MEASURE THEORY Course: 201.1.0081 Fall 2014-15 Lecture

ARIEL YADIN

Course: 201.1.0081 Fall 2014-15 Lecture notes updated: September 28, 2014

Contents

Lecture 1. Introduction 4 1.1. Measuring things 4 1.2. Elementary measure 5 This lecture has 6 exercises.8

Lecture 2. Jordan measure 9 2.1. Jordan measure 9 This lecture has 14 exercises. 14

Lecture 3. Lebesgue outer measure 15 3.1. From ﬁnite to countable 15 This lecture has 5 exercises. 19

Lecture 4. Lebesgue measure 20 4.1. Deﬁnition of Lebesgue measure 20 4.2. Lebesgue measure as a measure 26 This lecture has 12 exercises. 29

Lecture 5. Abstract measures 30 5.1. σ-algebras 30 5.2. Measures 32 5.3. Fatou’s Lemma and continuity 35 This lecture has 13 exercises. 38

Lecture 6. Outer measures 39 1 2

6.1. Outer measures 39 6.2. Measurability 40 6.3. Pre-measures 42 This lecture has 12 exercises. 46

Lecture 7. Lebesgue-Stieltjes Theory 47 7.1. Lebesgue-Stieltjes measure 47 7.2. Regularity 51 7.3. Non-measureable sets 52 7.4. Cantor set 52 This lecture has 6 exercises. 53

Lecture 8. Functions of measure spaces 54 8.1. Products 54 8.2. Measurable functions 55 8.3. Simple functions 60 This lecture has 20 exercises. 62

Lecture 9. Integration: positive functions 63 9.1. Integration of simple functions 63 9.2. Integration of positive functions 65 This lecture has 7 exercises. 69

Lecture 10. Integration: general functions 70 10.1. Real valued functions 70 10.2. Complex valued functions 71 10.3. Convergence 73 10.4. Riemann vs. Lebesgue integration 75 This lecture has 10 exercises. 76

Lecture 11. Product measures 77 11.1. Sections 80 11.2. Product integrals 81 3

11.3. The Fubini-Tonelli Theorem 85 This lecture has 18 exercises. 88

Lecture 12. Change of variables 89 12.1. Linear transformations of Lebesgue measure 89 12.2. Change of variables formula 91 This lecture has 4 exercises. 95

Lecture 13. Lebesgue-Radon-Nykodim 96 13.1. Signed measures 96 13.2. The Lebesgue-Radon-Nikodym Theorem 102 This lecture has 27 exercises. 109

Lecture 14. Convergence 110 14.1. Modes of convergence 110 14.2. Uniform and almost uniform convergence 114 This lecture has 10 exercises. 118

Lecture 15. Differentiation 119 15.1. Hardy-Littlewood Maximal Theorem 119 15.2. The Lebesgue Differentiation Theorem 122 15.3. Differentiation and Radon-Nykodim derivative 123 This lecture has 6 exercises. 125

Lecture 16. The Riesz Representation Theorem 126 16.1. Compactly supported functions 126 16.2. Linear functionals 128 16.3. The Riesz Representation Theorem 129 This lecture has 5 exercises. 134

Total number of exercises: 175 134 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) = n(b a) + 2. ≤ { ∈ b b≤ ≤ d d} ≤ − − − Similarly,

#(I 1 Z) # z Z : na < z < nb n(b a) 2. ∩ 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 1.2.

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. Give an example of a set A [0, 1] and x R such that the limits ⊂ ∈ 1 1 lim #(A 1 ) and lim #((A + x) 1 ) n n Z n n Z →∞ n ∩ →∞ n ∩ both exist but are not equal.

This last exercise shows that we do not want to use the discretisation 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 properties.

m : [0, ). • E0 → ∞ (Additivity) m(E F ) = m(E) + m(F ). • ] 8

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). • ∈

Exercise 1.5. Show that if m : [0, ) is additive and translation invariant, 0 E → ∞ 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.

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 ). × ·

Number of exercises in lecture: 6 Total number of exercises until here: 6 9

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. ∗

Note that every elementary set E is Jordan measurable and m(E) = J ∗(E) = J (E). ∗ 10

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.

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. ⊂ ∗ \

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. • { ∈ } 11

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

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

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. ⊂

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 12

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. • ∗

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.

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). × ·

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. 13

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 j | i,j| | i,j| B = I I we have that M (B) = I cI I and for i < j, 1 × · · · × d j 1 × · · · × j × · · · × d S (B) = I I I I . Also, i,j 1 × · · · j × · · · × i × · · · × d 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 − . So Ri,j(B) is Jordan measurable for all i, j and boxes B. d Note that if U = [0, 1) then in all cases one may compute that m(Mj(U)) = c , m(S (U)) = 1 and m(R (U)) = 1. This last computation due to the fact that | | i,j i,j in two dimensions m( (x, x + y): x, y [0, 1) ) { ∈ } = m( (x, x + y): x, x + y [0, 1) ) + m( (x, x + y): x [0, 1), x + y [1, 2) ), { ∈ } { ∈ ∈ } which is the sum of the measure of two right-angle triangles of side length 1. 14

So we conclude that m(L(U)) = det L for any invertible linear map L. | | Since L is bijective, we have that L(A B) = L(A) L(B). So we conclude that for ] ] any elementary set E, the set L(E) is a ﬁnite union of disjoint Jordan measurable sets, and thus Jordan measurable.

