Chapter 4. Product Measures §1. Product Measures Let (X, S, µ) and (Y, T, ν) be two measure spaces. In order to define the product of µ and ν, we first define an outer measure on X × Y in terms of µ and ν. For each subset E of X × Y , we define ∞ X ∞ τ(E) := inf µ(Aj)ν(Bj): Aj ∈ S, Bj ∈ T ∀ j ∈ IN and E ⊆ ∪j=1Aj × Bj . j=1 Theorem 1.1. The set function τ is an outer measure on X × Y . Proof. Clearly, τ(∅) = 0. Moreover, if E ⊆ F ⊆ X × Y , then τ(E) ≤ τ(F ). It remains ∞ to prove that τ is countably subadditive. Suppose E ⊆ ∪k=1Ek ⊆ X × Y . We wish to P∞ P∞ P∞ show τ(E) ≤ k=1 τ(Ek). This is true if k=1 τ(Ek) = ∞. Suppose k=1 τ(Ek) < ∞. k k Fix ε > 0. For each k ∈ IN, there exist sets Aj ∈ S and Bj ∈ T for j ∈ IN such that ∞ k k Ek ⊆ ∪j=1Aj × Bj and ∞ X ε µ(Ak)ν(Bk) < τ(E ) + . j j k 2k j=1 ∞ ∞ k k Thus, E ⊆ ∪k=1 ∪j=1 Aj × Bj and ∞ ∞ ∞ ∞ X X k k X k X τ(E) ≤ µ(Aj )ν(Bj ) ≤ (τ(Ek) + ε/2 ) = ε + τ(Ek). k=1 j=1 k=1 k=1 P∞ Since ε > 0 could be arbitrary, we conclude that τ(E) ≤ k=1 τ(Ek). Let A denote the σ-algebra of τ-measurable subsets of X × Y and let µ × ν = τ|A. We call µ × ν the product measure of µ and ν. A subset E of X × Y is said to be µ × ν-measurable if and only if it is τ-measurable. Theorem 1.2. If A ∈ S and B ∈ T , then A × B is τ-measurable and (µ × ν)(A × B) = µ(A)ν(B). Proof. In order to show that A × B is τ-measurable, it suffices to prove that for any G ⊆ X × Y with τ(G) < ∞ the following inequality holds: τ(G) ≥ τ(G ∩ (A × B)) + τ(G ∩ (A × B)c). 1 ∞ Fix ε > 0. There exist sets Aj ∈ S and Bj ∈ T for j ∈ IN such that G ⊆ ∪j=1Aj × Bj P∞ and j=1 µ(Aj)ν(Bj) < τ(G) + ε. It follows that ∞ G ∩ (A × B) ⊆ ∪j=1(Aj ∩ A) × (Bj ∩ B) and c ∞ c c G ∩ (A × B) ⊆ ∪j=1 ((Aj ∩ A) × (Bj ∩ B )) ∪ ((Aj ∩ A ) × Bj) . Consequently, τ(G ∩ (A × B)) + τ(G ∩ (A × B)c) ∞ ∞ ∞ X X c X c ≤ µ(Aj ∩ A)ν(Bj ∩ B) + µ(Aj ∩ A)ν(Bj ∩ B ) + µ(Aj ∩ A )ν(Bj) j=1 j=1 j=1 ∞ ∞ ∞ X X c X = µ(Aj ∩ A)ν(Bj) + µ(Aj ∩ A )ν(Bj) = µ(Aj)ν(Bj) < τ(G) + ε. j=1 j=1 j=1 Thus, τ(G ∩ (A × B)) + τ(G ∩ (A × B)c) < τ(G) + ε for all ε > 0. Therefore, τ(G ∩ (A × B)) + τ(G ∩ (A × B)c) ≤ τ(G) ∀ G ⊆ X × Y. This shows that A × B is τ-measurable. We have (µ × ν)(A × B) ≤ µ(A)ν(B). To prove µ(A)ν(B) ≤ (µ × ν)(A × B), for ∞ j ∈ IN we let Aj and Bj be sets in S and T respectively such that A × B ⊆ ∪j=1Aj × Bj. It follows that ∞ X χA(x)χB(y) ≤ χAj (x)χBj (y), x ∈ X, y ∈ Y. j=1 Integrating both sides of the above inequality first with respect to ν and then with respect to µ, we obtain Z Z µ(A)ν(B) = χA(x)χB(y) dν(y) dµ(x) X Y ∞ ∞ X Z Z X ≤ χAj (x)χBj (y) dν(y) dµ(x) = µ(Aj)ν(Bj). j=1 X Y j=1 Therefore, µ(A)ν(B) ≤ (µ × ν)(A × B). ∞ A subset F of X×Y is said to be a σ-set, if F can be represented as ∪j=1Aj ×Bj, where Aj ∈ S and Bj ∈ T for j ∈ IN. By Theorem 1.2, a σ-set is (µ × ν)-measurable. Suppose 2 that E is a (µ × ν)-measurable subset of X × Y . By the definition of the product measure, for any ε > 0 there exists a σ-set F such that E ⊆ F and (µ × ν)(F ) < (µ × ν)(E) + ε. Let E be a subset of X × Y . For x ∈ X, the x-section of E is the subset of Y given by Ex := {y ∈ Y :(x, y) ∈ E}. For y ∈ Y , the y-section of E is the subset of X given by Ey := {x ∈ X :(x, y) ∈ E}. Regarding sections of sets, the following identities hold: y y (a) (∪i∈I Ai)x = ∪i∈I (Ai)x and (∪i∈I Ai) = ∪i∈I (Ai) ; y y (b) (∩i∈I Ai)x = ∩i∈I (Ai)x and (∩i∈I Ai) = ∩i∈I (Ai) ; y y y (c) (A \ B)x = Ax \ Bx and (A \ B) = A \ B . In what follows, (X, S, µ) and (Y, T, ν) are assumed to be complete measure spaces. Theorem 1.3. Let E be a µ × ν-measurable subset of X × Y with (µ × ν)(E) < ∞. Then for µ-almost every x ∈ X the set Ex is ν-measurable and the function x 7→ ν(Ex) is integrable over X, and Z (µ × ν)(E) = ν(Ex) dµ(x). X Similarly, for ν-almost every y ∈ Y the set Ey is µ-measurable and the function y 7→ µ(Ey) is integrable over Y , and Z (µ × ν)(E) = µ(Ey) dν(y). Y Proof. It suffices to establish the first formula. The proof goes by steps. Step 1. Assume E = A × B, where A ∈ S and B ∈ T . Then Ex = B if x ∈ A and Ex = ∅ if x∈ / A. Hence, ν(Ex) = ν(B)χA(x). It follows that Z Z ν(Ex) dµ(x) = ν(B)χA(x) dµ(x) = µ(A)ν(B) = (µ × ν)(E). X X ∞ Step 2. Suppose that E is a σ-set, that is, E = ∪n=1(An × Bn), where An ∈ S and Bn ∈ T for n ∈ IN. Let En := An × Bn. We may assume that En (n ∈ IN) are mutually ∞ P∞ disjoint. Then Ex = ∪n=1(En)x is ν-measurable and ν(Ex) = n=1 ν((En)x). Therefore, ∞ ∞ Z X Z X ν(Ex) dµ(x) = ν((En)x) dµ(x) = (µ × ν)(En) = (µ × ν)(E). X n=1 X n=1 3 ∞ Step 3. Suppose that E = ∩n=1En, where each En is a σ-set, En+1 ⊆ En for n ∈ IN and (µ × ν)(E1) < ∞. By Step 2 we have Z ν((E1)x) dµ(x) = (µ × ν)(E1) < ∞. X Hence, ν((E1)x) < ∞ for µ-almost every x ∈ X. Moreover, (En+1)x ⊆ (En)x for n ∈ IN. ∞ Thus, Ex = ∩n=1(En)x is ν-measurable and Z Z ν(Ex) dµ(x) = lim ν((En)x) dµ(x) = lim (µ × ν)(En) = (µ × ν)(E). X n→∞ X n→∞ Step 4. Suppose (µ × ν)(E) = 0. There exists a sequence (En)n=1,2,... of σ-sets such ∞ that E ⊆ ∩n=1En, En+1 ⊆ En for n ∈ IN, (µ × ν)(E1) < ∞, and limn→∞(µ × ν)(En) = 0. ∞ ∞ Let F := ∩n=1En. Then Fx = ∩n=1(En)x is ν-measurable for each x ∈ X and Z Z ν(Fx) dµ(x) = lim ν((En)x) dµ(x) = lim (µ × ν)(En) = 0. X n→∞ X n→∞ Hence, ν(Fx) = 0 for µ-almost every x ∈ X. But Ex ⊆ Fx and (X, S, µ) is complete. Therefore, for µ-almost every x ∈ X, Ex is ν-measurable and ν(Ex) = 0. Consequently, Z ν(Ex) dµ(x) = 0 = (µ × ν)(E). X Step 5. The general case. Since (µ×ν)(E) < ∞, There exists a sequence (En)n=1,2,... ∞ of σ-sets such that E ⊆ F := ∩n=1En,(µ × ν)(F \ E) = 0, En+1 ⊆ En for n ∈ IN, and (µ × ν)(E1) < ∞. By Step 3 and Step 4, for µ-almost every x ∈ X, Fx and (F \ E)x are ν-measurable, so Ex = Fx \ (F \ E)x is ν-measurable. Moreover, Z Z (µ × ν)(F ) = ν(Fx) dµ(x) and (µ × ν)(F \ E) = ν(Fx \ Ex) dµ(x). X X Subtracting the second equality from the first one, we obtain Z (µ × ν)(E) = ν(Ex) dµ(x). X The proof is complete. 4 §2. The Fubini-Tonelli Theorem Throughout this section, (X, S, µ) and (Y, T, ν) are assumed to be two complete mea- sure spaces. Let f be a function from X × Y to IR. For each fixed x ∈ X, we define fx(y) := f(x, y), y ∈ Y, and for each fixed y ∈ Y define f y(x) := f(x, y), x ∈ X. Theorem 2.1. Let f : X × Y → IR be a (µ × ν)-integrable function. Then fx is ν- integrable for µ-almost every x ∈ X, f y is µ-integrable for ν-almost every y ∈ Y , the R R y function g : x 7→ Y fx dν is µ-integrable, the function h : y 7→ X f dµ is ν-integrable, and Z Z Z Z Z f d(µ × ν) = f(x, y) dν(y) dµ(x) = f(x, y) dµ(x) dν(y). X×Y X Y Y X Proof. The proof goes by steps. Step 1. If f = χE, where E is (µ × ν)-measurable and (µ × ν)(E) < ∞, then our assertions are true, by Theorem 1.3.