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 j j j j ≤ p2 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 1 f g = (a; a). The length or measure of such I is defined to be I = m(I) = b a. ; j j − 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 . j j j 1j · · · j dj 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, • n f 2 62 g \ E F := (E F ) (F E), • 4 n [ n 6 E + x := y + x : y E . • f 2 g 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 !1 n \ Consequently, if B Rd is a box then, ⊂ 1 m(B) = lim #(B 1 d); n d n Z !1 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 ··· !1 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 2 n 2 ≤ f 2 ≤ ≤ g # z Z : na z nb nb + 1 (na 1) + 1 = n(b a) + 3: ≤ f 2 b c ≤ ≤ d eg ≤ − − − Similarly, #(I 1 Z) # z Z : na < z < nb n(b a) 1: \ n ≥ f 2 g ≥ − − Dividing by n and taking n we obtain the formula for intervals. ! 1 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. ! 1 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 ··· !1 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 !1 n \ does not exist. 1 1 lim #(A 1 ) and lim #((A + x) 1 ) n n Z n n Z !1 n \ !1 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 \ !1 !1 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 \ ≥ 2 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 !1 \ ≤ !1 \ 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.