Product measures, Tonelli s and Fubini s theorems For use in MAT4410, autumn 2017 Nadia S. Larsen. 17 November 2017.

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1 Product measures, Tonelli s and Fubini s theorems For use in MAT4410, autumn 017 Nadia S. Larsen 17 November Construction of the product measure The purpose of these notes is to prove the main theorems, Tonelli s and Fubini s theorems. These notes expand upon the material on product measure in []. The overall goal is to be able to integrate functions of two or several variables. For this we need to make sense of multiple integrals and show that we can compute them by changing the order of integration in each variable of the given function. Given two measure spaces, A 1, µ 1 and X, A, µ, the goal is to construct a σ-algebra A 1 A of subsets of X and a measure ω on A 1 A in such a way that the above mentioned change of order of integration is possible when we consider A 1 A -measurable functions nonnegative or real-valued. For motivation, consider the case of Lebesgue measure on R, and let I, J be intervals. Then the area of the rectangle I J is 1 areai J = lilj = λiλj. We can think of area as an appropriate choice for a measure ω assigned to a rectangle I J. Next we formalise this idea for two general measure spaces. We define the collection C of measurable rectangles in A 1 A to be C = {A B A A 1, B A }. Lemma 1.1. The collection C of measurable rectangles is closed under finite intersections. Proof. Exercise, or see Proposition 6.1 in [1]. For the existence of the product measure see Theorem 6.4 in [1]. For uniqueness see Theorem 6.5 in [1]. For a subset A of a set X we let A c denote the complement X \ A. Definition 1.. Let X be a set. A collection D of subsets of X is a Dynkin system if the following are satisfied: D1 X D, D A D A c D, and D3 if {A n } n is a pairwise disjoint family in D, then n A n D. Note the following: in a Dynkin system D, if A, B D are so that A B, then B \ A = B c A c D because the union is disjoint. Given a set X and a collection E of subsets of X, recall that σe denotes the σ- algebra generated by E. Since there exists at least one Dynkin system that contains E, namely the collection PX, then by taking the intersection of all Dynkin systems which contain E we obtain the Dynkin system generated by E: we denote this by DE and point out that this is the smallest Dynkin system that contains E. 1

2 Lemma 1.3. If D is a Dynkin system over X that is closed under finite intersections or finite unions, then D is a σ-algebra. Proof. Let {A n } n be an arbitrary sequence in D. We make this into a disjoint sequence, thus we let B 1 = A 1, B = A \A 1 = A A c 1, and for all m > let B m = A m \ m k=1 A k = A m m k=1 Ac k. Then {B m} m is a disjoint union in D, so m B m = n A n D. Lemma 1.4. Dynkin Let X be a set and E a collection of subsets of X with the property that A, B E A B E. Then DE = σe. Proof. Since any σ-algebra is also a Dynkin system and since E σe, we have that DE σe. To prove the converse inclusion it suffices to show that DE is a σ- algebra. For simplicity of notation, write D instead of DE. By Lemma 1.3, it is enough to show that D is closed under intersections. For A D we let D A = {B D A B D}. We claim that D A is a Dynkin system that contains E. Since A X = A is in D, we have X D A. Let B D A. Then A B c = A \ B A, which is in D by earlier observation. So B c D A. Let {B m } m be a pairwise disjoint sequence in D A. Then A m B m = m A B m, which is in D because {A B m } m is a pairwise disjoint sequence in D. This shows that m B m D A, so D A is a Dynkin system. If A E, then for every B E we have A B E by the hypothesis. Thus A B D, so B D A. Since B was arbitrary, we have E D A each time we start with A in E. Since D is the smallest Dynkin system on X that contains E, we have D D A, so D = D A when A E. It is important to realise that this argument is valid for any A E. Now let A D. Take B E. Then A B D by the previous paragraph. This means that B D A. Since B was arbitrary, E D A for A D, hence D = D A for all A D. This means that D is closed under intersections. Definition 1.5. A measure space X, A, µ is σ-finite if there exists a sequence {X n } n 1 in A such that X n X n+1 for n 1, X = n X n and µx n < for all n. Let, A 1, µ 1 and X, A, µ be two measure spaces, and let A 1 A be the product σ-algebra on X : this is the smallest σ-algebra on X that contains the collection C of measurable rectangles. If A A 1 A and x 1, x X, we consider the following sets 3 4 A x = {x 1 x 1, x A}, A x 1 = {x x 1, x A} X. Note that if A = A 1 A C, then A x = A 1 if x A and A x = if x / A. Therefore A x A 1 for each x X. Similarly, A x 1 A. We have the following result. Lemma 1.6. For all A A 1 A, x 1 and x X we have A x A 1 and A x 1 A.

