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with notations as previous question. Consider any two $X$ valued random variable $\eta_1$ and $\eta_2$ such that $\eta_1\sim \mu$ and $\eta_2\sim \nu$. Let $X=A^+\cup A^-$ be the Hahn decomposition of $\mu-\nu$. then

$\|\mu-\nu\|_{TV}=2(\mu(A^+)- \nu(A^+))$ (I don't understand how, could any one explain me a bit)

and $\|\mu-\nu\|_{TV}=2(\mu(A^+)- \nu(A^+))= 2\mathbb E(I_{\eta_1\in A^+}- I_{\eta_2\in A^+})= $, also quet not clear how.

Thanks!

I know that Hahn decomposition says for any signed measure $m$, there exists unique sets $P$ and $N$ such that $X=P\cup N$ with the fact that $m(A)\ge 0\forall A\subseteq P$ and $m(A)\le 0\forall A\subseteq N$.

Myshkin
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  • Something seems wrong in the first equality you say you don't understand. If $\mu = \nu \neq 0$ then any decomposition of $X$ is a Hahn-decomposition for $\mu - \nu = 0$ so we can suppose $A^+ = X, A^- = \emptyset$. Then $|\mu - \nu|_{TV} = 0$ but $\mu(A^+) - \nu(A^-) = \mu(X) > 0$. – Rhys Steele Sep 10 '19 at 12:19
  • Oops, i have edited it now. – Myshkin Sep 10 '19 at 12:25
  • This still isn't right. Consider the case $\mu = 0$, $\nu \neq 0$. Then $A^+ = \emptyset$ but $|\mu - \nu|{TV} \neq 0$. Do you mean to write $|\mu - \nu|{TV} = 2(\lambda(A^+) - \lambda(A^-))$ where $\lambda = \mu - \nu$? If so, you might be interested in my answer here as related material; it proves that result by an indirect route but also contains the ideas you'd need to make a direct argument. – Rhys Steele Sep 10 '19 at 20:47
  • Thanks! I will look into it. – Myshkin Sep 11 '19 at 10:38

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