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Radon-Nikodym定理“推广”(Ex 2.12.2)

Foundations of Modern Analysis, Avner Friedman)Problem 2.12.2. The Radon-Nikodym theorem remains ture in case μ is a σ-finite signed measure.

Radon-Nikodym theorem is about saying that (X,a,μ) is a σ-finite measure, μ is a measure, ν is a signed measure and νμ. Then there exists a measurable function f such that ν(E)=Efdμ. Problem 2.12.2 extended the condition “μ is a measure” to that “μ is a signed measure”.

My idea is that, consider the Jordan decomposition μ+ and μ (where μ=μ+μ) over the Hahn decomposition (A,B) of X. Thus μ+ and μ are both measures, we then can apply Radon-Nikodym theorem. Consider the positive set A, easily, we get νμ+, hence there exists a measurable function f1 such that v(E)=Ef1dμ+,EA

Similarily, there exists a measurable function f2 on B such that v(E)=Ef2dμ,EB. (I am not sure if it's true)

Then for all Ea, ν(E)=ν(EA)+ν(EB)=EAf1dμ+EBf2dμ=EχAf1dμ+EχBf2dμ(a)=E(χAf1+χBf2)dμ+E(χAf1+χBf2)dμ(b)=E(χAf1+χBf2)dμΔ=Efdμ

where equality (a) is because μ+(μ) to B(A) is 0; equality (b) is by definition that Efdμ=Efdμ+Efdμ.

posted @   Jun-Hui Li  阅读(602)  评论(0编辑  收藏  举报
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