3 Proof. Let x X be arbitrary. If we let C 1 = {A A 1 A A x A 1 }, it suffices to prove that C 1 is a σ-algebra such that C C 1. Now check that A c x = A x c for every A A 1 A and n E n x = n Ex n for an arbitrary sequence {E n } in A 1 A. Then the claims follow using the definitions. Theorem 1.7. Assume that, A 1, µ 1 and X, A, µ are two finite measure spaces. Let A A 1 A. Then, 1 The function [0, ] given by x 1 χ A x 1, x is A 1 -measurable for each x in X. The function X [0, ] given by x χ A x 1, x is A -measurable for each x 1 in. 3 The function g A : [0, ] given by x 1 X χ A x 1, x dµ x is A 1 - measurable. 4 The function h A : X [0, ] given by x χ A x 1, x dµ 1 x 1 is A - measurable. Moreover, the following holds: 5 χ A x 1, x dµ x dµ 1 x 1 = χ A x 1, x dµ 1 x 1 dµ x. X Proof. Let D denote the collection of measurable sets A in A 1 A such that the statements 1-5 hold. We claim that D contains C. Let A = A 1 A C. Since the function in 1 is χ A x for each x X, it is A 1 -measurable by Lemma 1.6. Similarly for the function in. Next note that g A x 1 = χ A x 1 x dµ x = µ A x 1 = χ A1 x 1 µ A. X Thus it is A 1 -measurable because it is a simple function, and similarly for h A. Finally, note that both sides in 5 equal µ 1 A 1 µ A = µ 1 µ A 1 A. Since C generates A 1 A and is closed under finite intersections by Lemma 1.1, it will follow from Lemma 1.4 that the theorem is true if we show that D is a Dynkin system. By definition, X D. Let A D. Since χ A c = 1 χ A, it is easy to verify that 1- hold for χ A c. Since µ is finite, the function in 3 is g A cx 1 = µ A µ A x 1 = µ A g A x 1, and so is A 1 -measurable because g A is by our assumption that 1-5 hold for A. A similar argument shows that the function in 4 is measurable. To verify that 5 holds for χ A c, use that it holds for A as follows: g A cx 1 dµ 1 x 1 = µ X µ A x 1 dµ 1 x 1 = µ 1 µ X χ A x 1, x dµ x dµ 1 x 1 X = µ 1 µ X χ A x 1, x dµ 1 x 1 dµ x X X 1 = h A cx dµ x. X X 3

4 4 It remains to prove that D is closed under disjoint countable unions. For this, we prove first that D is closed under finite disjoint unions. Let E i D, i = 1,..., n such that E i E j = whenever i j. Then also E i x E j x = whenever i j, for all x X. Since i E i x = i E i x for each x X, additivity of µ 1 implies that the function in 4 corresponding to n i=1 E i in place of A is x n i=1 h E i, and so is A -measurable by 4 for each E i. Similarly with 3. By additivity of the integral, 5 for i E i follows from n n h i E dµ i = h Ei dµ = g Ei dµ 1 = g i E dµ i 1. X X i=1 Now we show that D is closed under increasing unions. Let C i D for i 1 such that C i C i+1 for i 1. Note then that C i x C i+1 x for each x. Since µ 1 is a finite measure, h Ci+1 x h Ci x = µ 1 C i+1 x µ 1 C i x = µ 1 C i+1 x \ C i x 0. Thus {h Ci } i is a nondecreasing sequence of A -measurable functions. The pointwise limit of {h Ci } i is also A -measurable, and by the monotone convergence theorem we have lim i X h Ci dµ = X lim i h Ci dµ. Continuity of µ 1 implies that lim i h Ci x = lim i µ 1 C i x = µ 1 i C i x. Similarly, by applying the monotone convergence theorem to g Ci and by using assumption 5 for all C i we get that g Ci dµ 1 = lim g Ci dµ 1 = lim h Ci dµ 1 = h Ci dµ, X i 1 X i 1 X X which shows that i C i D, as claimed. It now follows that D is a Dynkin system use the previous two steps to show that 1-5 hold for the union of a countable disjoint family E i by forming an increasing sequence C m = m i=1e i, each being a finite disjoint union in D. Lemma 1.4 gives DC = σc. Since σc = A 1 A, it follows that A 1 A D, so the theorem is proved. Theorem 1.8. Let, A 1, µ 1 and X, A, µ be σ-finite measure spaces. Then statements 1-5 of Theorem 1.7 hold. Proof. By the hypothesis, there are sets X i,n A i for n 1 and i = 1, such that {X i,n } n is increasing, µ i X i,n < for all i = 1, and n 1, and X i = n X i,n for i = 1,. We reduce the problem to the finite case by considering, for each n 1, the measure spaces,n, A 1 X1,n, µ 1,n and similarly for j =, where µ 1,n B,n = µ 1 B,n and µ,n C X,n = µ C X,n for all B A 1 and C A. Clearly µ 1,n and µ,n are finite measures on A 1 X1,n and A X,n, respectively. Note that,n h dµ 1,n = hχ X1,n dµ 1 for any nonnegative A 1 -measurable function. First check for h a simple function, then approximate by nonnegative simple functions. Use that h X1,n :,n R is A 1 X1,n -measurable. See Exercise 5.60 in [1]. Similarly for µ,n and all n. i=1

