 there exists large enough n such that we have for all y x < 1 that f(y) f(x) < ε. Thus, for | − | n | − | all ε > 0 there exists n such that if n > n we have Θ (x) θ (x) < ε. Since 0 0 | n − n | Θ (x) Θ(x), θ (x) θ(x), we get that Θ(x) = θ(x). n → n → m 1 Now, consider a partition P = (xk)k=0. Let P = maxk xk xk 1 . If 2 P < , for | | | − − | | | n any x [xk 1, xk], we have that ∈ −

sup f(y) Θn(x) and inf f(y) θn(x). y [xk−1,xk] ≤ y [xk−1,xk] ≥ ∈ ∈ Thus, m Z Z b X Iaf SP f = sup f(y) dλ Θndλ, ≤ y [xk−1,xk] [xk−1,xk] ≤ [a,b] k=1 ∈ and similarly, m Z Z b X iaf sP f = inf f(y) dλ θndλ. ≥ y [xk−1,xk] ≥ k=1 ∈ [xk−1,xk] [a,b] 1 1 On the other hand, if n is very large and x [xk 1 + , xk ] then ∈ − n − n

Θn(x) sup f(y) and θn(x) inf f(y). ≤ y [xk−1,xk] ≥ y [xk−1,xk] ∈ ∈ So since f is bounded, say by f M, we have that | | ≤ Z 2 Θndλ sup f(y) (xk xk 1) + M , − n [xk−1,xk] ≤ y [xk−1,xk] · − · ∈ and Z 2 Θndλ SP f + m M n , [a,b] ≤ · · where m is the number of intervals in the partition. Similarly, Z 2 θndλ sP f m M n . [a,b] ≥ − · · By Dominated Convergence, taking n we have that for any partition P , → ∞ Z Z Θdλ SP f and [a, b]θdλ sP f. [a,b] ≤ ≥ Taking infimum and supremum respectively, we have that Z Z b b Θdλ = Iaf and θdλ = iaf. [a,b] [a,b] 100

Finally, note that f is Riemann integrable if and only if R (Θ θ)dλ = 0 which is [a,b] − if and only if Θ = θ a.e. on [a, b], which is if and only if f is continuous λ-a.e. on [a, b]. :) X

R Exercise 10.10. Let f L1(R, , µ) and F (x) := fdµ. ∈ B ( ,x]  −∞ Show that if f 0 then F is right continuous and F is continuous at x if and • ≥ only if f(x)µ( x ) = 0. { } Show that if µ( x ) = 0 then F is continuous at x. • { }

Solution to ex:10.10. :( ♣ For any x and ε, note that Z F (x + ε) F (x) = fdµ. − (x,x+ε]

First assume that f is non-negative. Then, ν(A) := R fdµ is a measure. If x x A n & then the sets An = (x, xn] are decreasing and

\ F (x ) F (x) = ν(A ) ν( A ) = ν( ) = 0. n − n → n ∅ n

So F is right continuous. If x x then n % Z F (x) F (xn) ν( x ) = fdµ = f(x)µ( x ). − → { } x { } { } because T (x , x] = x . So F is left continuous as well if and only if f(x)µ( x ) = 0. n n { } { } This resolves the case where f is non-negative. For general f L1(µ), write f = f f + i(f f ) for non-negative f . Then ∈ 1 − 2 3 − 4 j R F = F1 F2 + i(F3 F4) where Fj(x) = ( ,x] fjdµ. − − −∞ If µ( x ) = 0 then for every j the function F is continuous at x, and thus so is F .:) { } j X 101

Number of exercises in lecture: 10 Total number of exercises until here: 106 102

Measure Theory

Ariel Yadin

Lecture 11: Product measures

Recall that if (X, ), (Y, ) are measurable spaces then = σ(A B : A F G F ⊗ G × ∈ ,B ) is a σ-algebra on X Y . F ∈ G × An example is = . B1 ⊗ B1 B2 When we constructed Lebesgue measure in Rd “hands on” we took boxes as basic building blocks; these are just products of intervals which where the basic building blocks in dimension 1. We now give a general construction of a product measure on (X, ) (Y, ). F ⊗ G

Definition 11.1. For measurable spaces (X, ), (Y, ), a box is a set of the form • F G A B where A ,B . × ∈ F ∈ G

Exercise 11.1. Show that for A, C X and B,D Y we have ⊂ ⊂  (A B) (C D) = (A C) (B D) and (A B)c = (Ac Y ) (A Bc) = (X Bc) (Ac B). × ∩ × ∩ × ∩ × × ] × × ] ×

Exercise 11.2. Show that the family of all finite disjoint unions of boxes (from (X, ), (Y, ) forms an algebra. F G Show that the σ-algebra generated by this algebra is . F ⊗ G

Exercise 11.3. Show that 1A B(x, y) = 1A(x)1B(y). ×  103

Proposition 11.2. Let (X, , µ), (Y, , ν) be measure spaces. • F G If A B = U (A B ) where A, (A ) are -measurable and B, (B ) are - × n n × n n n F n n G measurable, then X µ(A)ν(B) = µ(An)ν(Bn). n

Proof. This is actually just poor man’s Fubini. We have by the assumption A B = U (A B ) that × n n × n X 1A(x)1B(y) = 1An (x)1Bn (y). n Integrate this function over X for fixed y to get that for all y,

X µ(A)1B(y) = µ(An)1Bn (y), n where we have used poor man’s Fubini once. Integrate this again over Y to get by poor man’s Fubini again, X µ(A)ν(B) = µ(An)ν(Bn). n

ut

Exercise 11.4. Let (X, , µ), (Y, , ν) be measure spaces. F G Show that if ] ] (A B ) = (C D ) n × n n × n n n where A ,C ,B ,D , then n n ∈ F n n ∈ G X X µ(An)ν(Bn) = µ(Cn)ν(Dn). n n 104

Solution to ex:11.4. :( ♣ We have (A B ) (C D ) = (A C ) (B D ), j × j ∩ k × k j ∩ k × j × k so ] A B = (A C ) (B D ), j × j j ∩ n × j × n n which implies that X µ(A )ν(B ) = µ(A C )ν(B D ). j j j ∩ n j ∩ n n Similarly, X µ(C )ν(D ) = µ(A C )ν(B D ). k k n ∩ k n ∩ k n Thus, X X X µ(A )ν(B ) = µ(A C )ν(B D ) = µ(C )ν(D ). n n n ∩ k n ∩ k k k n n,k k :) X

Exercise 11.5. Let E = Un (A B ) . Define ρ(E) = Pn µ(A )ν(B ). j=1 j × j ∈ A j=1 j j Then ρ is well defined, and defines a pre-measure on . A

By Charath´eodory’s Extension Theorem we may extend ρ above to a measure on all of σ( ) = . Also note that if X and Y are σ-finite with respect to µ and ν, then A F ⊗ G X Y is σ-finite with respect to ρ. So in this case the extension, denoted µ ν, is the × ⊗ unique measure on (X, ) (Y, ) satisfying F ⊗ G µ ν(A B) = µ(A)ν(B) A ,B . ⊗ × ∀ ∈ F ∈ G

Exercise 11.6. Show that if X and Y are σ-finite with respect to µ and ν, then X Y is σ-finite with respect to ρ. × 105

Solution to ex:11.6. :( ♣ If X = U A ,Y = U B where µ(A ) < , ν(B ) < , then X Y = U (A B ) n n n n n ∞ n ∞ × n,k n × k and ρ(A B ) = µ(A )ν(B ) < .:) n × k n k ∞ X

Exercise 11.7. In this exercise the product measure is generalized to further prod- ucts. Let (X , , µ ) be σ-finite measure spaces, j = 1, 2,.... j Fj j Show that (µ µ ) µ = µ (µ µ ). 1 ⊗ 2 ⊗ 3 1 ⊗ 2 ⊗ 3 Show that for all n there exists a unique measure Nn µ on Nn (X , ) satisfying j=1 j j=1 j Fj that for all A , j = 1, 2, . . . , n, j ∈ Fj n n O Y µ (A A ) = µ (A ). j 1 × · · · × n j j j=1 j=1

11.1. Sections

Definition 11.3. Let E X Y . Define the sections of E for every x X, y Y • ⊂ × ∈ ∈ as

E = y Y :(x, y) E = π (E ( x Y )) x { ∈ ∈ } Y ∩ { } × Ey = x X :(x, y) E = π (E (X y )), { ∈ ∈ } X ∩ × { } where π : X Y X, π : X Y Y are the natural projections. X × → Y × → For a function f : X Y Z define the sections of f for every x X, y Y as the × → ∈ ∈ functions f : Y Z, f y : X Z by the formula f (y) = f y(x) = f(x, y). x → → x

y y Exercise 11.8. Show that (1E)x = 1Ex and (1E) = 1E .  1 1 y 1 1 y Show that (fx)− (A) = (f − (A))x and (f )− (A) = (f − (A)) . 106

Y

y

Ex E

Ey x X

y Figure 4. Depiction of Ex,E .

Solution to ex:11.8. :( ♣ We have for all x, y,

(1E)x(y) = 1 (x,y) E = 1 y Ex = 1Ex (y), { ∈ } { ∈ }

y and similarly for (1E) = 1Ey . Also,

1 1 1 y (f )− (A) f (y) A f(x, y) A (x, y) f − (A) y (f − (A)) , ∈ x ⇐⇒ x ∈ ⇐⇒ ∈ ⇐⇒ ∈ ⇐⇒ ∈ x y 1 1 y and similarly for (f )− (A) = (f − (A)) .:) X

S S S y S y Exercise 11.9. Show that ( n En)x = n(En)x and ( n En) = n(E) .  c c c y y c Show that (E )x = (Ex) and (E ) = (E ) .

Proposition 11.4. Let (X, ), (Y, ) be measurable spaces. If E then for • F G ∈ F ⊗ G all x X, y Y , Ey and E . ∈ ∈ ∈ F x ∈ G 107

Also if f is a function on X Y that is -measurable, then f y is -measurable × F ⊗ G F and f is -measurable. x G

Proof. Let

= E X Y : x X, y Y,Ey ,E . H { ⊂ × ∀ ∈ ∈ ∈ F x ∈ G} One may verify that is a σ-algebra (exercise!). H If A B is a box then (A B) = B for x A, and (A B)y = A for × ∈ F ⊗ G × x ∈ × y B, and (A B) = = (A B)y if x A or y B. So all boxes are in . Thus, ∈ × x ∅ × 6∈ 6∈ H , which proves the first assertion. F ⊗ G ⊂ H If f is -measurable then for any set A in the target σ-algebra, (f ) 1(A) = F ⊗ G x − (f 1(A)) which is in because f 1(A) . Similarly for (f y) 1(A) = (f 1(A))y − x G − ∈ F ⊗G − − ∈ . F ut

11.2. Product integrals

11.2.1. Monotone classes and algebras. We require some technical notions.

A monotone class on X is a family 2X that is closed under countable X C ⊂ increasing unions and countable decreasing intersections; i.e. for (E ) , if E n n ⊂ C n ⊂ E for all n then S , and if E E for all n then T . n+1 n ∈ C n ⊃ n+1 n ∈ C

Exercise 11.10. Show that an intersection of monotone classes is a monotone class. Show that for any collection of set 2X there is a monotone class ( ) that is the K ⊂ C K minimal monotone class containing ; i.e. ( ) and for any monotone class such K K ⊂ C K C that we have ( ) . K ⊂ C C K ⊂ C

Exercise 11.11. Show that any σ-algebra is a monotone class. Show that if is a monotone class and an algebra, then is a σ-algebra. C C 108

Solution to ex:11.11. :( ♣ A σ-algebra is closed under countable unions and intersections, so this is much more than is needed for it to be a monotone class. This proves the first assertion. Now suppose that is a monotone class and an algebra. Let (A ) be a sequence in C n n . We need to prove that S A . C n n ∈ C Indeed, set B = Sn A . Because is an algebra, (B ) is an increasing sequence n j=1 j C n n in . Because is a monotone class, S A = S B .:) C C n n n n ∈ C X

Exercise 11.12. Let be a monotone class. Fix A and define C ∈ C  = B : A B,B A, A B are all in . CA { ∈ C \ \ ∩ C}

Show that is a monotone class. CA ⊂ C

Solution to ex:11.12. :( ♣ If (B ) is an increasing sequence in then (B A) , (B A) are increasing sequences n n CA n\ n n∩ n in . So, C [ [ [ [ B A = (B A) and B A = (B A) . n \ n \ ∈ C n ∩ n ∩ ∈ C n n n n

Also, (A B ) is a decreasing sequence in so \ n n C [ \ \ A B = A Bc = (A B ) . \ n ∩ n \ n ∈ C n n n

Thus, S B as well. Similarly, if (B ) is a decreasing sequence in , then n n ∈ CA n n CA (B A) , (B A) are decreasing sequences in and (A B ) is an increasing n \ n n ∩ n C \ n n sequence in , which leads to C \ \ \ \ B A = (B A) and B A = (B A) , n \ n \ ∈ C n ∩ n ∩ ∈ C n n n n 109 and \ [ [ A B = A Bc = (A B ) . \ n ∩ n \ n ∈ C n n n So T B in this case.:) n n ∈ CA X

Lemma 11.5 (Monotone Class Lemma). For an algebra the monotone class ( ) = • A C A σ( ). A

Proof. Since σ( ) is a monotone class continuing , we have that ( ) σ( ). So it A A C A ⊂ A suffices to prove the other inclusion. Since ( ) it suffices to show that ( ) is a A ⊂ C A C A σ-algebra. For A ( ) define ∈ C A

= B ( ): A B,B A, A B are all in ( ) . CA { ∈ C A \ \ ∩ C A }

We have that is a monotone class. CA If A then since is an algebra we have that . By the minimality of ( ) ∈ A A A ⊂ CA C A we get that = ( ) for all A . CA C A ∈ A Note that by definition, B if and only if A . This implies that for all A ∈ CA ∈ CB ∈ A and any B ( ) = we have that A . Thus, for any B ( ). By ∈ C A CA ∈ CB A ⊂ CB ∈ C A minimality of ( ) again, = ( ) for any B ( ). C A CB C A ∈ C A We conclude that if A, B ( ) then A B,B A, A B are all in ( ). Since ∈ C A \ \ ∩ C A X = c ( ) we have that ( ) is closed under complements as well. So ( ) is ∅ ∈ A ⊂ C A C A C A an algebra and a monotone class. Thus ( ) is a σ-algebra, and we are done. C A ut

Exercise 11.13. Let (X, ) be a measurable space such that = σ( ) where F F A A  is an algebra. Let L∞ be the vector space of all bounded measurable functions from X to R. Let V be a subspace of L∞ such that: 1 V and 1 V for all A . • ∈ A ∈ ∈ A If (f ) is a monotone increasing sequence in V such that f f and such that • n n n % f is bounded, then f V as well. ∈ 110

Show that V = L∞.

Solution to ex:11.13. :( ♣ Define = A : 1 V . By the assumptions, . Also, if (A ) is an increasing G { A ∈ } A ⊂ G n n sequence in , then (1 ) is an increasing sequence in V such that 1 1 where G An n An % A A = S A . Since 1 is bounded, we have that 1 V , which implies that A . Note n n A A ∈ ∈ G that is closed under complements: If A then 1 V and so 1 c = 1 1 V G ∈ G A ∈ A − A ∈ (because V is a vector space). Thus, if (A ) is a decreasing sequence in , then (Ac ) n n G n n is an increasing sequence in , and T A = S Ac
c . G n n n n ∈ G We conclude that is a monotone class. Thus, by the Monotone Class Lemma, G = ( ) . F C A ⊂ G So 1 V for all A . Since V is a vector space, all simple functions are also in A ∈ ∈ F V . Now, if f 0 is a bounded measurable function, then there exists a monotone se- ≥ quence of simple functions ϕ f. Since f is bounded, the assumptions on V tell us n % that f V . So V contains all non-negative bounded measurable functions. ∈ For a general real-valued bounded measurable function, note that f +, f V so also − ∈ f = f + f .:) − − X

11.2.2. Fubini-Tonelli for indicators. The next theorem tells us how to compute the product measure of a set. First measure each section in one axis, and then integrate all measures of sections over the other axis.

