1.7. Outer measures, pre-measures, product measures 165 This makes the theory much more subtle, and we will not discuss it in this text. Example 1.7.13. From Exercise 1.2.22, we see that the product md × md of the Lebesgue measures md,md on (Rd, L[Rd]) and (Rd, L[Rd ]), respectively, will agree with Lebesgue measure md+d on the product space L[Rd]×L[Rd ], which as noted in Exercise 1.7.19 is a subalgebra of L[Rd+d ]. After taking the completion md × md of this product measure, one obtains the full Lebesgue measure md+d . Exercise 1.7.20. Let (X, BX ), (Y, BY ) be measurable spaces. (i) Show that the product of two Dirac measures on (X, BX ), (Y, BY ) is a Dirac measure on (X × Y, BX × BY ). (ii) If X, Y are at most countable, show that the product of the two counting measures on (X, BX), (Y, BY ) is the counting measure on (X × Y, BX × BY ). Exercise 1.7.21 (Associativity of product). Let (X, BX,μX), (Y, BY , μY ), (Z, BZ,μZ) be σ-finite sets. We may identify the Cartesian products (X × Y ) × Z and X × (Y × Z) with each other in the obvious manner. If we do so, show that (BX × BY ) × BZ = BX × (BY × BZ ) and (μX × μY ) × μZ = μX × (μY × μZ). Now we integrate using this product measure. We will need the following technical lemma. Define a monotone class in X is a collection B of subsets of X with the following two closure properties: (i) If E1 E2 . . . are a countable increasing sequence of sets in B, then n=1 En B. (ii) If E1 E2 . . . are a countable decreasing sequence of sets in B, then n=1 En B. Lemma 1.7.14 (Monotone class lemma). Let A be a Boolean algebra on X. Then A is the smallest monotone class that contains A. Proof. Let B be the intersection of all the monotone classes that contain A. Since A is clearly one such class, B is a subset of A. Our task is then to show that B contains A. It is also clear that B is a monotone class that contains A. By replacing all the elements of B with their complements, we see that B is necessarily closed under complements. For any E A, consider the set CE of all sets F B such that F \E, E\F , F E, and X\(E F ) all lie in B. It is clear that CE contains A since B is a monotone class, we see that CE is also. By definition of B, we conclude that CE = B for all E A.
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