5 Applying Theorem 1.7 to µ 1,n and µ,n for all n 1 gives 6 g A x 1 dµ 1,n x 1 =,n h A x dµ,n x X,n for all A A 1 A. By the observation in the previous paragraph, we have g A χ X1,n dµ 1 = h A χ X,n dµ X for all n 1. Now we take the limit as n and use that {,n } increases to and {X,n } increases to X. Therefore the monotone convergence theorem, applied in both the right- and left-hand side, shows that 5 holds in this case. The other statements are verified in a similar way. Corollary 1.9. Let, A 1, µ 1 and X, A, µ be two σ-finite measure spaces. Then for each A A 1 A we have 7 µ 1 µ A = χ A x 1, x dµ x dµ 1 x 1 X 8 = χ A x 1, x dµ 1 x 1 dµ x. X 5. Tonelli s and Fubini s theorems Let X i, A i, µ i be σ-finite measure spaces for i = 1, and let µ 1 µ be the product measure on A 1 A given by Theorem 1.9. Given a function f : X R, define its sections at x 1 and x X by f x x 1 = fx 1, x and f x1 x = fx 1, x. Proposition.1. If f : X R is A 1 A -measurable, then f x : R is A 1 -measurable for every x X. Similarly, f x1 : X R is A -measurable for every x 1. Proof. Since f is measurable, for any r R the set A = {x 1, x X fx 1, x r, ]} is in A 1 A. But {x 1 f x x 1 r, ]} = {x 1 x 1, x A}, in other words fx 1 r, ] = A x, which lies in A 1 by Lemma 1.6. Since r R is arbitrary, the claim for f x follows. The proof in the case of f x1 is similar. Note: the proposition is also valid for a complex-valued measurable function f on X. The proof is similar to the above or see Proposition 6.10 in [1]. Theorem.. Tonelli s theorem Let X i, A i, µ i be σ-finite measure spaces for i = 1, and let µ 1 µ be the product measure on A 1 A. Suppose that f : X [0, ] is A 1 A -measurable. Then the functions x 1 X f x1 dµ and x f x dµ 1 are

6 6 measurable and 9 f dµ 1 µ = X 10 = X fx 1, x dµ 1 x 1 dµ x X 1 fx 1, x dµ x dµ 1 x 1. X Proof. The case when f = χ A for A A 1 A was done in Corollary 1.9 check that χ A x1 = χ A x 1 and similarly for j =. By linearity the result extends to simple functions on X. Next let {s n } n be a sequence of simple functions on X such that s n 0, s n s n+1 for all n 1, and s n f pointwise. Thus we have for every n 1 that 11 s n dµ 1 µ = s n x 1, x dµ 1 x 1 dµ x X X X 1 1 = s n x 1, x dµ x dµ 1 x 1. X For every fixed x the sequence {s n x } of functions R is increasing and convergent to f x at every point x 1. By the MCT, f x dµ 1 = lim n s n x dµ 1. Since the theorem is true for simple functions, the function F n x := s n x dµ 1 is A -measurable for all n 1. Moreover, F n F n+1 for all n 1 by properties of the integral of nonnegative functions. It follows that lim n F n is measurable, which means precisely that x f x dµ 1 is A -measurable. Another application of the MCT gives lim F n x dµ x = lim F n x dµ x, n X X n which rewrites as lim s n x dµ 1 dµ x = f x x 1 dµ 1 x 1 dµ x. n X X By 11, the term in the left-hand side is lim n X s n dµ 1 µ, which converges to the integral X f dµ 1 µ again by the monotone convergence theorem. Thus we obtain 9. The other equality is proved in a similar way. Theorem.3. Fubini s theorem Let X i, A i, µ i be σ-finite measure spaces for i = 1, and let µ 1 µ be the product measure on A 1 A. Suppose that f : X R or to C is integrable. Then the following assertions are true: 1 The function f x is in L 1, A 1, µ 1 for µ -a.e. x and f x1 is in L 1 X, A, µ for µ 1 - a.e. x 1. The function gx 1 = X fx 1, x dµ x is defined µ 1 -a.e. on and hx = fx 1, x dµ 1 x 1 is defined µ -a.e. on X. Further, g L 1, A 1, µ 1 and h L 1 X, A, µ. Moreover, fx 1, x dµ 1 x 1 13 dµ x = f dµ 1 µ X X 14 = fx 1, x dµ x dµ 1 x 1. X