Theorem 11.6. Let (X, , µ), (Y, , ν) be σ-finite measure spaces. For any E ••• F G ∈ we have that the functions F ⊗ G x ν(E ) and y µ(Ey) 7→ x 7→ are measurable. Moreover, Z Z µ ν(E) = ν(E )dµ = µ(Ey)dν. ⊗ x 111

Proof. Step I. First assume that µ, ν are finite measures. Define to be the set of all set E such that the theorem holds for E. We C ∈ F ⊗ G need to show that = . C F ⊗ G If E = A B for A ,B then E = B and Ey = A for (x, y) E and × ∈ F ∈ G x ∈ E = = Ey for (x, y) E. So x ν(E ) is the function ν(B)1 and y µ(Ey) is x ∅ 6∈ 7→ x A 7→ the function µ(A)1 . These are of course measurable. Also, µ ν(A B) = µ(A)ν(B) B ⊗ × by definition, which is

Z Z µ ν(A B) = ν(B)1 dµ = µ(A)1 dν. ⊗ × A B

Thus, all boxes A B are in . × C We have already seen that if E = Un (A B ) then j=1 j × j

n X µ ν(E) = µ(A )ν(B ). ⊗ j j j=1

Since in this case

n n ] ] E = (A B ) and Ey = (A B )y, x j × j x j × j j=1 j=1 the functions x ν(E ) and y µ(E ) are just Pn ν(B )1 and Pn µ(A )1 , 7→ x 7→ y j=1 j Aj j=1 j Bj which are measurable and satisfy the conclusion of the theorem by additivity of the integral. Thus all finite disjoint unions of boxes are in , which is to say that contains the C C algebra that generates . If was a monotone class then we would have by the F ⊗ G C Monotone Class Lemma that . F ⊗ G ⊂ C So we are left with showing that is a monotone class. C To this end, let (E ) be an increasing sequence in . Let E = S E . Define n n C n n f (x) = ν((E ) ). This is a measurable function satisfying R f dµ = µ ν(E ). Also, n n x n ⊗ n (f ) form an increasing sequence in L+(X, , µ), because ((E ) ) is an increasing n n F n x n sequence. Note that E = S (E ) which implies that f f for f(x) = ν(E ) by x n n x n % x continuity of measures. So f L+(X, , µ). By Monotone Convergence and continuity ∈ F 112 of the measure µ ν, ⊗ Z Z fdµ = lim fndµ = lim µ ν(En) = µ ν(E). n n →∞ →∞ ⊗ ⊗ Similarly, the functions g (y) = µ((E )y) converge monotonically g g for g(y) = n n n % µ(Ey), and using Monotone Convergence as before, Z gdν = µ ν(E). ⊗

Thus, E . ∈ C Now for the case of a decreasing sequence (E ) in . Let E = T E . Again we define n n C n n y fn(x) = ν((En)x) and gn(y) = µ((En) ). These are decreasing sequences that converge y to f(x) = ν(Ex) and g(y) = µ(E ) respectively. So f, g are measurable. Also, for all n, f ν(Y ) < and g µ(X) < . Since the measures are finite, constant | n| ≤ ∞ | n| ≤ ∞ functions are in L1 for both µ and ν. Thus, we may use Dominated Convergence of conclude that Z Z fdµ = lim fndµ = lim µ ν(En) = µ ν(E), n n →∞ →∞ ⊗ ⊗ and similarly, Z gdν = µ ν(E). ⊗ So E in the decreasing case as well. ∈ C This shows that is a monotone class, completing the proof in the case µ, ν are finite C measures. Step II. For general µ, ν which are σ-finite, let X Y = S (X Y ) where × n n × n (X ) , (Y ) are increasing sequences of measurable sets, and µ(X ) < , ν(Y ) < . n n n n n ∞ n ∞ Let E . For every n, consider E (X Y ). The functions µ (A) = ∈ F ⊗ G ∩ n × n n µ(A X ), ν (B) = ν(B Y ) are finite measures. For any box A B we have that ∩ n n ∩ n × µ ν (A B) = µ (A)ν (B) = µ ν(A B X Y ). Thus, ρ (C) := µ ν(C X Y ) n⊗ n × n n ⊗ × ∩ n× n n ⊗ ∩ n× n is the unique σ-finite product measure ρ = µ ν on . By the previous step, the n n ⊗ n F ⊗G functions x ν (E ) = ν(E Y ) and y µ (Ey) = µ(Ey X ) are measurable. Since 7→ n x x∩ n 7→ n ∩ n (X ) , (Y ) are increasing sequences, continuity of measures gives ν (E ) ν(E ) n n n n n x → x 113 and µ (Ey) µ(Ey), so that x ν(E ) and y µ(Ey) are measurable as limits of n → 7→ x 7→ measurable functions.

Exercise 11.14. Let (Z, , α) be a measure space. Let W and define H ∈ H β(A) = α(A W ) for all A . ∩ ∈ H Show that (Z, , β) is a measure space. H Show that for any f L+(Z, ) we have ∈ H Z Z fdβ = fdα. W

Solution to ex:11.14. :( ♣ β is a measure since β( ) = α( ) = 0 and ∅ ∅ ] ] X X β( A ) = α( (A W )) = α(A W ) = β(A ). n n ∩ n ∩ n n n n n

If f = 1A then Z Z fdβ = β(A) = α(A W ) = 1Adα. ∩ W R R If f is a simple function, then fdβ = W fdα by linearity. If f L+, then let ϕ f be a sequence of simple functions approximating f. Then, ∈ n % by monotone convergence, Z Z Z Z fdβ ϕndβ = ϕndα fdα. ← W → W

:) X

Using the above exercise, we conclude that Z Z µ ν(E X Y ) = µ ν (E) = ν(E Y )dµ = µ(Ey X )dν ⊗ ∩ n × n n ⊗ n x ∩ n n ∩ n n Z Z y = ν(Ex Yn)dµ = µ(E Xn)dν. Xn ∩ Yn ∩ 114

Since ν(E Y )1 (x) ν(E ) and µ(Ey Y )1 (y) µ(Ey), we use Monotone x ∩ n Xn % x ∩ n Yn % Convergence to get that Z Z y µ ν(E) lim µ ν(E Xn Yn) = ν(Ex)dµ = µ(E )dν. n ⊗ →∞ ⊗ ∩ ×

ut 11.3. The Fubini-Tonelli Theorem

??? THEOREM 11.7 (Fubini-Tonelli). Let (X, , µ), (Y, , ν) be σ-finite measure F G spaces.

Let f L+((X, , µ) (Y, , ν)). Then the functions x R f dν and y • ∈ F ⊗ G 7→ x 7→ R f ydµ are in L+. Let f L1((X, , µ) (Y, , ν)). Then the functions x R f dν and y • ∈ F ⊗ G 7→ x 7→ R f ydµ are defined a.e. and in L1. In both cases we have Z Z Z  Z Z  fd(µ ν) = f dν dµ(x) = f ydµ dν(y). ⊗ x

Proof. Start with f L+. If f = 1 this is just the previous theorem. Since (f + g) = ∈ E x f + g and (f + g)y = f y + gy, we have the theorem for all simple functions. If f L+ x x ∈ is general, let ϕ f be an increasing sequence of approximating simple functions. It n % may be easily verified that (ϕ ) f and (ϕ )y f y. So f , f y are indeed in L+. n x % x n % x y R R y Also, (ϕn)x, (ϕn) are simple functions. Note that (ϕn)xdν and (ϕn) dµ are also increasing sequences of functions in L+. So by Monotone Convergence, Z Z ZZ ZZ fd(µ ν) = lim ϕnd(µ ν) = lim (ϕn)xdνdµ = fxdνdµ. n n ⊗ →∞ ⊗ →∞ Similarly for Z ZZ fd(µ ν) = f ydµdν. ⊗ Now for f L1. Write f = f f + i(f f ) for f L+. Note that (f ) = (f ) ∈ 1 − 2 3 − 4 j ∈ x j j x y y and (f )j = (fj) . Since f is integrable, so are fj for all j. Thus, the first part of the theorem tells us that Z Z  Z Z  Z (11.1) (f ) dν dµ = (f )ydµ dν = f d(µ ν) < , j x j j ⊗ ∞ 115

R R y which gives that (fj)xdν, (fj) dµ are a.e. finite, and integrable as functions of x and y respectively. Since R f dν P4 R (f ) dν, we have that x R f dν is an L1 | x | ≤ j=1 j x 7→ x function. Similarly y R f ydµ is an L1 function. We also have by summing (11.1) that 7→ Z Z  Z Z  Z f dν dµ = f ydµ dν = fd(µ ν). x ⊗

ut

Exercise 11.15. [Folland, p.69, ex.51] Let (X, , µ), (Y, , ν) be measure spaces F G that are not necessarily σ-finite. Show that:

If f : X C is -measurable and g : Y C is -measurable then h(x, y) := • → F → G f(x)g(y) is -measurable. F ⊗ G If f L1(X, , µ) and g L1(Y, , ν) then h L1(X Y, , µ ν) and • ∈ F ∈ G ∈ × F ⊗ G ⊗ Z Z Z hd(µ ν) = fdµ gdν. ⊗ ·

Solution to ex:11.15. :( ♣ Note that he function φ : C2 C defined by φ(z, w) = zw is continuous and thus → measurable. Also, the function H(x, y) := (f(x), g(y)) from X Y to C2 is - × F ⊗ G 2 measurable since the Borel σ-algebra on C is generated by sets of the form B1 B2 × where B1,B2 are Borel in C, and for any such set B1 B2, ×

1 1 1 H− (B B ) = (x, y): f(x) B , g(y) B = f − (B ) g− (B ) . 1 × 2 { ∈ 1 ∈ 2} 1 × 2 ∈ F ×G ⊂ F ⊗G

Hence h(x, y) = φ(f(x), g(y)) = φ H(x, y) is -measurable as a composition of ◦ F ⊗ G measurable functions.

For the second assertion, suppose first that f = 1A, g = 1B. Then clearly h = 1A B × and Z Z Z hd(µ ν) = (µ ν)(A B) = µ(A)ν(B) = fdµ gdν. × × × · 116

Pn Pm If f = k=1 ak1Ak and g = j=1 bj1Bj are simple functions where (Ak)k and (Bj)j are both pairwise disjoint, then n m X X h = akbj1Ak Bj × k=1 j=1 is a simple function and Z Z Z hd(µ ν) = fdµ gdν × · follows by linearity. If f 0, g 0 then we may choose f f, g g where f , g are simple for all n. ≥ ≥ n % n % n n Since multiplication is continuous, f g h. Thus, by monotone convergence, n n % Z Z Z hd(µ ν) = fdµ gdν. × · If f, g are real-valued integrable functions, then

+ + h(x, y) = (f (x) f −(x)) (g (y) g−(y)) − · − + + + + = f (x)g (y) + f −(x)g−(y) f (x)g−(y) f −(x)g (y). − − This is a linear combination of 4 products of non-negative functions. Since Z Z Z f α(x)gβ(y)d(µ ν) = f αdµ gβdν × · for any choice of α, β +, , we have by linearity ∈ { −} Z X Z Z h d(µ ν) f αdµ gβdν < , | | × ≤ · ∞ α,β +, ∈{ −} so h is integrable. Also by linearity, and by the above for non-negative functions, Z X Z hd(µ ν) = αβf α(x)gβ(y)d(µ ν) × × α,β +, ∈{ −} X Z Z Z Z = αβ f αdµ gβdν = fdµ gdν. · · α,β +, ∈{ −} 1 Finally for complex valued functions in L write f = f1 + if2 and g = g1 + ig2, and note that if h(x, y) = f(x)g(y) then

h(x, y) = f (x)g (y) f (x)g (y) + i f (x)g (y) + f (x)g (y). 1 1 − 2 2 · 1 2 2 1 117

So the real part of h and imaginary part of h fall into the category of the previous 1 1 assertion, and so satisfy that they are in L (because f1, f2, g1, g2 are all in L ) and by linearity Z Z Z Z Z Z Z Z Z hd(µ ν) = f dµ g dν f dµ g dν + i f dµ g dν + i f dµ g dν × 1 · 1 − 2 · 2 · 1 · 2 · 2 · 1 Z Z = fdµ gdν. ·

:) X

Exercise 11.16. Let µ be the counting measure on ([0, 1], ([0, 1])) (for any Borel B set A, µ(A) = A ). Let λ be Lebesgue measure. Let D = (x, x) [0, 1]2 . Compute | | ∈ R R
R R
1D(x, y)dµ(x) dλ(y), 1D(x, y)dλ(y) dµ(x), and show that they are not equal. Use the fact that µ λ can be defined as an extension (although non-unique) of the ⊗ pre-measure on boxes to compute R 1 d(µ λ) and show this is also not equal to the D ⊗ other two integrals.

Solution to ex:11.16. :( ♣ For any y [0, 1], ∈ Z 1 (x, y)dλ(y) = λ( x ) = 0, D { } and Z 1 (x, y)dµ(x) = µ( y ) = 1. D { } Thus, Z Z  Z Z  1D(x, y)dµ(x) dλ(y) = 1 and 1D(x, y)dλ(y) dµ(x) = 0.

Also, R 1 d(µ λ) = (µ λ) (D), where (µ λ) is the outer measure induced by D ⊗ ⊗ ∗ ⊗ ∗ the pre-measure on the algebra of finite disjoint unions of boxes. This pre-measure is defined by (µ λ)(A B) = µ(A)λ(B). Thus, ⊗ × ( ) X [ (µ λ)∗(D) = inf µ(A )λ(B ): D A B . ⊗ n n ⊂ n × n n n 118

If D S A B then for any x [0, 1] we have that there exists some n for which ⊂ n n × n ∈ (x, x) A B , so x B A . Thus, [0, 1] S (B A ), so P λ(B A ) ∈ n × n ∈ n ∩ n ⊂ n n ∩ n n n ∩ n ≥ λ([0, 1]) = 1. So there must exist k such that B A has λ(B A ) > 0, and | k ∩ k| k ∩ k specifically B A is infinite. Now, for this k we have that µ(A ) µ(A B ) = . k ∩ k k ≥ k ∩ k ∞ Hence, for any D S A B there exists k such that ⊂ n n × n X µ(A )λ(B ) µ(A )λ(B ) = . n n ≥ k k ∞ n

Thus, (µ λ) (D) = .:) ⊗ ∗ ∞ X

Exercise 11.17. [Folland p.69] Let X be a linearly ordered set that is uncountable, but for any x X the set y X : y < x is countable. Let be the countable-co- ∈ { ∈ } F countable σ-algebra on X; i.e. A if A is countable or if Ac is countable. Define ∈ F E = (x, y) X X : x < y . { ∈ × } Show that E ,Ey are measurable for all x, y X. x ∈ For A define µ(A) = 0 if A is countable and µ(A) = 1 if Ac is countable. Show ∈ F that µ is a measure on (X, ). F R R
R R
Show that 1E(x, y)dµ(x) dµ(y), 1E(x, y)dµ(y) dµ(x) exist and are not equal.

Solution to ex:11.17. :( ♣ For any x X let C = y : y < x . By assumption C is countable for all x and ∈ x { } x c (Cx) is uncountable for all x. For any x, y X we have E = y : y > x = (C x )c, so (E )c is countable, ∈ x { } x ]{ } x and thus Ex is measurable and µ(Ex) = 1. Also, Ey = x : x < y = C which is countable, so Ey is measurable and µ(Ey) = 0. { } y µ is easily verified to be a measure. Finally, Z Z  Z y 1E(x, y)dµ(x) dµ(y) = µ(E )dµ(y) = 0, 119 Z Z  Z 1E(x, y)dµ(y) dµ(x) = µ(Ex)dµ(x) = µ(X) = 1.

:) X

Exercise 11.18. [Folland p.69] Consider (N, 2N, µ) where µ is the counting measure. Define  1 n = k   f(n, k) = 1 n = k + 1 −  0 otherwise.

Show that R f d(µ µ) = and that R R f(x, y)dµ(x)
dµ(y), R R f(x, y)dµ(y)
dµ(x) | | ⊗ ∞ exist but are not equal.

Solution to ex:11.18. :( ♣ f(n, k) = 1 n=k 1 n=k+1 . { } − { } f(n, k) = 1 n=k + 1 n=k+1 so | | { } { } Z X f d(µ µ) = (1 n=k + 1 n=k+1 ) = # (n, n), (n + 1, n): n N = . | | ⊗ { } { } { ∈ } ∞ (n,k) N2 ∈ For any k N, ∈ Z X f(n, k)dµ(n) = 1 n=k 1 n=k+1 = 0. { } − { } n

However, for any n N, ∈  Z X 0 n > 0 f(n, k)dµ(k) = 1 n=k 1 n=k+1 = { } − { } k 1 n = 0.

