1.7. Outer measures, pre-measures, product measures 167 additivity, we conclude that C also contains as an element a disjoint finite union S = E1 ×F1 ∪... ∪Ek ×Fk of such products. This implies that C also contains the Boolean algebra B0 in the proof of Proposition 1.7.11, as such sets can always be expressed as the disjoint finite union of Cartesian prod- ucts of measurable sets. Applying the monotone class lemma, we conclude that C contains B0 = BX × BY , and the claim follows. Remark 1.7.16. Note that Tonelli’s theorem for sums (Theorem 0.0.2) is a special case of the above result when μX,μY are counting measure. In a similar spirit, Corollary 1.4.45 is the special case when just one of μX , μY is counting measure. Corollary 1.7.17. Let (X, BX , μX) and (Y, BY , μY ) be σ-finite measure spaces, and let E BX ×BY be a null set with respect to μX ×μY . Then for μX-almost every x X, the set Ex := {y Y : (x, y) E} is a μY -null set and similarly, for μY -almost every y Y , the set Ey := {x X : (x, y) E} is a μX-null set. Proof. Applying the Tonelli theorem to the indicator function 1E, we con- clude that 0 = X ( Y 1E(x, y) dμY (y)) dμX(x) = Y ( X 1E(x, y) dμX(x)) dμY (y) and thus 0 = X μY (Ex) dμX(x) = Y μX(Ey) dμY (y), and the claim follows. With this corollary, we can extend Tonelli’s theorem to the completion (X ×Y, BX × BY , μX × μY ) of the product space (X ×Y, BX ×BY , μX ×μY ), as constructed in Exercise 1.4.26. But we can easily extend the Tonelli theorem to this context: Theorem 1.7.18 (Tonelli’s theorem, complete version). Let (X, BX,μX) and (Y, BY , μY ) be complete σ-finite measure spaces, and let f : X × Y [0, +∞] be measurable with respect to BX × BY . Then: (i) For μX-almost every x X, the function y f(x, y) is BY - measurable, and in particular, Y f(x, y) dμY (y) exists. Further- more, the (μX -almost everywhere defined) map x Y f(x, y) dμY is BX -measurable. (ii) For μY -almost every y Y , the function x f(x, y) is BX - measurable, and in particular, X f(x, y) dμX(x) exists. Further- more, the (μY -almost everywhere defined) map y X f(x, y) dμX is BY -measurable.
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