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M ATH 531 R EAL A NALYSIS I FALL 2009 (Some) Solutions to Homework # 2 Definition: Let (X, M, µ) be a measure space. If for each E ∈ M with µ(E) = ∞ there exists F ∈ M with F ⊆ E and 0 < µ(F ) < ∞, µ is called semifinite. Folland, p.27, Exercise 14: If µ is a semifinite measure and µ(E) = ∞, for any C > 0 there exists F ⊆ E with C < µ(F ) < ∞. Solution. Set α := sup {µ(F ) : F ∈ I(E)} where I(E) = {F ∈ M : F ⊆ E, µ(F ) < ∞} . Since µ is semifinite, α > 0. Let us see that α = ∞. For otherwise, for each n ∈ N, there exists Fn ∈ I(E) such that µ(Fn ) ≥ α − 1/n. Since µ is subadditive, for all n ∈ N, ∪n j=1 Fn ∈ I(E) and thus ´ ` n α ≥ µ ∪j=1 Fj ≥ µ(Fn ) ≥ α − 1/n. Accordingly, for F = ∪n∈N Fn ⊂ E, the continuity of µ from below implies that ´ ` µ(F ) = lim µ ∪n j=1 Fj = α n→∞ and F ∈ I(E). But then µ(E\F ) = ∞, and since µ is semifinite, there exists F ′ ⊆ E\F with 0 < µ(F ′ ) < ∞. This provides a contradiction since F ∪ F ′ ∈ I(E) and µ(F ∪ F ′ ) = α + µ(F ′ ) > α. Folland, p.27, Exercise 15: Given a measure µ on (X, M), define µ0 on M by µ0 (E) = sup {µ(F ) : F ∈ I(E)} where I(E) = {F ∈ M : F ⊆ E, µ(F ) < ∞} . a. µ0 is a semifinite measure. It is called the semifinite part of µ. Solution. Clearly µ0 (∅) = 0. Let us now see that µ0 is countably-additive. To this end, let {En }n∈N ⊆ M be a sequence of disjoint sets. Set E = ∪n∈N En . Note that F ∈ I(E) ⇒ ⇒ ⇒ ⇒ F ⊆ E & µ(F ) < ∞ F ∩ En ⊆ E ∩ En = En F ∩ En ∈ I(En ) µ(F ∩ En ) ≤ µ0 (En ) & µ(F ∩ En ) ≤ µ(F ) < ∞ so that, since {F ∩ En }n∈N is a sequence of disjoint sets, we get X X µ0 (En ); µ(F ∩ En ) ≤ µ(F ) = µ(F ∩ E) = µ (∪n∈N F ∩ En ) = n∈N n∈N this shows that µ0 (E) ≤ X µ0 (En ). n∈N The reverse inequality clearly holds when µ0 (E) = ∞. We may thus assume that µ0 (E) < ∞. Since I(En ) ⊆ I(E), we then have that µ0 (E) ≤ µ0 (E) < ∞. Now given n ∈ N and ε > 0, choose −j Fj ∈ I(EP < µ(Fj )[≤ µ0 (Ej ) < ∞]. Then F = ∪n j ) (j = 1,. . . ,n) such that µ0 (Ej ) − ε2 j=1 Fj ⊆ E and Pn µ(F ) = n µ(F ) ≤ µ (E ) < ∞ so that F ∈ I(E). Accordingly, j 0 j j=1 j=1 µ0 (E) ≥ µ(F ) = n X j=1 µ(Fj ) > n “ X j=1 n ” X µ0 (Ej ) − ε2−j > µ0 (Ej ) − ε. j=1 Letting first ε → 0, and then n → ∞, we thus get µ0 (E) ≥ X µ0 (En ). n∈N Therefore, µ0 is a measure on (X, M). As is clear from its definition, µ0 is semifinite. b. If µ is semifinite, then µ = µ0 . (Use Exercise 14.) Solution. Since µ is monotone, we have that µ0 (E) = µ(E) when µ(E) < ∞. On the other hand, if µ(E) = ∞, then Exercise 14 implies that µ0 (E) = ∞. c. There is a measure ν on M (in general, not unique) which assumes only the values 0 and ∞ such that µ = µ0 + ν. 1 M ATH 531 R EAL A NALYSIS I FALL 2009 Solution. Let us agree to say that a set E ∈ M is semifinite for µ if for all F ⊆ E with µ(F ) = ∞, there exists F ′ ⊆ F such that 0 < µ(F ′ ) < ∞. Define ν : M → [0, ∞] by setting 0, if E is semifinite for µ, ν(E) = ∞, otherwise. Note that any E ∈ M with µ(E) < ∞ is semifinite for µ. Thus in particular ν(∅) = 0. Let us see that ν is countably additive. To this end, let {En }n∈N ⊆ M be a sequence of disjoint sets, and let E = ∪n∈N En . We have two cases: (i) For some n ∈ N, En fails to be semifinite for µ: Clearly, in this case, E cannot be semifinite for µ. Accordingly, we have X ν(En ). ν(E) = ∞ = n∈N (ii) Each En is semifinite for µ: In this case, we have to show that E too is semifinite so that the identity X ν(En ) ν(E) = 0 = n∈N holds. Indeed, if F ⊆ E is any set with µ(F ) = ∞, then the identity X ∞ = µ(F ) = µ(F ∩ E) = µ(F ∩ En ) n∈N shows that µ(F ∩ En0 ) > 0 for at least one n0 ∈ N. If µ(F ∩ En0 ) < ∞, we set F ′ = F ∩ En0 . Otherwise, F ∩ En0 ⊆ En0 and µ(F ∩ En0 ) = ∞ so that, since En0 is semifinite, there exists F ′ ⊆ F ∩ En0 with 0 < µ(F ′ ) < ∞. Therefore E is semifinite for µ as claimed. We have thus shown that ν is a measure on (X, M). Let us now see that µ = µ0 + ν. For let E ∈ M. If µ(E) < ∞, then µ0 (E) = µ(E), and E is semifinite for µ so that ν(E) = 0. Thus, in this case, µ(E) = µ0 (E) + ν(E). If µ(E) = ∞ and E is semifinite for µ, then Exercise 14 shows that µ0 (E) = ∞, and this entails µ(E) = µ0 (E)+ν(E). If µ(E) = ∞ and E is not semifinite for µ, then the identity µ(E) = µ0 (E) + ν(E) holds trivially. Exercise: Show that ν is not the unique such measure. 2