So ZZ ZZ Z f(n, k)dµ(n)dµ(k) = 0 and f(n, k)dµ(k)dµ(n) = 1 0 (n)dµ(n) = 1. { }

:) X 120

Exercise 11.19. [Folland p.69] Let (X, , µ) be a σ-finite measure space. Let F f L+(X, , µ). Set Γ := (x, y) X [0, ]: y f(x) . ∈ F f { ∈ × ∞ ≤ } Show that Γ ([0, ]). Show that (µ λ)(Γ ) = R fdµ. f ∈ F ⊗ B ∞ ⊗ f

Solution to ex:11.19. :( ♣ The functions ϕ(x, y) = f(x) y is ([0, ])-measurable, as a composition of the − F ⊗ B ∞ continuous function subtraction on the measurable function (x, y) (f(x), y). Thus, 7→ 1 Γ = (x, y): f(x) y 0 = ϕ− ([0, ]) ([0, ]). f { − ≥ } ∞ ∈ F ⊗ B ∞ Note that (Γ ) = y : y f(x) = 1 . f x { ≤ } [0,f(x)] Now using Fubini-Tonelli, ZZ Z Z (µ λ)(Γ ) = (Γ ) (y)dλ(y)dµ(x) = λ([0, f(x)])dµ(x) = f(x)dµ(x). ⊗ f f x :) X

Number of exercises in lecture: 19 Total number of exercises until here: 125 121

Measure Theory

Ariel Yadin

Lecture 12: Change of variables

12.1. Linear transformations of Lebesgue measure

Recall that we showed that if L is an invertible linear transformation on Rd, for any A Rd that is Lebesgue measurable, we have that L(A) is also Lebesgue measurable and ⊂ λ(L(A)) = det L λ(A). The proof of this fact was not so simple and kind of technical. | | We now give a more general statement, and also a shorter proof using the tools we have developed.

Theorem 12.1. Let L be an invertible linear transformation of Rd and let f : Rd ••• → C. If f is Lebesgue measurable then so is f L. Also, if f 0 or if f L1, then so is ◦ ≥ ∈ f L and ◦ Z Z fdλ = det L f Ldλ. | | ◦ Specifically, if A Rd is Lebesgue measurable then L(A) is Lebesgue measurable and ⊂ λ(L(A)) = det L λ(A). | | Proof. The main idea is as in the first proof we gave for Lebesgue measure of sets. First suppose that f is Borel. Then f L is also because L is continuous. ◦ If the theorem holds for linear maps L, M then it also holds for L M as: ◦ Z Z Z Z f = det L f L = det L det M (f L) M = det L M f L M. | | ◦ | | · | | ◦ ◦ | ◦ | ◦ ◦ Thus, we want to decompose linear transformations into elementary types that we know how to deal with. Every invertible linear transformation can be written as composition of the following three types: M which is multiplication of coordinate j by a non-zero scalar α, A which is adding coordinate k to coordinate j, and S which is swapping coordinates k and j. So we only need to prove the theorem for M, A, S. 122

Now det M = α, det A = 1, det S = 1. Thus, using the Fubini-Tonneli Theorem, we − can integrate the coordinates in any order, so Z Z ZZ d d d 1 f M(x , . . . , x )dλ = f(x , . . . , αx , . . . , x )dλ = g (αx )dλ(x )dλ − (x), ◦ 1 d 1 j d x j j where x = (x1, . . . , xj 1, xj+1, . . . , xd) and gx(xj) = f(x1, . . . , xd). The statement − Z Z g(x)dλ(x) = α g(αx)dλ(x), | | · is just the d = 1 case, which we will prove shortly. Given the d = 1 case we have Z Z Z d d 1 α f Mdλ = α g (αx )dλ(x )dλ − (x) | | ◦ | | x j j ZZ Z d 1 d = gx(xj)dλ(xj)dλ − (x) = fdλ .

Similarly for A and S: Given the d = 1 case we have Z ZZ d d 1 f Adλ = g (x + x )dλ(x )dλ − (x) ◦ x j k j ZZ Z d 1 d = gx(xj)dλ(xj)dλ − (x) = fdλ .

For S we get by changing the order of integration (Fubini-Tonelli), Z f S(x , . . . , x , . . . , x , . . . , x )dλd ◦ 1 j k d Z Z = f(x , . . . , x , . . . , x , . . . , x )dλ(x ) dλ(x ) dλ(x ) dλ(x ) = fdλd. 1 k j d 1 ··· j ··· k ··· d So we are left with verifying the d = 1 cases of M and A: Z Z Z Z gdλ = α g(αx)dλ(x) and gdλ = g(x + a)dλ(x). | | · If g = 1 then this is just α λ(α 1A) = λ(A) and λ(A + a) = λ(A). By additivity A | | − this extends to simple functions. Monotone Convergence gives this for non-negative functions. Decomposing g into real and imaginary parts and those into positive and negative parts gives the general case. This completes the theorem for Borel functions. If f is Lebesgue then there exists a Borel function g and a Borel set N such that

λ(N) = 0 and g1N c = f1N c . We have essentially already proved this in an exercise, but 123 the next exercise also reproves this. Since

1 1 (f L)(x) (1 c L− )(x) = f(L(x)) 1 c (L(x)) = g(L(x)) 1 c (L(x)) = (g L)(x) (1 c L− )(x), ◦ · N ◦ · N · N ◦ · N ◦

1 we have that (f L) (1 c L ) is Borel. So f L is Lebesgue. Moreover, because ◦ · N ◦ − ◦ 1 λ(N) = λ(L− (N)) = 0, Z Z Z Z fdλ = gdλ = det L (g L)dλ = det L (f L)dλ. | | ◦ | | ◦

ut

Exercise 12.1. Show that a function f is Lebesgue if and only if there exists a  Borel function g and a Borel set N such that λ(N) = 0 and g1N c = f1N c .

Solution to ex:12.1. :( ♣ If there exists a Borel function g and a Borel set N such that λ(N) = 0 and g1N c = f1N c , then for any Borel set B in the image, f(x) B if and only if g(x) B, x N or ∈ ∈ 6∈ f(x) B, x N. Thus f 1(B) = g 1(B) N c f 1(B) N. Since f 1(B) N N ∈ ∈ − − ∩ ] − ∩ − ∩ ⊂ which has measure 0, we get that f 1(B) N is a Lebesgue null set, and thus Lebesgue − ∩ measurable. Since g 1(B) N c is Borel, we have that f 1(B) is Lebesgue. So in this − ∩ − case f is a Lebesgue function.

For the other direction, if f is Lebesgue, consider the case f = 1A for a Lebesgue set A. Since A = B F for a Borel set B and a null set F (not necessarily Borel). Since ∪ λ (F ) = 0, there exists a Borel set N F such that λ(N) = 0. Thus, 1 1 c = 1 1 c , ∗ ⊃ A N B N and we are done taking g = 1B. Pn By additivity this extends to simple functions: If f = j=1 aj1Aj for Lebesgue

(Aj)j, then for every j there exist a Borel set Bj and a Borel null set Nj such that n n c c S P 1Aj 1(Nj ) = 1Bj 1(Nj ) . Taking N = j=1 Nj and g = j=1 aj1Bj , we have that g is

Borel and f1N c = g1N c . Now if f 0, let ϕ f be a monotone sequence of Lebesgue simple functions ≥ k % approximating f. For every k there exists a Borel null set Nk and a Borel (simple) 124

c c S function gk such that ϕk1(Nk) = gk1(Nk) . Taking N = k Nk which is a Borel null set, we have that (gk1N c = ϕk1N c )k form an increasing sequence of Borel functions such that g 1 c f1 c . So g = f1 c is a Borel function. k N % N N Finally, if f is a general Lebesgue function, write f = f f + i(f f ) where f 1 − 2 3 − 4 j are all non-negative Lebesgue functions. Then, there exist Borel functions gj and Borel null sets N , j = 1, 2, 3, 4, such that g 1 c = f 1 c . Taking N = N N N N j j (Nj ) j (Nj ) 1 ∪ 2 ∪ 3 ∪ 4 which is a Borel null set, and g = g g + i(g g ) we have that g1 c = f1 c .:) 1 − 2 3 − 4 N N X

Exercise 12.2. Complete the details of the previous theorem; that is, show that  Z Z Z Z gdλ = α g(αx)dλ(x) and gdλ = g(x + a)dλ(x). | | ·

Exercise 12.3. Show that if λ is Lebesgue measure on Rd and U is a unitary  d operator on R then λ(U(A)) = λ(A). Specifically, Lebesgue measure is invariant under rotations.

12.2. Change of variables formula

Let U Rd be an open set. Suppose that Ψ : Rd Rd is a map such that for ⊂ → Ψ = (ψ , . . . , ψ ) all the maps ψ have continuous partial derivatives ∂ψi on U, in all 1 d i ∂xj 1 d coordinates xj.(i.e. ψi are all C in U). For any x R define a matrix DΨ(x) by ∈ (DΨ(x)) = ∂ψi (x). i,j ∂xj 125

Ψ is called a C1 diffeomorphism (on U) if Ψ is 1-1 and DΨ(x) is invertible for all x U. In this case by the inverse function theorem, Ψ 1 : Ψ(U) U is also a C1 ∈ − → 1 1 1 diffeomorphism, and DΨ− (y) = (DΨ(Ψ− (y)))− .

Exercise 12.4. Show that if L is an invertible linear map then it is a diffeomor-  d phism with DL(x) = L(x) for all x R . ∈ Also, show that for any diffeomorphism Ψ, D(L Ψ)(x) = L(DΨ(x)). ◦

Theorem 12.2. Let Ψ: U Rd be a C1 diffeomorphism for an open set U Rd. ••• → ⊂ If f : Ψ(U) C is Lebesgue, then f Ψ is Lebesgue on U. If in addition f 0 or f is → ◦ ≥ integrable, then Z Z fdλ = (f Ψ)(x) det DΨ(x) dλ(x). Ψ(U) U ◦ · | |

Proof. Start with Borel f. Because Ψ is continuous, f Ψ is Borel. ◦ We consider the sup-norm x = maxj xj and for a matrix M GLd(R), the || || | | ∈ operator norm M = max P M . So Mx M x . || || i j | i,j| || || ≤ || |||| || For a Rd and ε > 0 write Q(a, ε) = x : x a ε . Note that Q(a, ε) is a box. ∈ { || − || ≤ } Step I. We show that for Q = Q(a, ε), Z λ(Ψ(Q)) det DΨ(x) dλ(x). ≤ Q | |

1 Since ψj are all C functions the mean value theorem tells us that

d X ∂ψi ψi(x) ψi(a) = (xj aj) (y), − − ∂xj j=1 for some y on the line segment between x and a. Thus, for any x Q = Q(a, ε), ∈ X Ψ(x) Ψ(a) ε sup max (DΨ(y))i,j = ε sup DΨ(y) . || − || ≤ · y Q i · y Q || || ∈ j ∈ Thus, Ψ(Q) Q(a, ε sup DΨ(y) ). ⊂ · y Q || || ∈ 126

By the scaling of Lebesgue measure

λ(Ψ(Q)) λ(Q(a, ε sup DΨ(y) )) = λ(sup DΨ(y) Q) = (sup DΨ(y) )d λ(Q). ≤ · y Q || || y Q || || · y Q || || · ∈ ∈ ∈ For any invertible linear map L GLd(R), ∈ 1 1 d λ(Ψ(Q)) = det L λ(L− Ψ(Q)) det L (sup L− DΨ(y) ) λ(Q). | | ◦ ≤ | | · y Q || || · ∈ Now, y DΨ(y) is a continuous function, and so uniformly continuous on the com- 7→ pact set Q. Thus, for any η > 0 there exists k > 0 such that for all z y 1 , z, y Q || − || ≤ k ∈ we have (DΨ(z)) 1DΨ(y) d 1 + η. Write Q = Snk Q where Q = Q(x , 1 ) || − || ≤ j=1 k,j k,j k,j k and all have disjoint interiors. Then, n Xk λ(Ψ(Q)) = λ(Ψ(Qk,j)) j=1

nk X 1 d det DΨ(xk,j) ( sup (DΨ(xk,j))− DΨ(y) ) λ(Qk,j) ≤ | | · y Qj || || · j=1 ∈ n Z Xk (1 + η) det DΨ(x ) 1 (x)dλ(x). ≤ | k,j | Qk,j j=1

Snk 1 This holds for any Q = j=1 Qk,j where Qk,jQ(xk,j, k ), and all have disjoint interiors. The integrand Pnk det DΨ(x ) 1 (x) is continuous, and on Q tends uniformly j=1 | k,j | Qk,j to det DΨ(x) 1 (x) as k . By Dominated Convergence that | | Q → ∞ n Xk Z Z λ(Ψ(Q)) (1 + η) lim det DΨ(xk,j) dλ(x) = (1 + η) det DΨ(x) dλ(x). ≤ k | | | | →∞ j=1 Qk,j Q Thus, sending η 0, → Z λ(Ψ(Q)) det DΨ(x) dλ(x). ≤ Q | | Step II. We show that for any open set U, Z λ(Ψ(U)) det DΨ(x) dλ(x). ≤ U | | S Indeed, if U is an open set, we may write U = n Qn where Qn are almost disjoint boxes (i.e. have disjoint interiors). Thus, X X Z Z λ(Ψ(U)) λ(Ψ(Q )) det DΨ(x) dλ(x) = det DΨ(x) dλ(x). ≤ n ≤ | | | | n n Qn U 127

Step III. We show that for any Borel set A, Z λ(Ψ(A)) det DΨ(x) dλ(x). ≤ A | | Indeed, for a Borel set A with λ(A) < , there exist open sets (U ) such that A U ∞ n n ⊂ n for all n and λ(U A) 0. By replacing U with Tn U we may assume that (U ) n \ → n j=1 j n n T is a decreasing sequence. Thus, continuity of measure gives that for U = n Un we have A U and λ(U A) = 0. So 1 1 a.e., and the Dominated Convergence Theorem ⊂ \ Un → A now gives that

λ(Ψ(A)) λ(Ψ(U)) = lim λ(Ψ(Un)) ≤ n Z →∞ Z lim det DΨ(x) dλ(x) = det DΨ(x) dλ(x). n ≤ →∞ Un | | A | | U Now if A is a general Borel set then A = n An where (An)n all have finite measure (this is just σ-finiteness). So X X Z Z λ(Ψ(A)) λ(Ψ(A )) det DΨ(x) dλ(x) = det DΨ(x) dλ(x). ≤ n ≤ | | | | n n An A Pn Step IV. If f = j=1 aj1Aj is a Borel simple function defined on Ψ(U), n n Z X X Z fdλ = a λ(A ) a 1 −1 (x) det DΨ(x) dλ(x) j j ≤ j Ψ (Aj ) · | | Ψ(U) j=1 j=1 n Z X Z = det DΨ(x) a 1 (Ψ(x))dλ(x) = (f Ψ)(x) det DΨ(x) dλ(x). | | · j Aj ◦ · | | U j=1 U Taking limits with the monotone convergence theorem gives that for any Borel f 0 ≥ defined on Ψ(U), Z Z fdλ (f Ψ)(x) det DΨ(x) dλ(x). Ψ(U) ≤ U ◦ · | | Replacing f with the function (f Ψ)(x) det DΨ(x) which is defined on U we have ◦ · | | Z Z 1 1 1 (f Ψ)(x) det DΨ(x) dλ(x) (f Ψ Ψ− )(x) det DΨ(Ψ− (x)) det DΨ− (x) dλ(x). U ◦ · | | ≤ Ψ(U) ◦ ◦ · | | · | | Since DΨ(Ψ 1(x)) = (DΨ 1(x)) 1, we have that det DΨ(Ψ 1(x)) det DΨ 1(x) = 1, − − − | − |·| − | giving Z Z Z (f Ψ)(x) det DΨ(x) dλ(x) fdλ (f Ψ)(x) det DΨ(x) dλ(x). U ◦ · | | ≤ Ψ(U) ≤ U ◦ · | | 128

This establishes the theorem for all non-negative Borel f. To get general integrable Borel f just decompose into real, imaginary, positive and negative parts. To get Lebesgue f find a Borel function that is a.e. identical to f. We omit the details. ut

R x2 Example 12.3. Let us compute I := ∞ e− dx. Consider −∞ ZZ Z 2 x2 y2 2 I = e− e− dxdy = f(x, y)dλ (x, y), R2 (x2+y2) where f(x, y) = e− , by Fubini. Let U = (0, ) ( π, π). Let Ψ : U R2 be Ψ(r, θ) = (r cos θ, r sin θ). So Ψ(U) = ∞ × − → R R2 (( , 0] 0 ). Since λ2(( , 0] 0 ) = 0 we have that I2 = fdλ2. \ −∞ × { } −∞ × { } Ψ(U) Note that   cos θ r sin θ DΨ(r, θ) =  −  . sin θ r cos θ So det DΨ(r, θ) = r. | | r2 We get using Fubini again, since f(Ψ(r, θ)) = e− , Z Z I2 = fdλ2 = f(Ψ(r, θ)) rdλ(r, θ) Ψ(U) U · Z π Z Z ∞ r2 ∞ 1 d r2 = re− drdθ = 2π ( 2 ) dr e− dr = π. π 0 · 0 − − So I = √π. 4 5 4 Number of exercises in lecture: 4 Total number of exercises until here: 129 129

Measure Theory

Ariel Yadin

Lecture 13: Lebesgue-Radon-Nykodim

13.1. Signed measures

Definition 13.1. Let (X, ) be a measurable space. A signed measure on (X, ) • F F is a function ν : [ , ] such that F → −∞ ∞ ν( ) = 0; • ∅ ν( ) ( , ] or ν( ) [ , ); that is, ν takes on at most one of the • F ⊂ −∞ ∞ F ⊂ −∞ ∞ values , (but not both); −∞ ∞ If (A ) is a sequence of pairwise disjoint sets in then ν(U A ) = P ν(A ), • n n F n n n n and if ν(U A ) < then P ν(A ) < . | n n | ∞ n | n | ∞

When we say positive measure we stress that the measure in question is a measure that is not signed; i.e. in the usual sense. We really only require signed measures in order to prove differentiation theorems.

