# Total variation distance for probability measures

I'm having troubles solving this problem, any help will be appreciate :)

Let $(\Omega,\mathcal{F})$ a measurable space, and let $\mu, \ \nu$ probability measures on $\mathcal{F}$. It's well know that $d(\mu,\nu)=2\sup_{A\in F}\{|\mu(A)-\nu(A)|\}$ is a distance on the space of probability measures. Let $\nu=\nu_a + \nu_s$ (the Lebesgue decomposition of $\nu$ w/r $\mu$, i.e. $\nu_a\ll \mu$ and $\nu_s\perp\mu$).

Prove that $d(\mu,\nu)=2\int_{\Omega} \left(1-h\right)_+ d\mu$, where $h=\frac{d\nu_a}{d\mu}$

Thanks

EDIT: my attempt

Let $C\in \mathcal{F}$ be the set where $\mu(\cdot)=\mu(\cdot \cap C)$, and let $B=\{h\leq1\}\in\mathcal{F}$ then

$2\int_{\Omega} \left(1-h\right)_+ d\mu = 2\int_C \left(1-h\right)_+ d\mu= 2\int_{C\cap B} \left(1-h\right) d\mu=2|\int_{C\cap B} \left(1-h\right)d\mu|= \\ 2|\mu(C\cap B)-\nu_a(C\cap B)|=2|\mu(C\cap B)-\nu_a(C\cap B)- \underbrace{\nu_s(C\cap B)}_{=0}|=\\ 2|\mu(C\cap B)-\nu(C\cap B)|\leq2\sup_{A\in F}\{|\mu(A)-\nu(A)|\}=d(\mu,\nu)$

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