Now, consider the function m0(E) := m(L(E)) on elementary sets. This is a non- negative additive translation invariant function on . Indeed, L(E F ) = L(E) L(F ) E0 ] ] and L(E + x) = L(E) + Lx. We conclude that there exists a constant D > 0 such that m(L(E)) = Dm(E) for all E . Taking U = [0, 1)d as the unit box, we get that ∈ E0 D = m(L(U)) = det L . | | Now let A be Jordan measurable. For any ε > 0 there exist E A F , E,F ⊂ ⊂ ∈ E0 such that m(F E) < ε. So L(E) L(A) L(F ). Since E,F are elementary we have \ ⊂ ⊂ that m(L(E)) = Dm(E), m(L(F )) = Dm(F ), and so

J ∗(L(A)) m(L(F )) = Dm(F ) < Dm(E) + Dε Dm(A) + Dε, ≤ ≤ J (L(A)) m(L(E)) = Dm(E) > Dm(F ) Dε Dm(A) Dε. ∗ ≥ − ≥ −

Taking ε 0 gives J ∗(L(A)) = J (L(A)) = Dm(A). → ∗ ut

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.

Number of exercises in lecture: 14 Total number of exercises until here: 20 15

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. 16

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 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 ∗ ] ∗ ∗ 17 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 monotonicity 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. → 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. 18

The next proposition shows that the notation “m” is an appropriate one, as an extension 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. 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 19

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).

Number of exercises in lecture: 5 Total number of exercises until here: 25 20

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. 21

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 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 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) has diameter − ∈ n n ⊂ ⊂ n n √d2− ). Since for every n there exists some dyadic cube of scale n that contains x d S (because R = z Zd Dn(z) for every ﬁxed n), we get that for some large enough n ∈ d 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). Since this holds ∈ n 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.

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). | | ∗ 22

Figure 1. 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.

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. ∗ ≤ ∗ ⊃ 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| − 23 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. ∅ ∗ ∅ ≤ ∗ ∅ 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 24 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. 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 25

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

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

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. 26

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 27

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

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

Proposition 4.8 (Charathéodory 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). • ∗ ∩ ∗ \ 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 28

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 have that A is Lebesgue measurable as a countable union of Lebesgue measurable sets.

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

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

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. 29

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).

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 m is Lebesgue measure.

Number of exercises in lecture: 12 Total number of exercises until here: 37 30

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 31

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 32

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(B : B is an open box ). B B B

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

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 33

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 ]

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 34

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). \ ≥ 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

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 35

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

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 36

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 ≥ ≥ 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. 37

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 ≥ ≥ 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 →∞ 38

Number of exercises in lecture: 13 Total number of exercises until here: 50 39

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 ,X , 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

Show that µ∗ is an outer measure.

Compare this to the case that is the collection of boxes in Rd. E 40

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.

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. 41

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 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 42

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

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. 43

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

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 44

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 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 ≤ 45

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

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 ∗ (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.

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). ∗ ≤ ∗ 46

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 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: 62 47

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 ádroite). 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. Well deﬁned. To show that µ0 is well deﬁned: Note that if I,J are intervals of the form (a, b] for a < b, or of the form ( , b] or (a, ), or , then also I J is one −∞ ∞ ∅ ∩ of these forms. Now, if (a, b] = (a , b ] (a , b ], then without loss of generality 1 1 ]···] n n a = a < b = a < b = = a < b = b. So, 1 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 ]···] 48

(I J ), so i ∩ k 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 ). i i ∩ j 0 j i=1 i,j j=1

This shows that µ0 is well defined, and also finitely additive, a fact we will use 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 (A ) so that A = (a , b ] and a < b a n n n n n n n ≤ n+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, we may ﬁnd n1, . . . , nm such that

[a + δ, b] Sm (a , b + δ ), • ⊂ k=1 nk nk nk and for all k, we have b + δ (a , b + δ ). • nk nk ∈ nk+1 nk+1 nk+1 49

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 − − ≤ nm nm − n1 m 1 X− = F (b + δ ) F (a ) + F (a ) F (a ) + ε nm nm − nm nj+1 − nj j=1 m 1 X− F (b + δ ) F (a ) + F (b + δ ) F (a ) + ε ≤ nm nm − nm nj nj − nj j=1 m m X X µ (A ) + F (b + δ ) F (b ) + ε ≤ 0 nj nj nj − nj 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 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 − 50

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 Charathódory’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

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 − − −

Exercise 7.3. Show that for F (x) = x the associated measure µF is Lebesgue measure on R. 51

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). ∩ ≥ −

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. By the definition of the outer measure from the pre-measure µ0, for every ε > 0 there exists intervals ((a , b ]) such that A S (a , b ] and P µ((a , b ]) µ(A)+ε. n n n ⊂ n n n n n n ≤ n For every n there exists δn > 0 such that F (bn + δn) < F (bn) + ε2− . Thus, taking S 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, ≤ ≤ so µ(K) µ(A) ε. Taking supremum of such K A and ε 0 we have inner ≥ − ⊂ → regularity for bounded A. 52

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 continuity. r ≥ ∩ − − ∩ − → So sup µ(K) lim µ(A ( r, r)) ε = µ(A) ε. A K compact ≥ r infty ∩ − − − ⊃ → 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

Exercise 7.6. Show that C is Lebesgue measurable and has Lebesgue measure 0. (*) Show that C has cardinality of [0, 1]. 53

Number of exercises in lecture: 6 Total number of exercises until here: 68 54

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

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 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 55 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( E : E n) = σ(π− (E ): E ). Fn ⊗n n n ∈ En ∀ n n n ∈ En n

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 56

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

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 → 57

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.

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

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

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. 58

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. →∞

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 → | | → 59 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.

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 Y = F −∞ ∈ F Y Y F → F A Y : A . { ∩ ∈ 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. 60

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.

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 61

[0, ) ∞

n 2− (k + 1)

n 2− k

n ϕn 1 2− − ϕn

An,k

Figure 2. Approximating functions with simple functions j. Show that in fact, we have the standard representation

X f = a1f −1(a). 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. • − − − 62

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 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