Exercise 13.1. Show that if µ , µ are measures on (X, ) and at least one of 1 2 F them is a finite measure, then µ µ is a signed measure. 1 − 2

Exercise 13.2. Let f be a measurable function f :(X, ) [ , ]. Assume F → −∞ ∞  R + R R that one of the integrals f dµ, f −dµ is finite, in which case the integral fdµ exists. R Show that ν(A) = A fdµ is a signed measure. 130

Exercise 13.3. Let ν be a signed measure on (X, ). Let (A ) be a sequence in F n n . F Show that if (An)n is increasing, then [ ν( An) = lim ν(An). n n →∞ Show that if (A ) is decreasing and ν(A ) < then n n | 1 | ∞ \ ν( An) = lim ν(An). n n →∞

Definition 13.2. Let ν be a signed measure on (X, ). We say that A is • F ∈ F positive (respectively negative) if for every B A such that B we have ν(B) 0 ⊂ ∈ F ≥ (respectively ν(B) 0). ≤ If A is both positive and negative we say A is null.

Exercise 13.4. Let µ be a positive measure on (X, ). Let f be a measurable F function f :(X, ) [ , ] such that the integral R fdµ exists. Let ν(A) = R fdµ, F → −∞ ∞ A which we know is a signed measure. Show that A is ν-positive (respectively negative, null) if and only if f1 0 a.e. ∈ F A ≥ (respectively f1 0, f1 = 0 a.e.). A ≤ A

Solution to ex:13.4. :( ♣ Let A . ∈ F If A is positive then fix ε > 0 and let B = f1 ε which is measurable because { A ≤ − } f1 is measurable. Note that B = A f ε A, so ν(B) 0. Also, f1 A ∩ { ≤ − } ⊂ ≥ B ≤ 131

( ε)1B, and − Z 0 ν(B) = f1 dµ ε µ(B). ≤ B ≤ − · So µ(B) = 0 (which also implies that ν(B) = 0.) On the other hand, if f1 0 a.e. then for any B A we have that f1 = A ≥ F 3 ⊂ B f1A1B 0 a.e., so ≥ Z ν(B) = f1 dµ 0. B ≥ For the negative case, note that A is ν-negative if and only if A is ( ν)-positive, ∈ F − where ν is the signed measure defined by ν(B) = R fdµ = R ( f)dµ, we have − − − B B − that A is negative if and only if ( f)1 0 a.e., which is if and only if f1 0 a.e. − A ≥ A ≤ For the null case, A is null if and only if it is both positive and negative, which ∈ F is if and only if 0 f1 0 a.e.:) ≤ A ≤ X

Exercise 13.5. Show that if A is positive then any B such that B A is ∈ F ⊂ also positive. Show that a countable union of positive sets is positive.

Solution to ex:13.5. :( ♣ The first assertion is easy. S For the second, let (An)n be a sequence of positive sets and let A = n An. Then, Sn 1 U B = A − A A is also positive. Since A = B , for any measurable C A n n \ j=1 j ⊂ n n n ⊂ we have that X ν(C) = ν(C B ) 0. ∩ n ≥ n :) X

Theorem 13.3 (Hahn Decomposition Theorem). If ν is a signed measure on (X, ) ••• F then there exist disjoint measurable sets P N = such that X = P N and P is positive ∩ ∅ ] and N is negative. This is called a Hahn decomposition of X. 132

If P ,N are another Hahn decomposition of X we have that P P = N N is a 0 0 4 0 4 0 null set.

Proof. By possible considering ν instead of ν, we may assume that ν(A) < for all − ∞ A . ∈ F Let m = sup ν(P ): P is positive . Let (P ) be a sequence of positive sets such { } n n that ν(P ) m. Let P = S P . P is positive and n → n n n [ m ν(P ) = lim ν( Pj) lim ν(Pn) = m. ≥ n ≥ n →∞ j=1 →∞

So m = ν(P ) < . ∞ Let N = X P . \ For any A N, if A is positive then for any B A, also B P is positive, F 3 ⊂ F 3 ⊂ ] so ν(P ) = m ν(B P ) = ν(B) + ν(P ), and so ν(B) = 0, which implies that A is null. ≥ ] That is, any positive subset of N is null. Now let A N. If ν(A) > 0 then A is not null, and since A is not positive F 3 ⊂ there exists B A such that ν(B) < 0. So for C = A B A we have that F 3 ⊂ \ ⊂ ν(C) = ν(A) ν(B) > ν(A). So for any A N such that ν(A) > 0 define − ⊂

 1 n := inf n 1 : B A , ν(B) > ν(A) + n− , A ≥ ∃ ⊂

1 and let BA A be a set such that ν(BA) > ν(A) + . ⊂ nA So assume for a contradiction that N is not negative. So there exists A N F 3 0 ⊂ such that ν(A0) > 0. Given Ak define inductively nk+1 = nAk and Ak+1 = BAk . So for 1 all k 1 we have that ν(Ak) > ν(Ak 1) + and Ak Ak 1. Note that ≥ − nk ⊂ −

k 1 X 1 ν(Ak) > ν(Ak 1) + > > . − nk ··· nj j=1

Since ν(A ) < , for A = T A , 0 ∞ k k

X 1 > ν(A) = lim ν(Ak) > 0. ∞ k ≥ nk →∞ k 133

P 1 Thus, < so nk . Also, ν(A) > 0. Take k large enough so that nk > 2nA. k nk ∞ → ∞

Then, B A Ak 1. By the definition of nk = nAk−1 we have that ⊂ ⊂ −

ν(A) + 1 < ν(B) ν(A ) + 1 ν(A ) + 1 . nA k 1 nk 1 k 1 2nA ≤ − − ≤ − 1 So ν(Ak 1) > ν(A) + for all k such that nk > 2nA. Thus, − 2nA

1 ν(A) = lim ν(Ak) ν(A) + , k ≥ 2nA →∞ a contradiction! Thus, N is negative. Finally if P ,N is a Hahn decomposition, then P P N P , so must be both 0 0 \ 0 ⊂ 0 ∩ positive and negative. Similarly, P P N P . 0 \ ⊂ ∩ 0 ut

Theorem 13.4 (Jordan Decomposition Theorem). If ν is a signed measure on ••• (X, ) then there exist unique positive measures ν+, ν such that ν = ν+ ν and such F − − − that X = P N where ν+(N) = ν (P ) = 0. ] −

Proof. Let X = P N be a Hahn decomposition. Define ]

+ ν (A) = ν(A P ) and ν−(A) = ν(A N). ∩ − ∩

These are positive measures because P is positive and N is negative. It is immediate that ν = ν+ ν and that ν+(N) = ν (P ) = 0. − − − Now for uniqueness: Suppose that ν = µ+ µ for positive measure µ+, µ and X = − − − P N where µ+(N ) = µ (P ) = 0. For any A P we have that ν(A) = µ+(A) 0 0 ] 0 0 − 0 ⊂ 0 ≥ and for any B N we have ν(B) = µ (B) 0. So P ,N for a Hahn decomposition, ⊂ 0 − − ≤ 0 0 and so P P is ν-null. For any A , since (A P ) (A P ) P P , 4 0 ∈ F ∩ 4 ∩ 0 ⊂ 4 0

+ + + + µ (A) = µ (A P 0) = ν(A P 0) = ν(A P ) = ν (A P ) = ν (A), ∩ ∩ ∩ ∩ and similarly,

µ−(A) = µ−(A N 0) = ν(A N 0) = ν(A N) = ν−(A N) = ν−(A). ∩ − ∩ − ∩ ∩

ut 134

Definition 13.5. ν = ν+ ν is called the Jordan decomposition of ν. ν+ is the • − − positive part and ν− is the negative part. We also define the total variation or absolute value of ν as ν := ν+ + ν . | | − ν is called finite (respectively, σ-finite) if ν is a finite (respectively, σ-finite) measure. |

+ Exercise 13.6. Show that A is ν-null if and only if ν (A) = ν−(A) = 0, which is if and only if ν (A) = 0. | |

+ Exercise 13.7. Show that if P,N is a Hahn decomposition then ν (N) = ν−(P ) = 0.

Exercise 13.8. Show that for all A , ∈ F  Z ν(A) = (1P 1N )d ν , A − | | where P,N are any Hahn decomposition.

Solution to ex:13.8. :( ♣ + If P,N are a Hahn decomposition, then ν (N) = ν−(P ) = 0. So Z Z + ν(A) = ν (A P ) ν−(A N) = ν (A P ) ν (A N) = 1P d ν 1N d ν . ∩ − ∩ | | ∩ − | | ∩ A | | − A | | We may combine the integrals into one since at least one of them is finite, because either + ν or ν− is a finite measure.:) X

Exercise 13.9. Show that L1(ν+) L1(ν ) = L1( ν ). ∩ − | |  135

Definition 13.6. For a signed measure and f L1( ν ) we define • ∈ | | Z Z Z + fdν = fdν fdν−. −

Exercise 13.10. Show that for f L 1( ν ), ∈ | |  Z Z

fdν f d ν . ≤ | | | |

Exercise 13.11. Show that for measurable A,   Z 

ν (A) = sup fdν : f 1 . | | A | | ≤

Exercise 13.12. Show that if ν = µ ρ where µ, ρ are positive measures (of which − one is finite), then µ ν+, ρ ν . ≥ ≥ −

Exercise 13.13. Let ν , ν be signed measures such that ν < , ν < . 1 2 1 ∞ 2 ∞ Show that ν + ν ν + ν . | 1 2| ≤ | 1| | 2| Show that ν = ν . | − 1| | 1| 136

13.2. The Lebesgue-Radon-Nikodym Theorem

Definition 13.7. Let ν, µ be signed measures on (X, ). We say that ν, µ are mu- • F tually singular (or just singular), denoted ν µ, if X = A B where A is ν-null and ⊥ ] B is µ-null.

Exercise 13.14. Show that ν+ ν . ⊥ − Show that ν µ if and only if ν µ if and only if ν+ µ, ν µ. ⊥ | | ⊥ ⊥ − ⊥

Definition 13.8. Let ν, µ be signed measures on (X, ). We say that ν is absolutely • F continuous with respect to µ, denoted ν µ, if every µ-null set is also a ν-null set. 

Exercise 13.15. Show that ν µ if and only if ν µ if and only if  | |  | | ν+ µ , ν µ .  | | −  | |

Exercise 13.16. Show that if ν µ and ν µ then ν 0.  ⊥ ≡ 

Exercise 13.17. Let ν(A) = R fdµ for some measurable f such that f < (but A ∞ perhaps f can take the value ). Show that ν µ. Show that ν is finite if and only −∞  if f is integrable.

Solution to ex:13.17. :( ♣ R If µ(A) = 0 then f1A = 0 µ-a.e. So ν(A) = f1Adµ = 0.:) X 137

Proposition 13.9. Let ν be a finite signed measure and µ a positive measure on • (X, ). ν µ if and only if for every ε > 0 there exists δ > 0 such that ν(A) < ε for F  | | all A such that µ(A) < δ. ∈ F

Proof. The ε, δ condition implies that for all A such that µ(A) = 0, we have that ∈ F for any ε > 0, ν(A) < ε. Thus, µ(A) = 0 implies ν(A) = 0. So if A is a µ-null set, then | | for any B A we have µ(B) = 0 so also ν(B) = 0. This implies that A is a ν-null set. ⊂ For the other direction, assume that ν is a positive measure. Suppose the ε, δ condition fails. So there exists ε > 0 such that for any n there exists a set A with µ(A ) < n ∈ F n 2 n but ν(A ) ε. Define − n ≥ \ [ F = lim sup An = Ak. n k n ≥ S Since k n Ak is a decreasing sequence, and since ν is a finite measure, ≥ [ ν(F ) = lim ν( Ak) lim ν(An) ε. n ≥ n ≥ →∞ k n →∞ ≥ Also, for any n,

[ X X k n+1 µ( A ) µ(A ) 2− = 2− . k ≤ k ≤ k n k n k n ≥ ≥ ≥ Thus, µ(F ) inf 2 n+1 0. So F is a µ-null set and ν(F ) > 0, which implies that it ≤ n − ≤ is not the case that ν µ.  Now, if ν is a signed measure then ν µ if and only if ν+, ν µ. So for any  −  + ε > 0 there exists δ > 0 such that if µ(A) < δ then ν (A), ν−(A) < ε. This implies that ν(A) = ν+(A) + ν (A) < ε. | | − ut

Exercise 13.18. Show that for any integrable f, for every ε > 0 there exists δ > 0 R such that if µ(A) < δ then fdµ < ε. A |

We are almost ready to prove the basic differentiation theorem. A technical lemma first: 138

Lemma 13.10. Let ν, µ be finite positive measures on (X, ). Either ν µ or there • F ⊥ exist some ε > 0 and A such that µ(A) > 0 and A is positive for the signed measure ∈ F ν εµ. − Proof. Let X = P N be a Hahn decomposition for the signed measure ν 1 µ. Let n ] n − n P = S P and N = T N = P c. For any n we have that N N is negative for n n n n ⊂ n ν 1 µ. That is, 0 ν(N) 1 µ(N) for all n, and so ν(N) = 0. − n ≤ ≤ n If µ(P ) = 0 then ν µ. ⊥ Otherwise, there exists some n for which µ(P ) > 0. Since P is positive for ν 1 µ, n n − n we can take ε = 1 and A = P . n n ut We have seen that measures of the form A R fdµ are absolutely continuous with 7→ A respect to µ. The next theorem tells us that these are the only examples.

??? THEOREM 13.11 (Lebesgue-Radon-Nikodym Theorem). Let ν be a σ-finite signed measure and let µ be a σ-finite positive measure on (X, ). There exist unique F signed measures σ, ρ such that σ µ, ρ µ, and ν = σ + ρ. Moreover, there exists ⊥  a measurable function f : X [ , ] such that for all A , ρ(A) = R fdµ → −∞ ∞ ∈ F A (specifically the integral exists). Moreover, ρ is positive if and only if f is positive. ρ is finite if and only if f is integrable. σ, ρ are positive if and only if ν is positive. σ, ρ are finite if and only if ν is finite.

X The decomposition ν = σ + ρ is called the Lebesgue decomposition. We use

dρ the notation dµ to denote the function f given by the theorem. This function is µ-a.e. unique, and called the Radon-Nikodym derivative of ρ with respect to µ.

