164 1. Measure theory In a similar spirit, by breaking each Sn up into component product sets and using finite additivity again, we may assume without loss of generality that each Sn takes the form Sn = En × Fn for some En BX and Fn BY . By definition of μ0, our objective is now to show that μX(E)μY (F ) = n=1 μX (En)μY (Fn). To do this, first observe from construction that we have the pointwise iden- tity 1E(x)1F (y) = n=1 1En (x)1Fn (y) for all x X and y Y . We fix x X, and integrate this identity in y (noting that both sides are measurable and unsigned) to conclude that Y 1E(x)1F (y) dμY (y) = Y n=1 1En (x)1Fn (y) dμY (y). The left-hand side simplifies to 1E(x)μY (F ). To compute the right-hand side, we use the monotone convergence theorem (Theorem 1.4.43) to inter- change the summation and integration, and soon see that the right-hand side is ∑∞ n=1 1En (x)μY (Fn), thus 1E(x)μY (F ) = n=1 1En (x)μY (Fn) for all x. Both sides are measurable and unsigned in x, so we may integrate in X and conclude that X 1E(x)μY (F ) dμX = X n=1 1En (x)μY (Fn) dμX(x). The left-hand side here is μX(E)μY (F ). Using monotone convergence as before, the right-hand side simplifies to n=1 μX(En)μY (Fn), and the claim follows. Now that we have established that μ0 is a pre-measure, we may apply Theorem 1.7.8 to extend this measure to a countably additive measure μX × μY on a σ-algebra containing B0. By Exercise 1.7.18(ii), μX × μY is a countably additive measure on BX × BY , and as it extends μ0, it will obey (1.36). Finally, to show uniqueness, observe from finite additivity that any measure μX × μY on BX × BY that obeys (1.36) must extend μ0, and so uniqueness follows from Exercise 1.7.7. Remark 1.7.12. When X, Y are not both σ-finite, then one can still con- struct at least one product measure, but it will, in general, not be unique.
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