170 1. Measure theory (iii) We have X×Y f(x, y) dμX × μY (x, y) = X ( Y f(x, y) dμY (y)) dμX(x) = X ( Y f(x, y) dμY (y)) dμX(x). Proof. By taking real and imaginary parts we may assume that f is real by taking positive and negative parts we may assume that f is unsigned. But then the claim follows from Tonelli’s theorem note from (1.37) that X ( Y f(x, y) dμY (y)) dμX(x) is finite, and so Y f(x, y) dμY (y) for μX-almost every x X, and similarly X f(x, y) dμX(x) for μY -almost every y Y . Exercise 1.7.23. Give an example of a Borel measurable function f : [0, 1]2 R such that the integrals [0,1] f(x, y) dy and [0,1] f(x, y) dx exist and are absolutely integrable for all x [0, 1] and y [0, 1], respectively, and that [0,1] ( [0,1] f(x, y) dy) dx and [0,1] ( [0,1] f(x, y) dy) dx exist and are absolutely integrable, but such that [0,1] ( [0,1] f(x, y) dy) dx = [0,1] ( [0,1] f(x, y) dy) dx. are unequal. (Hint: Adapt the example from Remark 0.0.3.) Thus we see that Fubini’s theorem fails when one drops the hypothesis that f is absolutely integrable with respect to the product space. Remark 1.7.22. Despite the failure of Tonelli’s theorem in the σ-finite setting, it is possible to (carefully) extend Fubini’s theorem to the non-σ- finite setting, as the absolute integrability hypotheses, when combined with Markov’s inequality (Exercise 1.4.35(vi)), can provide a substitute for the σ-finite property. However, we will not do so here, and indeed I would recommend proceeding with extreme caution when performing any sort of interchange of integrals or invoking of product measure when one is not in the σ-finite setting. Informally, Fubini’s theorem allows one to always interchange the order of two integrals, as long as the integrand is absolutely integrable in the product space (or its completion). In particular, specialising to Lebesgue measure, we have Rd+d f(x, y) d(x, y) = Rd ( Rd f(x, y) dy) dx = Rd ( Rd f(x, y) dx) dy whenever f : Rd+d C is absolutely integrable. In view of this, we often write dxdy (or dydx) for d(x, y).
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