Proof. Case I. ν, µ are finite and positive. Define  Z  M = f : X [0, ]: fdµ ν(A) A . → ∞ A ≤ ∀ ∈ F 0 M so M = . If f, g M then for B = f > g and for any A , ∈ 6 ∅ ∈ { } ∈ F Z Z Z (f g)dµ = fdµ + gdµ ν(A B) + ν(A Bc) = ν(A). A ∨ A B A Bc ≤ ∩ ∩ ∩ ∩ 139

So f g M as well. ∨ ∈ R R Set m = supf M fdµ ν(X) < . Let (fn)n M such that fndµ m. Let ∈ ≤ ∞ ⊂ → g = max f , . . . , f . So (g ) is an increasing sequence such that g f := sup f . n { 1 n} n n n % n n Note that g M for all n. So n ∈ Z Z m = lim fndµ lim gndµ m. n n →∞ ≤ →∞ ≤ R R By Monotone Convergence, m = limn gndµ = fdµ. So by perhaps modifying f →∞ on a set of measure 0, we may assume that f < . Also, for any A , by Monotone ∞ ∈ F Convegence, Z Z fdµ = lim gndµ ν(A), n A →∞ A ≤ so f M. ∈ Consider the signed measure σ(A) := ν(A) R fdµ. Note that σ is positive because − A f M. Since σ, µ are finite positive measures, if σ is not singular with respect to µ then ∈ there exist ε > 0 and A such that µ(A) > 0 and σ(B) εµ(B) for all measurable ∈ F ≥ B A. This implies that for any C , ⊂ ∈ F Z Z Z Z (f + ε1A)dµ = fdµ + εµ(A C) fdµ + ν(A C) fdµ C C ∩ ≤ C ∩ − A C Z ∩ = fdµ + ν(C A) ν(C Ac) + ν(C A) = ν(C). C Ac ∩ ≤ ∩ ∩ ∩ So f + ε1 M which implies that A ∈ Z Z m (f + ε1 )dµ = fdµ + εµ(A) > m, ≥ A a contradiction. So σ µ. Setting ρ(A) = R fdµ, since ρ µ, we have the decompo- ⊥ A  sition for positive finite ν, µ. Uniqueness of the decomposition is proved in an exercise following. U Case II. ν, µ are σ-finite positive measures. In this case we can write X = n Xn where ν(X ) < , µ(X ) < . For every n set ν (A) = ν(A X ), µ (A) = µ(A X ). n ∞ n ∞ n ∩ n n ∩ n These are finite measures, so we have th? Lebesgue decomposition νn = σn + ρn where σ µ and ρ (A) = R f dµ for some integrable f . We also know that µ (Xc ) = n ⊥ n n A n n n n n c νn(Xn) = 0, so Z Z

fndµn = fn1Xn dµn, A A 140

and by replacing fn with fn1Xn we may assume that fn = 0 off Xn. It is an exercise to R R c c R show that fndµ = fndµn. Also, σn(Xn) = νn(Xn) c fndµn = 0. A A − Xn Set σ = P σ It is another exercise to show that σ µ. Set f = P f and n n ⊥ n n R P R P ρ(A) = A fdµ = n A fndµn = n ρn(A). So

X X ν(A) = νn(A) = (σn(A) + ρn(A)) = σ(A) + ρ(A). n n

This completes the proof for σ-finite positive measures. Case III. For σ-finite signed measure ν, write the Jordan decomposition ν = ν+ ν . − − We have the Lebesgue decomposition ν = σ +ρ and then ν = (σ+ σ )+(ρ+ ρ ) ± ± ± − − − − which satisfy the conditions of the theorem. ut

Exercise 13.19. Show that the Lebesgue decomposition is unique. 

Solution to ex:13.19. :( ♣ Suppose that ν = σ + ρ = σ + ρ where σ, σ µ and ρ, ρ µ. If ν is a finite measure, 0 0 0 ⊥ 0  then σ σ = ρ ρ are well defined signed measures. Since ρ ρ µ and σ σ µ − 0 0 − 0 −  − 0 ⊥ we have that σ σ = ρ ρ 0. − 0 0 − ≡ Now, if ν is σ-finite, write X = U X where ν(X ) < . Then, define ν (A) = n n ∞ n ν(A X ). In this case, if ν = σ + ρ is a Lebesgue decomposition of ν with respect to ∩ n µ, then for any measurable A,

ν (A) = ν(A X ) = σ(A X ) + ρ(A X ). n ∩ n ∩ n ∩ n

Since the signed measure ρ (A) := ρ(A X ) is absolutely continuous with respect to n ∩ n µ, and since the signed measure σ (A) := σ(A X ) is singular with respect to µ, this n ∩ n forms a Lebesgue decomposition of νn with respect to µ, which is unique because νn is a finite measure. 141

Thus, if ν = σ+ρ = σ +ρ where σ, σ µ and ρ, ρ µ, then σ(A X ) = σ (A X ) 0 0 0 ⊥ 0  ∩ n 0 ∩ n and ρ(A X ) = ρ(A X ), which implies that ∩ n ∩ n X X σ(A) = σ(A X ) = σ0(A X ) = σ0(A), ∩ n ∩ n n n X X ρ(A) = ρ(A X ) = ρ0(A X ) = ρ0(A). ∩ n ∩ n n n :) X

Exercise 13.20. Let µ be a positive measure and let Y be a measurable set of finite measure. Set µ (A) = µ(A Y ) for all measurable A. Let f be a measurable Y ∩ function such that f = 0 off Y . R R Show that A fdµ = A fdµY .

Exercise 13.21. Let (νn)n be a sequence of positive measures. Let µ be a positive measure. Show that if ν µ for all n then P ν µ. n ⊥ n n ⊥ Show that if ν µ for all n then P ν µ. n  n n 

Proposition 13.12 (Chain rule). Suppose that ν µ ρ for σ-finite positive •   measures ν, µ, ρ. If f L1(ν) then f dν L1(µ) and ∈ dµ ∈ Z Z dν fdν = f dµ. dµ Also, ν ρ and dν = dν dµ , ρ-a.e.  dρ dµ · dρ R R dν Proof. If f = 1A then fdν = f dµ dµ holds by definition of the Radon-Nikodym derivative. This extends by additivity to simple functions and by Monotone convergence to non-negative functions. Taking real, imaginary, positive and negative parts proves this for all f L1(ν). ∈ 142

For the second assertion note that for any measurable A,

Z dν Z dν Z dν dµ dρ = ν(A) = dµ = dρ. A dρ A dµ A dµ · dρ

So the functions inside the integrals on both sides must be equal ρ-a.e. ut

Exercise 13.22. Show that if ν µ and µ ν then dν dµ = 1 for ν-a.e. and   dµ · dν µ-a.e.

Exercise 13.23. Let ν µ for a positive σ-finite µ and signed σ-finite ν , j = 1, 2. j  j  d(ν1+ν2) dν1 dν2 Show that dµ = dµ + dµ .

Exercise 13.24. Let ν µ be σ-finite measures on (X , ) for j = 1, 2. Then j  j j Fj ν ν µ µ and for a.e. (x, y) X X , 1 ⊗ 2  1 ⊗ 2 ∈ 1 × 2 d(ν ν ) dν dν 1 ⊗ 2 (x, y) = 1 (x) 2 (y). d(µ µ ) dµ · dµ 1 ⊗ 2 1 2

Exercise 13.25. Let µ be the counting measure on ([0, 1], ([0, 1])) and λ Lebesgue B measure. Show that λ µ but dλ does not exist. (What fails in the Radon-Nikodym Theorem?)  dµ Show µ has no Lebesgue decomposition with respect to λ. 143

Solution to ex:13.25. :( ♣ Note that µ(A) = 0 if and only if A = . So µ(A) = 0 implies A = which implies that ∅ ∅ λ(A) = 0. That is, λ µ.  Assume for a contradiction dλ exists. Then, for any x [0, 1], since x is Borel, dµ ∈ { } Z dλ dλ dλ 0 = λ( x ) = dµ 1 x dµ = dµ (x)µ( x ) = dµ (x). { } { } { }

So dλ 0 which is impossible. dµ ≡ The Radon-Nikodym theorem requires σ-finiteness, and µ is not σ-finite. Now, assume that µ = σ + ρ where σ λ and ρ λ. Note that for any Borel set ⊥  A, if A = and x A then σ(A) σ( x ) = µ( x ) ρ( x ) = 1 because λ( x ) = 0. 6 ∅ ∈ ≥ { } { } − { } { } However this implies that σ(A) > 0 for any non-empty Borel set A. Since σ λ there ⊥ exist Borel sets [0, 1] = A B such that σ(A) = 0 and λ(B) = 0. But then A = and ] ∅ B = [0, 1] which contradicts λ(B) = 0.:) X

Exercise 13.26. Let µ, ν be σ-finite measures on (X, ) such that ν µ. Let F  λ= ν + µ.

(a) Show that ν λ.  (b) Show that λ µ.  (c) Show that 0 dν (x) < 1 for µ-a.e. x X. ≤ dλ ∈ (d) Show that µ-a.e. dν dν = dλ . dµ 1 dν − dλ

Solution to ex:13.26. :( ♣

(a) If A is such that ν(A) > 0 then µ(A) > 0, because ν µ. So λ(A) = ∈ F  µ(A) + ν(A) > 0. Thus, if λ(A) = 0 then ν(A) = 0. Since this holds for all A we get that ν λ. ∈ F  144

(b) If A is such that µ(A) = 0 then ν(A) = 0 because ν µ. Thus, λ(A) = ∈ F  µ(A) + ν(A) = 0. Since this holds for all A we have that λ µ. ∈ F  (c) Because λ, ν are positive measure we have that dν 0. Let A =  dν 1 . dλ ≥ dλ ≥ Then

Z dν ν(A) = dλ λ(A) = ν(A) + µ(A). A dλ ≥ So µ(A) = 0. dλ dν dµ dν (d) Since dµ = dµ + dµ = dµ + 1, we get that dν dν dλ dν dν dν = = + . dµ dλ · dµ dλ · dµ dλ So, dν  dν  dν 1 = . dµ · − dλ dλ By (c) we get that µ-a.e. 0 < 1 dν 1, so we can divide by this function to − dλ ≤ complete the proof.

:) X

Exercise 13.27. Let (X, , µ) be a σ-finite measure space. Let be a sub- F G ⊂ F

σ-algebra of . Let ν = µ . F G (a) Suppose that f L1(X, , µ). Show that there exists g L1(X, , ν) such that ∈ F ∈ G for every A , ∈ G Z Z fdµ = gdν. A A (b) Suppose that for f L1(X, , µ) there are two such functions g, g L1(X, , ν) ∈ F 0 ∈ G such that for all A , ∈ G Z Z Z fdµ = gdν = g0dν. A A A

Show that g = g0 ν-a.e. 145

Solution to ex:13.27. :( ♣

(a) Because f L1, we know that f < µ-a.e., so we may assume that f < . ∈ | | ∞ | | ∞ First assume that f is positive and µ is finite. In this case, consider the function Z ρ(A) := fdµ A defined for all A . First of all, we showed in class that this defines a finite ∈ G positive measure on (X, ). Moreover, if ν(A) = 0 for some A , since G ∈ G R ν = µ we have that µ(A) = 0, and so ρ(A) = A fdµ = 0. Since this holds G for all A , the signed measure ρ is absolutely continuous with respect to the ∈ G measure ν. Since µ is finite, so is ν. Thus, by the Radon-Nykodim Theorem there exists a positive integrable g = dρ L1(X, , ν) such that dρ = gdν; that dν ∈ G is, for all A , ∈ G Z Z Z fdµ = ρ(A) = dρ = gdν. A A A Now, if µ is only σ-finite, then write X = U X with µ(X ) < . Consider n n n ∞ ν (A) := µ(A X ) for all A . So ν = P ν . Define ρ (A) := R fdµ. n n n n n A Xn ∩ ∈ G ∩ Since n X f1A Xj f1A, ∩ % j=1 by monotone convergence we get that Z X Z X ρ(A) := fdµ = fdµ = ρn(A). A A Xn n ∩ n Also, as above, if ν (A) = 0 then µ(A X ) = 0 and so ρ (A) = 0. So n ∩ n n dρn 1 ρn νn. Since νn is finite, gn := exists and is in L (X, , νn). Specifically,  νn G g is -measurable. Also, since ρ (A) = 0 for A X = , we have that g can n G n ∩ n ∅ n P be chosen such that it is supported on Xn. Define g = n gn. Since (Xn)n are

disjoint and so gn have disjoint support, we get that g is always finite and well

defined. Also, since gn = gn1Xn , by monotone convergence again Z Z X X Z X gdν = gn1A Xn dν = gndνn = ρn(A) = ρ(A). A ∩ A Xn n n ∩ n 146

Specifically, Z Z gdν = ρ(X) = fdµ < , X X ∞ so g L1(X, , ν). ∈ G Now, for the case that f is a general (not necessarily positive) function in L1. Write f = (f f ) + i(f f ) for f L1 positive. By the previous case, 1 − 2 3 − 4 j ∈ there exist real-valued functions g L1(X, , ν) such that for any A and j ∈ G ∈ G j = 1, 2, 3, 4 we have Z Z fjdµ = gjdν. A A By linearity of the integral we get that for all A , ∈ G Z Z Z Z Z Z fdµ = f1dµ f2dµ + i f3dµ i f4dµ = (g1 g2) + i(g3 g4)dν. A A − A · A − · A A − − So we may choose g = g g + i(g g ) which is a function in L1(X, , ν). 1 − 2 3 − 4 G (b) Suppose that g, g are as in the question. Then for all A , 0 ∈ G Z Z gdν = g0dν. A A

We have shown in class that this implies that g = g0 ν-a.e.

:) X

Number of exercises in lecture: 27 Total number of exercises until here: 156 147

Measure Theory

Ariel Yadin

Lecture 14: Convergence

Until now we have only considered the convergence of a sequence of functions (fn)n to a limit f is a pointwise sense: f f means that fn(x) f for all x; we also extended → → this a bit to include a.e. convergence. We had a glimpse at some points of uniform convergence. The introduction of measure gives us many other topologies to consider.

14.1. Modes of convergence

Definition 14.1. Let (f ) , f be a sequence of measurable functions on a measure • n n space (X, , µ). F a.e. We say that (fn)n converges to f a.e., denoted fn f, if µ fn f = 0. • −→ 1 { 6→ } We say that (f ) converges to f in L1, denoted f L f, if R f f dµ 0 • n n n −→ | n − | → as n . → ∞ We say that (f ) is Cauchy in measure if for every ε > 0 µ f f > ε • n n {| n − m| } → 0 as m, n . → ∞ µ We say that (f ) converges to f in measure, denoted f f, if for every • n n n −→ ε > 0 µ f f > ε 0 as n . {| n − | } → → ∞

1 Example 14.2. Let us consider a few examples on (R, , λ). fn = 1[0,n], gn = 1[n,n+1], B n `n = n1[0,1/n] and

k k h = 1 −k −k for n = 2 + j , 0 j < 2 . n [j2 ,(j+1)2 ] ≤ (That is, k = log n and j = n 2k.) So that b 2 c −

h1 = 1[0,1] h2 = 1 1 h3 = 1 1 etc. [0, 2 ] [ 2 ,1] We have

f a.e. 0 g a.e. 0 ` a.e. 0. n −→ n −→ n −→ 148

However Z Z Z f dλ = g dλ = ` dλ = 1. | n| | n| | n| 1 1 1 So f L 0, g L 0, ` L 0. n −→6 n −→6 n −→6 R L1 As for h , note that h dλ = 2 log2 n . So h 0. However, (h ) does not n | n| −b c n −→ n n converge pointwise, because for any x (0, 1) there are infinitely many n such that ∈ hn(x) = 0 and infinitely many n for which hn(x) = 1. Note that (g ) is not Cauchy in measure. Indeed, g (x) g (x) > ε if and only if n n | n − m | x [n, n + 1] [m, m + 1]. So ∈ 4 λ g g > ε = λ([n, n + 1] [m, m + 1]) = 2. {| n − m| } 4 It is also a sequence that does not converge in measure, as we will see later. On the other hand, f λ 0, ` λ 0 and h λ 0: For all n > 1 we have that n −→ n −→ n −→ ε f 1 < ε, so λ f > ε = λ( ) = 0. ` (x) > ε if and only if x [0, 1/n] and n > ε. | n| ≤ n {| n| } ∅ n ∈ So

lim λ `n > ε = lim λ([0, 1/n]) = 0. n n →∞ {| | } →∞ h (x) > ε if and only if x [j2 k, (j + 1)2 k] for k = log n and j = n 2k. So | n | ∈ − − b 2 c − k k log n λ h > ε = λ([j2− , (j + 1)2− ]) = 2−b 2 c 0. {| n| } → To sum up: a.e. L1 measure

fn X X X

gn X X X 4 5 4 `n X X X

hn X X X

By these examples, it is not always true that a.e. convergence implies L1 convergence, or vice-versa. Recall however the Dominated Convergence Theorem which actually relates a.e. convergence to L1 convergence.

a.e. 1 1 Exercise 14.1. Show that if fn f and fn g L for all n then f L and 1 −→ | | ≤ ∈ ∈ f L f. n −→ 149

a.e. Show that if fn f are bounded functions and the measure space is finite, then 1 −→ f L f. n −→

1 Proposition 14.3 (L1 implies measure). If f L f then f µµf. • n −→ n → Proof. Set A = f f > ε . Then, n,ε {| n − | } Z Z ε µ(An,ε) fn f dµ fn f dµ 0. · ≤ An,ε | − | ≤ | − | →

ut

The converse is false as is seen by fn, `n above.

a.e. Proposition 14.4 (a.e. implies measure). If f f then f µµf. • n −→ n → Proof. For any ε > 0 let A = f f > ε . Note that if x A then f (x) f(x). n,ε {| n − | } ∈ n,ε n 6→ So A N where N = x : f (x) f(x) . Thus, µ(A ) µ(N) = 0. n,ε ⊂ { n 6→ } n,ε ≤ ut

The converse is false as shown by hn above. However, we can still relate convergence in measure to a.e. convergence of a subse- quence.