7 Proof. Suppose that f is C-valued. Since f = Ref + iimf, if we prove the theorem for R-valued functions then by linearity of integrals additivity and multiplication with the complex constant i we obtain the theorem for C-valued functions. Assume therefore that f : X R. The assumption that f is integrable is equivalent to knowing that X f ± dµ 1 µ is finite. Then each of the four iterated integrals obtained from applying Tonelli s theorem to f + and f is finite. Recall the fact that h dµ < if and only if h < µ a.e. for h nonnegative. We Ω will use this fact twice one for each integral, for example first over then over X in the iterated finite integrals for f ± given by Tonelli s theorem. Thus from f + x 1, x dµ x dµ 1 x 1 < X it follows that X f + x 1, x dµ x is finite except possibly on a set N 1 A 1 with µ 1 N 1 = 0. By redefining it to be zero on N 1, the function x 1 X f + x 1, x dµ x belongs to L 1, A 1, µ 1. Similarly, the function x 1 X f x 1, x dµ x is finite µ 1 a.e.x 1 and after redefining on a set of measure zero belongs to L 1, A 1, µ 1. Next note that f + x1 x = f + x 1, x = f x1 + x as functions on X. Similarly for f. Thus x 1 X f x1 + dµ < for µ 1 a.e.x 1 and is into L 1, A 1, µ 1, and x 1 X f x1 dµ < for µ 1 a.e.x 1 and is in L 1, A 1, µ 1. We have g = f x1 + dµ f x1 dµ, µ 1 a.e. X X implying that gx 1 = X f x1 dµ is in L 1, A 1, µ 1. The other claim in follows from a similar argument. Note for later use that by [1, Theorem 5.8a and b], 15 gx 1 dµ 1 x 1 = f x1 + dµ dµ 1 f x1 dµ dµ 1. X X Now, the fact that X f x1 ± dµ < for µ 1 a.e.x 1 implies that f x1 L 1 X, A, µ for µ 1 a.e.x 1. This proves one statement in 1, and the other is similar. Finally, to prove for example 14, note that by 15 in the first equality and Tonelli s theorem for f + in the third equality, we get fx 1, x dµ dµ 1 = f x1 + dµ dµ 1 f x1 dµ dµ 1 X X X = f + x1 dµ dµ 1 f x1 dµ dµ 1 X X = f + dµ 1 µ f dµ 1 µ X X = f dµ 1 µ. X Lemma.4. See []. If E j generate σ-algebras A j on X j for j = 1,, then the collection E 1 E of elements of the form C 1 C with C j E j, j = 1,, generates A 1 A. 7

8 8 Proof. Let π 1 : X be the projection onto the first coordinate π 1 x 1, x = x 1. Denote A = σe 1 E. Verify that the set {A 1 π 1 A 1 A} is a σ-algebra which contains E 1. Then this set contains A 1. Thus A 1 X A, and similarly A A. Then A 1 X A = A 1 A X = {A 1 A A j A j, j = 1, } A, and so A 1 A A. The other inclusion follows because E 1 E is contained in the collection C of rectangles that generates A 1 A. Lemma.5. See []. Let X j, A j, µ j, j = 1,, 3 be σ-finite measure spaces. Then A 1 A A 3 = A 1 A A 3 and µ 1 µ µ 3 = µ 1 µ µ 3. Proof. For the first claim, use rectangles A 1 A A 3 = A 1 A A 3 for i = 1,, 3 as generating sets and apply the previous lemma. For the second, use that µ 1 µ µ 3 and µ 1 µ µ 3 coincide on rectangles A 1 A A 3 generating A 1 A A 3, and apply Theorem 1.9. Remark.6. For n 1 we let B n be the Borel algebra on R n, i.e. the σ-algebra generated by the open subsets of R n. Lemma.5 allows us to define B B on R R, where we have n factors. We claim that B n = B B. It suffices to prove the case n =. Now, B is generated by the collection E of open intervals a, b with a, b Q. By Lemma.4, we have B B = BE E. But every open ball in R can be written as a countable union of elements in E E, so BE E = B and the claim follows. We define Lebesgue measure on R n to be λ n = λ λ n factors on B n. This is the unique measure on R n such that λ n a 1, b 1 a n, b n = b 1 a 1 b n a n. It is not a complete measure. However, it admits a completion λ n on the σ-algebra M n of n-dimensional Lebesque measurable subsets of R n, where M n contains B n. It is possible to prove a version of Fubini s theorem for complete measure spaces and the completion of the product measure, as constructed here. For details, see [1, Theorem 6.9]. References [1] J.N. McDonald and N.A. Weiss, A course in Real Analysis, nd edition, Academic Press, Amsterdam, 013. [] G. Teschl, Topics in real and functional analysis, gerald/ftp/bookfa/index.html.