1.7. Outer measures, pre-measures, product measures 169 (ii) Show that X ( Y f(x, y) d#(y))dm(x) = 1. (iii) Show that Y ( X f(x, y) dm(x))d#(y) = 0. (iv) Show that there is more than one measure μ on L([0, 1])×2[0,1] with the property that μ(E × F ) = m(E)#(F ) for all E ∈ L([0, 1]) and F ∈ 2[0,1]. (Hint: Use the two different ways to perform a double integral to create two different measures.) Remark 1.7.20. If f is not assumed to be measurable in the product space (or its completion), then of course the expression X×Y f(x, y) dμX ×μY (x, y) does not make sense. Furthermore, in this case the remaining two expres- sions in (1.37) may become different as well (in some models of set theory, at least), even when X and Y are finite measure. For instance, let us as- sume the continuum hypothesis, which implies that the unit interval [0, 1] can be placed in one-to-one correspondence with the first uncountable ordi- nal ω1. Let ≺ be the ordering of [0, 1] that is associated to this ordinal, let E := {(x, y) ∈ [0, 1]2 : x ≺ y}, and let f := 1E. Then, for any y ∈ [0, 1], there are at most countably many x such that x ≺ y, and so [0,1] f(x, y) dx exists and is equal to zero for every y. On the other hand, for every x ∈ [0, 1], one has x ≺ y for all but countably many y ∈ [0, 1], and so [0,1] f(x, y) dy exists and is equal to one for every y, and so the last two expressions in (1.37) exist but are unequal. (In particular, Tonelli’s theorem implies that E cannot be a Lebesgue measurable subset of [0, 1]2.) Thus we see that measurability in the product space is an important hypothesis. (There do, however, exist models of set theory (with the axiom of choice) in which such counterexamples cannot be constructed, at least in the case when X and Y are the unit interval with Lebesgue measure.) Tonelli’s theorem is for the unsigned integral, but it leads to an im- portant analogue for the absolutely convergent integral, known as Fubini’s theorem: Theorem 1.7.21 (Fubini’s theorem). Let (X, BX,μX) and (Y, BY , μY ) be complete σ-finite measure spaces, and let f : X × Y → C be absolutely inte- grable with respect to BX × BY . Then: (i) For μX-almost every x ∈ X, the function y → f(x, y) is absolutely integrable with respect to μY , and in particular, Y f(x, y) dμY (y) exists. Furthermore, the (μX -almost everywhere defined) map x → Y f(x, y) dμY (y) is absolutely integrable with respect to μX. (ii) For μY -almost every y ∈ Y , the function x → f(x, y) is absolutely integrable with respect to μX, and in particular, X f(x, y) dμX(x) exists. Furthermore, the (μY -almost everywhere defined) map y → X f(x, y) dμX(x) is absolutely integrable with respect to μY .

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