Proposition 14.5. (f ) is Cauchy in measure if and only if there is a measurable • n n µ function f such that f f. n −→ µ a.e. Moreover, if f f then there exists a subsequence (f ) such that f f a.e. n −→ nk k nk −→ as k . → ∞

Proof. Suppose that (fn)n is Cauchy in measure. For any k there exists nk such that for all n, m n , µ  f f > 2 k < 2 k. Let g = f . Set ≥ k | n − m| − − k nk n ko \ [ n ko F = lim sup f f > 2− = f f > 2− . | nk − nk+1| | nk − nk+1| k n k n ≥ Since for any n,

[ n ko
X k n+1 µ(F ) µ f f > 2− 2− = 2− , ≤ | nk − nk+1| ≤ k n k n ≥ ≥ 150 we have that µ(F ) = 0. Note that for any x F we have that (g (x)) is a Cauchy 6∈ k k sequence in C. So we may define f(x) = limk gk(x) for x F and f(x) = 0 for →∞ 6∈ x F . We had an exercise proving that such a function is measurable. Since µ(F ) = 0 ∈ µ we have that f = g a.e. f, and so g f. But then, nk k −→ k −→ [ µ f f > ε µ f g > ε/2 g f > ε/2
{| n − | } ≤ {| n − k| } {| k − | } µ f g > ε/2 + µ g f > ε/2 0 as n, k . ≤ {| n − k| } {| k − | } → → ∞ µ So f f. n −→ µ For the other direction, if f f then for any ε > 0, n −→ µ f f > ε µ f f > ε/2 + µ f f > ε/2 0 as m, n . {| n − m| } ≤ {| n − | } {| m − | } → → ∞ So (f ) is Cauchy in measure. n n ut

µ µ Exercise 14.2. Show that if f f and f g then f = g a.e. n −→ n −→  Conclude that if (fn)n converges to f and to g in any one of the three modes (a.e., L1, measure), possibly different modes for f and for g, then f = g a.e.

Solution to ex:14.2. :( ♣ µ µ If (f ) converges to f and g then f f and f g. So it suffices to work with n n n −→ n −→ this assumption. For any m > 0,

µ  f g > 1 µ  f f > 1 + µ  g f > 1 0 as n . | − | m ≤ | − n| m | − n| m → → ∞ So X µ f = g µ  f g > 1 = 0. { 6 } ≤ | − | m m :) X

Exercise 14.3. Let (X, , µ) be a measure space. Let (f ) be a sequence of F n n  151 non-negative measurable functions, and let f be a measurable function such that (fn)n converges to f in measure. Show that Z Z fdµ lim inf fndµ. n ≤ →∞

Solution to ex:14.3. :( ♣

Let (fnk )k be a subsequence such that Z Z

lim fnk dµ = lim inf fndµ. k n →∞ R R So we want to show that fdµ limk fnk dµ. Since (fnk )k is a subsequence, we ≤ →∞ have that for all ε > 0,

lim µ fnk f > ε lim sup µ fn f > ε = 0. k {| − | } ≤ n {| − | } →∞

So (fnk )k converges in measure to f.

Let gk := fnk for all k, which convergen in measure to f. By a theorem in class we now have that there is a further subsequence (gkj )j such that limj gkj = f a.e. Since →∞ these are all non-negative functions, Fatou’s Lemma tells us that Z Z fdµ lim inf gk dµ. ≤ j j R R However, the sequence ( gkj dµ)j is a subsequence of the converging sequence ( fnk dµ)k R which converges to lim infn fndµ. So the limit is Z Z Z Z Z

fdµ lim inf gkj dµ lim gkj dµ = lim fnk dµ = lim inf fndµ. ≤ j ≤ j k n →∞ →∞ :) X

Exercise 14.4. Let (X, , µ) be a measure space. Let (f ) be a sequence of F n n  measurable functions, and let f be a measurable function such that (fn)n converges to f in measure. Suppose that g L1(X, , µ) such that for all n, f g. ∈ F | n| ≤ 152

1 Show that (fn)n converges to f in L .

Solution to ex:14.4. :( ♣

Since (fn)n converges to f in measure, there is a subsequence (nk)k such that (fnk )k converges to f µ-a.e. Since f g for all n, we get that f f 0 µ-a.e. as k . | n| ≤ | nk − | → → ∞ Also f = lim f g. Thus, f f 2g for all n. | | k | nk | ≤ | n − | ≤ R Consider the sequence In := fn f dµ. We want to show that lim supn In = 0. | − | →∞ Let (f ) be a subsequence such that I lim sup I as k . Set h := f f . nk k nk → n n → ∞ k | nk − | Since this is a subsequence, we have that for all ε > 0,

lim µ hk > ε lim sup µ fn f > ε = 0. k { } ≤ n {| − | } →∞ →∞ So (hk)k converges to 0 in measure.

We saw in class that there exists a subsequence (hkj )j such that limj hkj = 0 µ-a.e. →∞ Because (h ) is a subsequence of (h ) = ( f f ) which is a subsequence of kj j k k | nk − | k ( f f ) , we have that h 2g L1 for all j. Since (h ) converges to 0 a.e. and is | n − | n kj ≤ ∈ kj j a dominated sequence, we have by the dominated convergence theorem that Z lim hk dµ = 0. j j →∞ But by definition, (h ) is a subsequence of ( f f ) ; so the sequence (R h dµ) is kj j | nk − | k kj j a subsequence of the converging sequence (Ink )k. Thus these sequences have the same limit which is Z

lim sup In = lim Ink = lim hkj dµ = 0. n k j →∞ →∞ :) X

Exercise 14.5. [Folland p.63] Let (X, , µ) be a measure space. Let (An)n be a F 1 sequence of measurable sets with finite measure. Suppose that 1 L f. An −→ Show that f = 1 a.e. for some A . A ∈ F 153

14.2. Uniform and almost uniform convergence

Definition 14.6. Let (f ) , f be a sequence of measurable functions on a measure • n n space (X, , µ). F We say that (f ) converges uniformly to f if sup f (x) f(x) 0 as n ; that n n x | n − | → → ∞ is, for every ε > 0 there exists n such that for all n > n and any x, f (x) f(x) < ε. 0 0 | n − | We say that (fn)n converges uniformly to f on A if (fn1A)n converges uniformly to f1A; that is supx A fn(x) f(x) 0 as n . ∈ | − | → → ∞ We say that (fn)n converges almost uniformly to f if for every ε > 0 there exists c a measurable A such that µ(A ) < ε and (fn)n converges uniformly to f on A.

Theorem 14.7 (Egoroff’s Theorem). Let (X, , µ) be a finite measure space. Sup- ••• F a.e. pose that f f. Then (f ) converges almost uniformly to f. n −→ n n

Proof. By augmenting fn, f on a set of measure 0 we may assume without loss of gen- erality that f (x) f(x) for every x. n → Let [ B =  f f > 1 . n,k | m − | k m n ≥ For any x B := T B we have that for all n there exists m n such that ∈ k n n,k ≥ 1 1 fm(x) f(x) > k . That is, lim supn fn(x) f(x) k > 0. So fn(x) f(x). | − | →∞ | − | ≥ 6→ That is, B f f . k ⊂ { n 6→ } For fixed k, the sequence (Bn,k)n is decreasing. Since µ is a finite measure,

lim µ(Bn,k) = µ(Bk) µ fn f = 0. n →∞ ≤ { 6→ } For any ε > 0 and any k 1 let n be large enough so that µ(B ) < ε2 k. Let ≥ k,ε nk,ε,k − B = S B . Then µ(B) P µ(B ) ε. For A = Bc we have that for any k nk,ε,k ≤ k nk,ε,k ≤ η > 0 taking k = η 1 , there exists n = n such that for all n n we have that for d − e 0 k,ε ≥ 0 all x A, f (x) f(x) 1 η. Thus, (f ) converges uniformly to f on A. ∈ | n − | ≤ k ≤ n n To conclude: for any ε > 0 we can find A such that (fn)n converges uniformly to f on A and µ(Ac) < ε. This is almost uniform convergence. ut

Exercise 14.6. Show that if (f ) converges almost uniformly to f then f a.e. f. n n n −→  154

Exercise 14.7. A version of Egoroff’s theorem in non-finite settings: Let (X, , µ) be a measure space. Suppose that f a.e. f. Suppose that f g L1. F n −→ | n| ≤ ∈ Show that (fn)n converges almost uniformly to f.

Exercise 14.8. Suppose that (X, , µ) is a σ-finite measure space. Suppose that F f a.e. f. n −→ T c Show that there exists a sequence of measurable sets (Ak)k, such that µ( k Ak) = 0 and such that for any k,(fn)n converges uniformly to f on Ak.

Exercise 14.9. Show that if (fn)n are continuous functions into C on some Borel  measure space, and if (fn)n converges to f uniformly, then f is continuous as well.

Solution to ex:14.9. :( ♣ Let ε > 0. Take n = n (ε) so that for all n > n we have sup f (x) f(x) < ε/2. 0 0 0 x | n − | Now, if (x ) is any sequence x x, then f (x ) f (x) for all n. For all n > n (ε), k k k → n k → n 0

f(x ) f(x) f(x ) f (x ) + f(x) f (x) + f (x ) f (x) | k − | ≤ | k − n k | | − n | | n k − n |

2 sup f(x) fn(x) + fn(xk) fn(x) < ε + fn(xk) fn(x) . ≤ x | − | | − | | − |

Taking k we have that for any ε > 0 lim sup f(x ) f(x) ε. Thus, f is → ∞ k | k − | ≤ continuous at x, for all x.:) X 155

Lemma 14.8. Let A R be a Lebesgue set of finite Lebesgue measure. For any ε > 0 • ⊂ there exists a continuous function ψ such that 0 ψ 1, ψ vanishes outside an interval ≤ ≤ and Z ψ 1 dλ < ε. | − A|

Proof. For any interval I = [a, b], let

  0 x (a, b),  6∈  1 x [a + ε, b ε], ψε,I (x) = ∈ − x a  − x [a, a + ε],  ε ∈   b x  − x [b ε, b]. ε ∈ −

So ψ is continuous and vanishes outside I, and is 1 in [a + ε, b ε] and 0 ψ(x) 1 ε,I − ≤ ≤ for x [a, b] [a + ε, b ε]. Thus, R 1 ψ dλ 2ε. ∈ \ − | I − ε,I | ≤ If A [a, b] then λ(A) = supA K compact λ(K). Fix ε > 0. Choose a compact K A ⊂ ⊃ ⊂ so that λ(K) λ(A) ε. Now we may find a sequence of almost disjoint closed intervals ≥ − (i.e. with disjoint interiors) (I ) such that λ(I ) λ(K) + ε and K S I . We may n n n ≤ ⊂ n n assume without loss of generality that I K = for all n. n ∩ 6 ∅ For every n let I be the open ε 2 n enlargement of I ; that is, if I = [x , y ] then n0 · − n n n n I = (x ε 2 n, y + ε 2 n). So K S I is an open cover. So there exists n such n0 n − · − n · − ⊂ n n0 that K Sn I . Note that ⊂ j=1 j0

n n [ X X∞ λ( I0 ) λ(I0 ) ε + λ(I ) λ(K) + 2ε. j ≤ j ≤ n ≤ j=1 j=1 j=1

R j −j For every j = 1, . . . , n let ψj = ψε 2 ,Ij . So ψj 1Ij dλ 2 2− . Thus for · | − | ≤ · Pn Pn ϕ = j=1 1Ij and ψ = j=1 ψj we have that

Z n Z n X X j ψ ϕ dλ ψ 1 dλ ε2− ε. | − | ≤ | j − Ij | ≤ ≤ j=1 j=1 156

Also,

n Z X Z Z 1A ϕ dλ 1K In 1In dλ + 1A 1K dλ | − | ≤ | ∩ − | | − | j=1 n n X [ λ(I K) + λ(A K) λ( I K) + λ(A K) ≤ n \ \ ≤ j \ \ j=1 j=1 n [ λ( I0 ) λ(K) + λ(A) λ(K) 3ε. ≤ j − − ≤ j=1

Sn Note that ψ is a function that vanishes outside j=1 Ij, which is contained in some interval I = [a, b]. ut

Exercise 14.10. Show that if f : R C is Lebesgue and integrable then for → every ε > 0 there exists a continuous function ψ such that ψ vanishes outside a bounded interval and Z f ψ dλ < ε. | − |

Solution to ex:14.10. :( ♣ Pn n If f = j=1 aj1Aj is a simple function, where (Aj)j=1 are pairwise disjoint, the for each j find a continuous function ψj vanishing outside some bounded interval such that R ε Pn ψj 1A dλ < . Then, for ψ = ajψj we have that | − j | naj j=1 n Z X Z f ψ dλ a 1 ψ dλ < ε. | − | ≤ j | Aj − j| j=1

Note that ψ vanishes outside some interval (which just is the smallest interval containing the union of the bounded intervals supporting the ψj’s). Now if f is any non-negative measurable function, let ϕ f be an approximating n % sequence of simple functions. By Dominated Convergence, R ϕ f dλ 0. So there | n − | → exist n large enough so that R ϕ f dλ < ε . for this n let ψ be a continuous function | n − | 2 n 157 supported in a bounded interval such that R ϕ ψ dλ < ε . Note that | n − n| 2 Z Z Z ψ f dλ ϕ f dλ + ψ ϕ dλ < ε. | n − | ≤ | n − | | n − n|

Now if f is a general Lebesgue measurable function (into C), write f = f1 f2 + − i(f f ), for non-negative Lebesgue f . Then choose continuous functions ψ that 3 − 4 j j are supported in a bounded interval and R f ψ dλ < ε/4. Summing we have that | j − j| ψ = ψ ψ + i(ψ ψ ) is supported in a bounded interval and continuous, and 1 − 2 3 − 4 Z Z p f ψ dλ f f (ψ ψ ) 2 + f f (ψ ψ ) 2dλ | − | ≤ | 1 − 2 − 1 − 2 | | 3 − 4 − 3 − 4 | Z Z f f (ψ ψ ) dλ + f f (ψ ψ ) dλ ≤ | 1 − 2 − 1 − 2 | | 3 − 4 − 3 − 4 | 4 X Z f ψ dλ < ε. ≤ | j − j| j=1 :) X

Theorem 14.9 (Lusin’s Theorem). Let f :[a, b] C be Lebesgue. Let ε > 0. Then ••• → there exists a compact set K [a, b] such that λ([a, b] K) < ε and f is continuous. ⊂ \ K

Proof. Let (ψn)n be a sequence of continuous functions each supported in a bounded 1 closed interval I [a, b] such that R f ψ dλ < 1 . Thus, ψ L f, and so ψ λ f. n ⊂ | − n| n n −→ n −→ Let (g = ψ ) be a subsequence such that g a.e. f. Since λ([a, b]) < , k nk k k −→ ∞ Egoroff’s Theorem tells us that (gk)k converges almost uniformly to f. That is, for any ε > 0 there exists a Lebesgue set A such that λ([a, b] A ) < ε/2 and (g ) converges ε \ ε k k uniformly to f on A . Let K A be a compact subset such that λ(A K ) < ε/2. ε ε ⊂ ε ε \ ε So λ([a, b] K ) λ([a, b] A ) + λ(A K ) < ε. Also, (g ) converges uniformly to f \ ε ≤ \ ε ε \ ε k k on K , so f is continuous on K . ε ε ut Lusin’s Theorem is informally the statement that Lebesgue functions are almost con- tinuous. Number of exercises in lecture: 10 Total number of exercises until here: 166 158

Measure Theory

Ariel Yadin

Lecture 15: Differentiation

We now turn to understanding differentiation in Rd. We will work throughout in the d d measure space (R , d, λ = λ ). B(x, r) denotes the open ball of radius r around x (in B L2-norm).

