168 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). (1.37) Proof. From Exercise 1.4.28, every measurable set in BX × BY is equal to a measurable set in BX × BY outside of a μX × μY -null set. This implies that the BX × BY -measurable function f agrees with a BX ×BY -measurable function ˜ f outside of a μX × μY -null set E (as can be seen by expressing f as the limit of simple functions). From Corollary 1.7.17, we see that for μX-almost every x X, the function y f(x, y) agrees with y ˜(x, f y) outside of a μY -null set (and is, in particular, measurable, as (Y, BY , μY ) is complete) and similarly for μY -almost every y Y , the function x f(x, y) agrees with x ˜(x, f y) outside of a μX-null set and is measurable, and the claim follows. Specialising to the case when f is an indicator function f = 1E, we conclude Corollary 1.7.19 (Tonelli’s theorem for sets). Let (X, BX , μX) and (Y, BY , μY ) be complete σ-finite measure spaces, and let E BX × BY . Then: (i) For μX-almost every x X, the set Ex := {y Y : (x, y) E} lies in BY , and the (μX -almost everywhere defined) map x μY (Ex) is BX -measurable. (ii) For μY -almost every y Y , the set Ey := {x X : (x, y) E} lies in BX , and the (μY -almost everywhere defined) map y μX(Ey) is BY -measurable. (iii) We have μX × μY (E) = X μY (Ex) dμX(x) (1.38) = X μX(Ey) dμX(x). Exercise 1.7.22. The goal of this exercise is to demonstrate that Tonelli’s theorem can fail if the σ-finite hypothesis is removed, and also that product measure need not be unique. Let X is the unit interval [0, 1] with Lebesgue measure m (and the Lebesgue σ-algebra L([0, 1])) and Y is the unit interval [0, 1] with counting measure (and the discrete σ-algebra 2[0,1]) #. Let f := 1E be the indicator function of the diagonal E := {(x, x) : x [0, 1]}. (i) Show that f is measurable in the product σ-algebra.
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