1.7. Outer measures, pre-measures, product measures 163 each of which has finite measure. On the other hand, Rd with counting measure is not σ-finite (why?). But most measure spaces that one actually encounters in analysis (including, clearly, all probability spaces) are σ-finite. It is possible to partially extend the theory of product spaces to the non-σ- finite setting, but there are a number of very delicate technical issues that arise and so we will not discuss such extensions here. As long as we restrict attention to the σ-finite case, product measure always exists and is unique: Proposition 1.7.11 (Existence and uniqueness of product measure). Let (X, BX,μX) and (Y, BY , μY ) be σ-finite measure spaces. Then there exists a unique measure μX × μY on BX × BY that obeys μX × μY (E × F ) = μX(E)μY (F ) whenever E ∈ BX and F ∈ BY . Proof. We first show existence. Inspired by the fact that Lebesgue measure is the Hahn-Kolmogorov completion of elementary (pre-)measure, we shall first construct an “elementary product pre-measure” that we will then apply Theorem 1.7.8 to. Let B0 be the collection of all finite unions S := (E1 × F1) ∪ . . . ∪ (Ek × Fk) of Cartesian products of BX -measurable sets E1,...,Ek and BY -measurable sets F1,...,Fk. (One can think of such sets as being somewhat analogous to elementary sets in Euclidean space, although the analogy is not perfectly exact.) It is not diﬃcult to verify that this is a Boolean algebra (though it is not, in general, a σ-algebra). Also, any set in B0 can be easily decomposed into a disjoint union of product sets E1 × F1,...,Ek × Fk of BX -measurable sets and BY -measurable sets (cf. Exercise 1.1.2). We then define the quan- tity μ0(S) associated such a disjoint union S by the formula μ0(S) := k j=1 μX(Ej)μY (Fj) whenever S is the disjoint union of products E1 × F1,...,Ek × Fk of BX - measurable sets and BY -measurable sets. One can show that this definition does not depend on exactly how S is decomposed, and gives a finitely addi- tive measure μ0 : B0 → [0, +∞] (cf. Exercise 1.1.2). Now we show that μ0 is a pre-measure. It suﬃces to show that if S ∈ B0 is the countable disjoint union of sets S1,S2,... ∈ B0, then μ0(S) = ∑ ∞ n=1 μ(Sn). Splitting S up into disjoint product sets, and restricting the Sn to each of these product sets in turn, we may assume without loss of generality (using the finite additivity of μ0) that S = E × F for some E ∈ BX and F ∈ BY .

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