154 1. Measure theory B0 = 2N to be the discrete algebra, and define μ0 separately for finite and infinite sets.) Theorem 1.7.8 (Hahn-Kolmogorov theorem). Every pre-measure μ0 : B0 [0, +∞] on a Boolean algebra B0 in X can be extended to a countably additive measure μ: B [0, +∞]. Proof. We mimic the construction of Lebesgue measure from elementary measure. Namely, for any set E X, define the outer measure μ∗(E) of E to be the quantity μ∗(E) := inf{ n=1 μ0(En): E n=1 En En B0 for all n}. It is easy to verify (cf. Exercise 1.2.3) that μ∗ is indeed an outer measure. Let B be the collection of all sets E X that are Carath´ eodory measurable with respect to μ∗, and let μ be the restriction of μ∗ to B. By the Carath´eodory extension theorem, B is a σ-algebra and μ is a countably additive measure. It remains to show that B contains B0 and that μ extends μ0. Thus, let E B0 we need to show that E is Carath´ eodory measurable with respect to μ∗ and that μ∗(E) = μ0(E). To prove the first claim, let A X be arbitrary. We need to show that μ∗(A) = μ∗(A E) + μ∗(A\E) by subadditivity, it suffices to show that μ∗(A) μ∗(A E) + μ∗(A\E). We may assume that μ∗(A) is finite, since the claim is trivial otherwise. Fix ε 0. By definition of μ∗, one can find E1,E2,... B0 covering A such that n=1 μ0(En) μ∗(A) + ε. The sets En E lie in B0 and cover A E and thus μ∗(A E) n=1 μ0(En E). Similarly, we have μ∗(A\E) n=1 μ0(En\E). Meanwhile, from finite additivity we have μ0(En E) + μ0(En\E) = μ0(En).
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