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(X, S, Μ) and (Y,T,Ν) Be Two Measure Spaces. in Order to Define the Prod 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.
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