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 ν≪μ. 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 ν≪μ+, hence there exists a measurable function f1 such that v(E)=∫Ef1dμ+,∀E⊆A
Similarily, there exists a measurable function f2 on B such that −v(E)=∫Ef2dμ−,∀E⊆B. (I am not sure if it's true)
Then for all E∈a, ν(E)=ν(E∩A)+ν(E∩B)=∫E∩Af1dμ+−∫E∩Bf2dμ−=∫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−.
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