If $μ$ is a signed measure, $Ω^±$ is a Hahn decomposition and $μ^±$ is the Jordan decomposition, then $\int_{A∩Ω^±}X{\rm d}μ=\pm\int_AX{\rm d}μ^±$

Question

Let

• $(\Omega,\mathcal A,\mu)$ be a finite signed measure space
• $(\Omega^+,\Omega^-)$ be a Hahn decomposition of $\Omega$ with respect to $\mu$ and $(\mu^+,\mu^-)$ denot the Jordan decomposition of $\mu$
• $X:\Omega\to\mathbb R$ be $(\mathcal A,\mathcal B(R))$-measurable with $$\int|X|\:{\rm d}|\mu|<\infty\tag1$$

I want to show that $$\int_{A\:\cap\:\Omega^\pm}X\:{\rm d}\mu=\pm\int_AX\:{\rm d}\mu^\pm\;\;\;\text{for all }A\in\mathcal A\tag2\;.$$

Note that $\Omega^\pm\in\mathcal A$ with $\Omega=\Omega^+\uplus\Omega^-$ and $$\mu^\pm(A)=\pm\mu(A\cap\Omega^\pm)\ge0\;\;\;\text{for all }A\in\mathcal A\tag3\;.$$ So, the statement should be easy to prove, but how do we need to argue?

Answer 1

$\displaystyle\int_{A\cap\Omega^{+}}Xd\mu=\displaystyle\int_{A\cap\Omega^{+}}Xd\mu^{+}-\int_{A\cap\Omega^{+}}Xd\mu^{-}=\int_{A}\chi_{\Omega^{+}}Xd\mu^{+}-\int_{A}\chi_{\Omega^{+}}d\mu^{-}$. We know that $\chi_{\Omega^{+}}d\mu^{+}=d\mu^{+}$ and $\chi_{\Omega^{+}}d\mu^{-}=0$.