15.1. Hardy-Littlewood Maximal Theorem

A technical lemma:

Lemma 15.1. Let U = S B be a union of a family of open balls B = B(x , r ) • α α α α α for every α. Then, for any c < λ(U) there exist finitely many pairwise disjoint balls Pn d Bj = Bαj , j = 1, . . . , n, such that j=1 λ(Bj) > 3− c.

Proof. There exists a compact K U such that λ(K) > c, by inner regularity. Since ⊂ K S B is an open cover, there exists finitely many A = B , j = 1, . . . , m, that ⊂ α α j αj cover the compact K, K Sm A . ⊂ j=1 j Reorder A ,...,A so that the radii are decreasing. Set B = A and for all j 2 1 m 1 1 ≥ let Bj be Ak for the smallest k such that Ak is disjoint from B1 Bj 1. ∪ · ∪ − Suppose that Bj = B(xj, rj) for all j. These are disjoint by definition. Now, if x A , the for some j we have that A B = . If j is the smallest such ∈ k k ∩ j 6 ∅ possible index, then Ak is disjoint from B1 Bj 1, so the radius of Ak is at most ∪ · ∪ − that of Bj; otherwise we would have wanted to choose Ak instead of Bj. Thus, since there is some x A B , we have that all elements in A are at distance at most ∈ k ∩ j j 2rad(A ) 2r from x, and so A B(x, 2r ) B(x , 3r ). k ≤ j j ⊂ j ⊂ j j Since K Sm A Sn B(x , 3r ) we have that ⊂ j=1 j ⊂ j=1 j j n n X X c < λ(K) λ(B(x , 3r )) = 3d λ(B ). ≤ j j j j=1 j=1

ut 159

Definition 15.2. A measurable function f : Rd C is called locally integrable if • → R for every bounded measurable set B Rd, f dλ < . ⊂ B | | ∞ 1 1 d L = L (R , d, λ) denotes the space of locally integrable functions. loc loc B For any x Rd and r > 0 we define the average of f L1 on B(x, r) by ∈ ∈ loc 1 Z (Arf)(x) = fdλ. λ(B(x, r)) B(x,r)

Proposition 15.3. For any f L1 , A f is jointly continuous in x, r. • ∈ loc r

Proof. Let x x, r r. n → n → Then 1 (x) 1 for all x B[x, r] B(x, r). Since λ(B[x, r] B(x, r)) = 0 B(xn,rn) → B(x,r) 6∈ \ \ we get that f1 a.e. f1 . Also, since f L1 , for all n such that r r < B(xn,rn) −→ B(x,r) ∈ loc | n − | 1, dist(x , x) < 1 we have that f1 f1 L1. So by Dominated n | B(xn,rn)| ≤ | B(x,r+2)| ∈ Convergence, Z Z fdλ fdλ and λ(B(xn, rn)) λ(B(x, r)). B(xn,rn) → B(x,r) →

Thus, A f(x ) A f(x). rn n → r ut

This implies that Arf is a measurable function.

Exercise 15.1. Let x x, r r. Show that 1 (x) 1 for all n → n → B(xn,rn) → B(x,r) x B[x, r] B(x, r). 6∈ \

Exercise 15.2. Show that λ(B[x, r]) = λ(B(x, r)). 

Definition 15.4. Given f L1 define the Hardy-Littlewood maximal function • ∈ loc 1 Z Mf(x) = sup Ar f (x) = sup f dλ. r>0 | | r>0 λ(B(x, r)) B(x,r) | | 160

Exercise 15.3. Show that Mf is measurable. 

Theorem 15.5 (Hardy-Littlewood Maximal Theorem). For all α > 0 and all ••• f L1, ∈ 3d Z λ(Hf > α) = λ( x : Hf(x) > α ) f dλ. { } ≤ α | | R Proof. Let U = Hf > α . For every x U there exists rx > 0 such that f dλ > { } ∈ B(x,rx) | | α λ(B(x, rx)). Since U x U B(x, rx), for any c < λ(U) there exist x1, . . . , xn such that · ⊂ ∈ n Pn d for rj = rxj the balls (Bj := B(xj, rj))j=1 are pairwise disjoint and j=1 λ(Bj) > 3− c. Thus, n n X 1 X Z 3d Z c < 3d λ(B ) 3d f dλ f dλ. j ≤ α | | ≤ α · | | j=1 j=1 B(xj ,rj ) Taking supremum of c < λ(U) we obtain the result. ut

Exercise 15.4. Recall for a function ψ : R R, the definitions →  lim sup ψ(x) := lim sup ψ(x) = inf sup ψ(x). x y ε 0 0< x y <ε ε>0 0< x y <ε → → | − | | − |

Show that lim supx y ψ(x) c if and only if for every sequence xn y we have → | − | → ψ(xn) c. (In this case we say that limx y ψ(x) = c.) → →

Theorem 15.6 (Basic Differentiation Theorem). For any f L1 , ••• ∈ loc

lim Arf(x) = f(x) for λ-a.e. x. r 0 → Proof. First assume that f = f1 for some R > 0. So f L1. B(0,R) ∈ For every ε > 0 we can find a continuous function g such that R f g dλ < ε. So for | − | any x and δ > 0 there exists r > 0 such that if y x < r then g(x) g(y) < δ. For | − | | − | 161 this r > 0, 1 Z Arg(x) g(x) g(y) g(x) dλ(y) δ. | − | ≤ λ(B(x, r)) B(x,r) | − | ≤

Thus, limr 0 Arg(x) = g(x) for all x. We conclude that →

lim sup Arf(x) f(x) lim sup( Ar(f g)(x) + Arg(x) g(x) + g(x) f(x) r 0 | − | ≤ r 0 | − | | − | | − | → → H(f g)(x) + f g (x). ≤ − | − | So setting   Eα = x : lim sup Arf(x) f(x) > α , r 0 | − | → we have that E Hf > α/2 f g > α/2 . We have α ⊂ { } ∪ {| − | } Z Z αλ( f g > α ) f g dλ f g dλ < ε, {| − | } ≤ f g >α | − | ≤ | − | {| − | } so by the Hardy-Littlewood Maximal Theorem, 3dε 2ε λ(E ) + 0 as ε 0. α ≤ α α → → S Set E = n E1/n. Then λ(E) = 0 and for any x E, limr 0 Arf(x) = f(x). 6∈ → This proves the theorem for f that vanishes outside B(0,R). For general f L1 , then for all x B(x, R) and r < R we have that A f(x) = ∈ loc ∈ r A g(x) where g = f1 . So for a.e. x B(0,R), r B(0,2R) ∈

lim Arf(x) = lim Arf(x) = lim Arg(x) = g(x) = f(x). r 0 R>r 0 r 0 → → → So for every R there is a set N B(0,R) such that λ(N ) = 0 and for x B(0,R) N R ⊂ R ∈ \ R S we have limr 0 Arf(x) = f(x). Set N = R NR. Then λ(N) = 0 and if x N then → 6∈ limr Arf(x) = f(x). →∞ ut

15.2. The Lebesgue Differentiation Theorem

Definition 15.7. For a function f L1 define the Lebesgue set of f by • ∈ loc ( Z ) 1 Lf = x : λ(B(x,r)) f(y) f(x) dλ(y) = 0 . B(x,r) | − |

Proposition 15.8. For any f L1 , λ(Lc ) = 0. • ∈ loc f 162

Proof. For any z Q + iQ C, we have a null set Nz such that for x Nz, by applying ∈ ⊂ 6∈ the Basic Differentiation Theorem to the function y f(y) z , 7→ | − | 1 Z lim f(y) z fλ(y) = 0. r 0 λ(B(x, r)) | − | → B(x,r) S Let N = z Q+iQ Nz. So λ(N) = 0. ∈ Let x N. Fix ε > 0. Let z Q + iQ be such that f(x) z < ε. So f(y) f(x) 6∈ ∈ | − | | − | ≤ f(y) z + ε, and | − | 1 Z 1 Z f(y) f(x) dλ(y) f(y) z dλ(y) + ε ε. λ(B(x, r)) B(x,r) | − | ≤ λ(B(x, r)) B(x,r) | − | →

Taking ε 0 we have that x L . → ∈ f So Lc N. f ⊂ ut

A family (A ) of sets is said to shrink nicely to x if for all r > 0, A B(x, r) X r r r ⊂ and λ(A ) αλ(B(x, r)) for some constant α. r ≥ For example, if A B(0, 1) is a Borel set of positive measure α = λ(A) > 0, then ⊂ rA + x B(x, r) and λ(rA + x) = αrd cαλ(B(x, r)), so (rA + x) shrink nicely to x. ⊂ ≥ r

Theorem 15.9 (Lebesgue Differentiation Theorem). Let f L1 and any x L , ••• ∈ loc ∈ f if (Ar)r shrink nicely to x then

1 Z lim f(y) f(x) dλ(y) = 0. r 0 λ(A ) | − | → r Ar Proof. For some α > 0,

1 Z 1 Z f(y) f(x) dλ(y) f(y) f(x) dλ(y) 0. λ(Ar) Ar | − | ≤ αλ(B(x, r) B(x,r) | − | →

ut

15.3. Differentiation and Radon-Nykodim derivative

We now make a connection between differentiation and the Radon-Nykodim deriva- tive. 163

d A measure µ on (R , d) is called regular if µ(K) < for every compact K and for B ∞ every Borel A, µ(A) = inf µ(U). A U open ⊂

Exercise 15.5. Let f : Rd [0, ] be a measurable function, and define dµ = fdλ; → ∞  R i.e. µ(A) = A fdµ. Show that µ is regular if and only if f L1 . ∈ loc

Solution to ex:15.5. :( ♣ If f L1 then since any compact set K is bounded, we have that µ(K) = R fdλ < . ∈ loc K ∞ Also, for any ball B, f1 L1, and so the measures µ, λ restricted to B are finite. Thus, B ∈ for any ε > 0 there exists δ > 0 so that if λ(A B) < δ then µ(A B) < ε. For any ∩ ∩ Borel A B, and any δ > 0 we can find an open set A U B such that λ(U A) < δ. ⊂ ⊂ ⊂ \ So µ(U) µ(A) = µ(U A) < ε. This proves outer regularity for bounded A. − \ S If A is unbounded, write A = n An where An is bounded for all n. Then, for any ε > 0 take an open set U A such that µ(U ) µ(A ) + ε2 n. So for U = S U n ⊃ n n ≤ n − n n we have µ(U) µ(A) + ε. ≤ For the other direction, if dµ = fdλ, then for any bounded Borel set A we may find a compact set K containing A. So R fdλ R fdλ = µ(K) < .:) A ≤ K ∞ X

d Theorem 15.10. Let µ be a regular Borel measure on (R , d). Let µ = σ + ρ be ••• B the Lebesgue decomposition with respect to Lebesgue measure λ, so σ λ and ρ λ. ⊥  Then, for λ-a.e. x Rd, ∈ µ(A ) dρ lim r = (x), r 0 λ(A ) dλ → r for any family (Ar)r that shrinks nicely to x.

Proof. First note that if x L , ∈ f ρ(A ) 1 Z dρ dρ lim r = lim dλ(y) = (x), r 0 λ(A ) r 0 λ(A ) dλ dλ → r → r Ar by the Lebesgue Differentiation Theorem. 164

So we only need to show that for a.e. x, σ(A ) lim r = 0. r 0 λ(A ) → r Note that since A B(x, r), r ⊂ σ(A ) σ(B(x, r)) r , λ(Ar) ≤ αλ(B(x, r)) so we may assume that Ar = B(x, r). Let E be a Borel set such that σ(E) = 0 and λ(Ec) = 0. For a positive integer k set   σ(B(x,r)) 1 Fk = x E : lim sup λ(B(x,r)) > k . ∈ r 0 → It suffices to prove that λ(Fk) = 0 for all k. µ is outer regular, so for any ε > 0 there exists an open set U E such that ε ⊃ σ(U ) + λ(U ) = µ(U ) µ(E) + ε = λ(E) + ε, ε ε ε ≤ which implies that σ(U ) ε. For every x F there exists an open ball B U ε ≤ ∈ k x ⊂ ε such that σ(B ) > 1 λ(B ). Set V = S B U . So for any c < λ(V ) there exist x k x ε x Fk x ε ε ∈ ⊂

finitely many pairwise disjoint balls Bx1 ,...,Bxn such that n n X X 3d 3d c < 3d λ(B ) 3d 1 σ(B ) σ(V ) ε. xj ≤ k xj ≤ k ε ≤ k j=1 j=1

d So σ(F ) 3 ε. Since ε > 0 was arbitrary, we get that σ(F ) = 0. k ≤ k k ut

Exercise 15.6. For f L1 , define ∈ loc   Z  1 H∗f(x) = sup λ(B) f dλ : B = B(y, r) , x B . B | | ∈ Show that Hf H f 2dHf. ≤ ∗ ≤

Number of exercises in lecture: 6 Total number of exercises until here: 172 165

Measure Theory

Ariel Yadin

Lecture 16: The Riesz Representation Theorem

16.1. Compactly supported functions

Let X be some topological space. Let Cc(X) denote the space of all functions f : X C that have supp(f) = cl( x : f(x) = 0 ) is a compact set. → { 6 } We will use the notation f U to denote f : X [0, 1], supp(f) U, f C (X). ≺ → ⊂ ∈ c

Definition 16.1. We say that a space X is Urysohn if for any x U where U is • ∈ open, there exists f U and an open set V such that x V supp(f) with 1 f. ≺ ∈ ⊂ V ≤

Example 16.2. Any locally compact Hausdorff space is Urysohn. (This is a conse- quence of what is known as Urysohn’s Lemma.) The notion of locally compact Hausdorff spaces is an important one in topology, but it is beyond the scope of this course. 4 5 4

Exercise 16.1. Show that Rd is Urysohn. 

Solution to ex:16.1. :( ♣ Let x U for U open. So there exists an open ball B = B(x, 3ε) such that x ∈ ∈ B(x, 3ε) U. ⊂ Define  1 if y x ε,   || − || ≤ 1 f(y) = 2 ε− y x if ε y x 2ε,  − · || − || ≤ || − || ≤  0 if y x 2ε. || − || ≥ Since the function y y x is continuous, so is f. 7→ || − || 166

Since x B(x, ε) clB(x, 2ε) = supp(f) U and f 1, we are done.:) ∈ ⊂ ⊂ B(x,ε) ≡ X

Exercise 16.2. Let X be a countable set. Consider X as a topological space with the discrete topology (i.e. all subsets are open; alternatively, consider the metric dist(x, y) = 0 for x = y and dist(x, y) = 1 for x = y. 6 Show that a subset K X is compact if and only if K is finite. ⊂ Show that X is Urysohn.

Solution to ex:16.2. :( ♣ S If K is compact then K x K x is an open cover, so there exists a finite sub-cover, ⊂ ∈ { } that is K Sn x for some x , . . . , x K. This implies that K = x , . . . , x is ⊂ j=1 { j} 1 n ∈ { 1 n} finite. For the other direction, if K is finite and K S U is an open cover, then for every ⊂ α α S x K there exists α = α(x) such that x Ux := Uα. So K x K Ux is a finite ∈ ∈ ⊂ ∈ sub-cover. Every open cover has a finite sub-cover, so K is compact.

We now show that X is Urysohn. If x U for open U, then define f = 1 x . Since ∈ { } x is open and compact we have that x x supp(f) U. Finally, f is continuous { } ∈ { } ⊂ ⊂ since for the discrete topology any function is continuous.:) X

Exercise 16.3. Show that if f C (X) is a non-negative function then g := 1 f ∈ c ∧ is compactly supported and continuous; g C (X), g : X [0, 1]. ∈ c →

Solution to ex:16.3. :( ♣ It is immediate that supp(g) supp(f), so we only need to show that g is continuous. ⊂ Two proofs: 167

1 1 Proof I. If U R is an open set, then so is V = U 1 . If 1 U then g− (U) = f − (U) ⊂ \{ } 6∈ which is open. If 1 U then g(x) U if and only if either f(x) U or f(x) > 1. So ∈ ∈ ∈ g 1(U) = f 1(U) S f > 1 which is also open. − − { } Proof II. Let x x. If f(x) < 1 then because f is continuous, there exists n such n → 0 that for all n > n0 we have f(xn) < 1. Thus, limn g(xn) = limn f(xn) = f(x) = →∞ →∞ g(x). If f(x) 1 then for any ε > 0 there exists n such that for all n > n we have ≥ 0 0 f(x ) f(x) < ε, which implies that f(x ) 1 ε. This implies that g(x ) 1 ε | n − | n ≥ − n ≥ − for all n > n . So lim inf g(x ) 1 ε. Since this holds for all ε > 0, we have that 0 n n ≥ − lim supn g(xn) 1 lim infn g(xn) so limn g(xn) = 1 = g(x).:) X ≤ ≤ →∞

Proposition 16.3. Let X be Urysohn. • Let K U for compact K and open U. Then there exists f C (X) such that ⊂ ∈ c 1 f U. K ≤ ≺ Proof. For any x K U we can find an open set V and a function f U such that ∈ ⊂ x x ≺ x V supp(f ) and 1 f. ∈ x ⊂ x Vx ≤ S Since K is compact and K x K Vx is an open cover, we have a finite sub-cover; ⊂ ∈ that is, there exist x , . . . , x such that K Sm V . 1 m ⊂ j=1 xj Pm Pm Sn Set f := fx . Then 1K 1V f and supp(f) supp(fx ) U. j=1 j ≤ j=1 xj ≤ ⊂ j=1 j ⊂ So f C (X) and non-negative. Thus, 1 f U and 1 1 f. ∈ c ∧ ≺ K ≤ ∧ ut Proposition 16.4. Let X be an Urysohn space. Let K Sn U for compact K • ⊂ j=1 j and open (U )n . Then, there exist functions f U such that Pn f (x) = 1 for all j j=1 j ≺ j j=1 j x K. ∈ Proof. For every x K there exists j such that x U . So we can find a function ∈ ∈ j g U and an open set V such that x V supp(g ) U and 1 g . x ≺ j x ∈ x ⊂ x ⊂ j Vx ≤ x S Since K is compact and K x K Vx is an open cover, we have that there exists a fi- ⊂ ∈ nite sub-cover; i.e. there exist x , . . . , x K such that K Sm V Sm supp(g ). 1 m ∈ ⊂ i=1 xi ⊂ i=1 xi For every j set [ K = supp(g ) : supp(g ) U . j { xi xi ⊂ j} Then K is a finite union of compact sets, so it is compact. Also, by definition, K U . j j ⊂ j 168

Thus, we may find a function g C (X) such that 1 g U . j ∈ c Kj ≤ j ≺ j Let g = Pn g . Since for every x there exists j such that supp(g ) U , then K j=1 j i xi ⊂ j ⊂ Sm supp(g ) Sn K . So for all x K we have that g(x) 1. Let V = g > 0 i=1 xi ⊂ j=1 j ∈ ≥ { } which is an open set because g is continuous. Since K V , we can find h C (X) such ⊂ ∈ c that 1 h V . K ≤ ≺ Define f = g +1 h. Since supp(h) g > 0 , we have that f(x) > 0 for all x. Thus, − ⊂ { } gj we can define fj := f . For x K, since h(x) = 1, then Pn f (x) = Pn gj (x) = 1. Also, supp(f ) ∈ j=1 j j=1 g(x) j ⊂ supp(g ) U and so f U. j ⊂ j ≺ ut

16.2. Linear functionals

Definition 16.5. Let F be some space of functions f : X C.A linear functional • → I on F is a function I : F C such that I(αf +g) = αI(f)+I(g) for all α C, f, g F . → ∈ ∈ A linear functional is called positive if for any f F that is non-negative, we have ∈ I(f) 0. ≥

Recall that f 0 means that f is non-negative. Also, f g means that f g 0. X ≥ ≥ − ≥

Exercise 16.4. Let (X, , µ) be a measure space. Show that the function F I(f) = R fdµ is a positive linear functional on L1(X, , µ). F

Proposition 16.6. Let X be an Urysohn topological space. Let I be a positive linear • functional on C (X). Then, for any compact K X there exists a constant C > 0 c ⊂ K such that for any f Cc(X) with supp(f) K we have I(f) CK f (recall ∈ ⊂ ≤ · || ||∞ f = supx X f(x) ). || ||∞ ∈ | |

Proof. For general f C (X), since f = Ref + iImf we have I(f) I(Ref) + ∈ c | | ≤ | | I(Imf) , so it suffices to consider real-valued f. | | 169

Given a compact K, Let ϕK Cc(X) be a function ϕK : X [0, 1] such that ϕ 1 ∈ → K ≡ (X is always open, so this is possible by Urysohn).

If supp(f) K then f(x) ϕK (x) f for any x X. So f ϕK f 0 ⊂ | | ≤ · || ||∞ ∈ || ||∞ · − ≥ and f ϕK ( f) 0 f I(ϕK ) I(f) 0 and f I(ϕK ) + I(f) 0, which || ||∞ · − − ≥ || ||∞ · − ≥ || ||∞ · ≥ is I(f) I(ϕK ) f . | | ≤ · || ||∞ ut

16.3. The Riesz Representation Theorem

Definition 16.7. Let X be a topological space. A Borel measure on X is a measure • on (X, (X)). B A Borel measure µ is called Radon if

For any compact K X we have µ(K) < . • ⊂ ∞ µ is outer regular; i.e. for all Borel sets A, • µ(A) = inf µ(U). A U open ⊂ µ is inner regular on open sets; i.e. for any open set U, • µ(U) = sup µ(K). U K compact ⊃

The notation f U is short for f C (X), f : X [0, 1] and supp(f) U. X ≺ ∈ c → ⊂

??? THEOREM 16.8 (Riesz Representation Theorem). Let X be an Urysohn space.

Let I be a positive linear functional on Cc(X). Then, there exists a unique Radon measure µ on X such that I(f) = R fdµ for all f C (X). ∈ c Moreover, the measure µ satisfies

µ(U) = sup I(f): f U open U, { ≺ } ∀ µ(K) = inf I(f): f C (X) , f 1 compact K. { ∈ c ≥ K } ∀

Proof. The statement of the theorem itself suggest how to define µ: Define

µ(U) := sup I(f): f U { ≺ } for any open U X. ⊂ 170 S Step I. µ is countably sub-additive on open sets; i.e. if U = n Un for open sets (U ) , we claim that µ(U) P µ(U ). n n ≤ n n Indeed, for any f C (X) such that f U, we have that K = supp(f) is compact ∈ c ≺ and K S U is an open cover. So there exists a finite sub-cover K Sn U . We ⊂ n n ⊂ j=1 j may now find g , . . . , g such that g U and Pn g (x) = 1 for all x K. This 1 n j ≺ j j=1 j ∈ implies that f = Pn fg and fg U . So by definition of µ, j=1 j j ≺ j n n X X X I(f) = I(fg ) µ(U ) µ(U ). j ≤ j ≤ n j=1 j=1 n Since this holds for any f U, we get that µ(U) P µ(U ). ≺ ≤ n n For an arbitrary subset A X define ⊂

µ∗(A) = inf µ(U). A U open ⊂

(*) Exercise: Show that µ∗(U) = µ(U) for all open U.

Step II. µ∗ is an outer measure. Recall the axioms of an outer measure: µ ( ) = 0, µ (A) µ (B) for all A B and ∗ ∅ ∗ ≤ ∗ ⊂ µ (S A ) P µ (A ). ∗ n n ≤ n ∗ n The first two are easy. Now for the third axiom. Since a countable union of open sets is open, we have that ( ) X [ µ∗(A) = inf µ(U ): A U ,U are all open . n ⊂ n n n n (*) Exercise: prove this. Let A = S A . If µ (A ) = for some n there is nothing to prove. So assume n n ∗ n ∞ that µ (A ) < for all n. Fix ε > 0 and for every n let U be an open set such that ∗ n ∞ n A U and µ(U ) µ (A ) + ε 2 n. Then, A S U and n ⊂ n n ≤ ∗ n · − ⊂ n n X X µ∗(A) µ(U ) µ∗(A ) + ε. ≤ n ≤ n n n Taking ε 0 completes the proof that µ is an outer measure (step II). → ∗ Step III. Every open set U is µ∗-measurable. Recall that U is µ -measurable if and only if for any set A such that µ (A) < , we ∗ ∗ ∞ have µ (A) µ (A U) + µ (A U c). If A is such a set, fix some ε > 0 and let V be ∗ ≥ ∗ ∩ ∗ ∩ 171 an open set such that A V and µ(V ) µ (A) + ε. Since V U is open, by definition ⊂ ≤ ∗ ∩ of µ there exists f C (X), f V U, such that I(f) µ(V U) ε. Also, the set ∈ c ≺ ∩ ≥ ∩ − K = supp(f) is closed so V Kc is open. So we may find g C (X), g V Kc, such ∩ ∈ c ≺ ∩ that I(g) µ(V Kc) ε. Now note that supp(f) supp(g) = because supp(g) Kc. ≥ ∩ − ∩ ∅ ⊂ So 0 f + g 1 and supp(f + g) K (V Kc) V . Thus, by definition of µ, ≤ ≤ ⊂ ∪ ∩ ⊂

µ∗(A) µ(V ) ε I(f + g) ε = I(f) + I(g) ε ≥ − ≥ − − µ(V U) + µ(V Kc) 3ε µ(A U) + µ(A U c) 3ε. ≥ ∩ ∩ − ≥ ∩ ∩ −

Taking ε 0 completes the proof that U is µ -measurable (step III). → ∗ Step IV. By Charath´eodory’s Theorem µ∗ restricted to the Borel sets is a measure. Denote this measure by µ. For any Borel set A we have a sequence of open sets (U ) such that µ(U ) µ(A) n n n → and A U for all n. So µ is outer-regular. ⊂ n Step V. We show that µ(K) = inf I(f): f 1 for all compact K. { ≥ K } If K is compact and f 1 then for any ε > 0 let U = f > 1 ε . Then U ≥ K ε { − } ε is open (because f is continuous). For any g U we have that f > (1 ε)g, so ≺ ε − I(f) (1 ε)I(g). Taking supremum over all g U we have that µ(K) µ(U ) ≥ − ≺ ε ≤ ε ≤ (1 ε) 1I(f). Since this holds for all ε, taking ε 0 we get that µ(K) I(f). This − − → ≤ was for any f 1 , so µ(K) inf I(f): f 1 ≥ K ≤ { ≥ K } For the other inequality, let U K be any open set. Then we can find f C (X), ⊃ U ∈ c such that f 1 and fU U. So fU 1K and I(fU ) µ(U). Taking infimum, we K ≡ ≺ ≥ ≤ have that inf I(f): f 1 inf I(f ): K U open inf µ(U): K U open = µ(K). { ≥ K } ≤ { U ⊂ } ≤ { ⊂ }

This proves µ(K) = inf I(f): f 1 for all compact K (step V). { ≥ K } Step VI. µ(K) < for all compact K. Indeed, there exists f C (X) such that ∞ ∈ c f : X [0, 1] and f 1. So f 1K , and I(f) CK f = CK for some constant → K ≡ ≥ ≤ · || ||∞ C > 0. Thus, µ(K) I(f) < . K ≤ ∞ 172

Step VII. µ is inner regular on open sets. Indeed, if U is an open set, for any α < µ(U) let f C (X), f U be such that I(f) α. Set K := supp(f) which is ∈ c ≺ ≥ compact and K U. ⊂ For any g C (X) such that g 1 we have that g f 0, so I(g) I(f) α. ∈ c ≥ K − ≥ ≥ ≥ Taking infimum over all such g we have that µ(K) α. ≥ Thus, for any α < µ(U) we have a compact K U such that µ(K) α. Taking ⊂ ≥ supremum over α gives inner regularity (step VII). Step VIII. We now show that I(f) = R fdµ for all f C (X). ∈ c If f C (X) write f = f f +i(f f ) for non-negative f , so it suffices to consider ∈ c 1 − 2 3 − 4 j non-negative f. Since f C (X) is continuous on a compact set, namely supp(f), it ∈ c attains a maximum there. So f < . So replacing f by 1 f, we may assume f ∞ || ||∞ ∞ || || without loss of generality that f : X [0, 1]. → Fix n > 0. For any 1 j n set K = f j/n and K = supp(f). For 1 j n ≤ ≤ j { ≥ } 0 ≤ ≤ define

+ 1  j 1  f = f − . j n ∧ − n

That is,

 j 1 0 f(x) − ,  n  ≤ j 1 j 1 j fj(x) = f(x) − f(x) [ − , ].  − n ∈ n n   1 f(x) j . n ≥ n

1 1 1 R 1 With this definition 1Kj fj 1Kj−1 , so µ(Kj) fjdµ µ(Kj 1). n ≤ ≤ n n ≤ ≤ n − Note that fj is continuous, and supp(fj) Kj 1, so fj Cc(X) (this includes the ⊂ − ∈ case j = 1 where K0 = supp(f)).

For any open set U Kj 1, we have that nfj U. So nI(fj) µ(U). Taking ⊃ − ≺ ≤ infimum over all such open sets U Kj 1, by outer regularity we have that I(fj) ⊃ − ≤ 1 µ(Kj 1). n − Also, nf 1 , so 1 µ(K ) I(f ). j ≥ Kj n j ≤ j 173

j n j 1 fj + −n j 1 −n

j 1 Figure 5. The function fj raised by −n .

Pn Now, note that f = j=1 fj. (Exercise!) So by additivity of the integral and the linear functional,

n n 1 X 1 X µ(Kj) I(f) µ(Kj 1) and n ≤ ≤ n − j=1 j=1 n n 1 X Z 1 X µ(Kj) fdµ µ(Kj 1). n ≤ ≤ n − j=1 j=1

Taking the differences between these inequalities,

Z n 1 X µ(K0) µ(Kn) 1 I(f) fdµ (µ(Kj 1) µ(Kj)) = − µ(supp(f)). − ≤ n − − n ≤ n j=1

µ is finite on compact sets (as a Radon measure), so µ(supp(f)) < . Thus, taking ∞ n we have that I(f) = R fdµ as required (step VIII). → ∞ Step IX. Uniqueness. Let ν be another Radon measure satisfying I(f) = R fdν for all f C (X). For any open set U, if f U then I(f) = R fdν ν(U). This holds ∈ c ≺ U ≤ for any f U so ≺ ν(U) sup I(f): f U . ≥ { ≺ } 174

Since ν is inner regular on open sets, we can find a sequence of compact sets (Kn)n such that K U and ν(K ) ν(U). For every n, Urysohn guaranties that there exists a n ⊂ n → function fn Cc(X) such that fn 1 and fn U. So ∈ Kn ≡ ≺ Z Z ν(Kn) fndν fndν = I(fn) sup I(f): f U . ≤ Kn ≤ ≤ { ≺ } Taking n we get that ν(U) = sup I(f): f U = µ(U). → ∞ { ≺ } So ν(U) = µ(U) for all open sets U. Since ν, µ are outer regular, this implies that they are equal on all Borel sets. ut

Exercise 16.5. Let X be an Urysohn space. Let F X be a closed subset. Let ⊂  R µ be a Radon measure on F . Define I(f) = f dµ for all f Cc(X). F ∈ Show that I is a positive linear functional on Cc(X). Show that if I(f) = R fdν for the Radon measure ν guarantied by the Riesz Repre- sentation Theorem, then ν(A) = µ(A F ) for all Borel A. ∩

Number of exercises in lecture: 5 Total number of exercises until here: 177