rzfansubs87

2022-07-10

I am trying to understand the following equality involving probability measures in Wikipedia:
$‖\mu -\nu ‖=|\mu -\nu |\left(X\right)=2sup\left\{|\mu \left(A\right)-\nu \left(A\right)|:A\in \mathrm{\Sigma }\right\}$
where the total variation norm $‖\cdot ‖$ and the total variation of a measure $|\cdot |$ are defined in the some articles. The article has been flagged for not citing sources or offering a proof.
I have found a proof of a similar result for discrete probability distributions (1). But I have not been able to adapt it to continuous probability distributions, which is the result above.

Valeria Wolfe

Expert

First we need to show $\mu -\nu \left(\cdot \right)$ is a (finite) signed measure. I am not sure if that is trivial or requires some clever trick. Then, we let $B$ be the measurable set $E\cap {D}^{+}$ using the Hahn decomposition (${D}^{+}\bigsqcup {D}^{-}$) of $X$ under $\mu -\nu$ s.t. $\mu -\nu$ is positive on ${D}^{+}$. It is then easy to prove that for all $A\in \mathrm{\Sigma }$, $\mu \left(A\right)-\nu \left(A\right)\le \mu \left(B\right)-\nu \left(B\right)$ and $\le \nu \left({B}^{c}\right)-\mu \left({B}^{c}\right)=\mu \left(B\right)-\nu \left(B\right)$. Then you go on to say that
$\underset{A\in \mathrm{\Sigma }}{sup}|\mu \left(A\right)-\nu \left(A\right)|\le \mu \left(B\right)-\nu \left(B\right)$
and, since $B\in \mathrm{\Sigma }$,
$\underset{A\in \mathrm{\Sigma }}{sup}|\mu \left(A\right)-\nu \left(A\right)|\ge \mu \left(B\right)-\nu \left(B\right).$
From this, it is possible to conclude
$\begin{array}{rl}\underset{A\in \mathrm{\Sigma }}{sup}|\mu \left(A\right)-\nu \left(A\right)|& =\mu \left(B\right)-\nu \left(B\right)\\ & =\frac{1}{2}\left(\mu \left(B\right)-\nu \left(B\right)+\nu \left({B}^{c}\right)-\mu \left({B}^{c}\right)\right)\\ & =\frac{1}{2}\left(\mu \left(A\cup {D}^{+}\right)-\nu \left(A\cup {D}^{+}\right)+\nu \left(A\cup {D}^{-}\right)-\mu \left(A\cup {D}^{-}\right)\right)\\ & =\frac{1}{2}\left(\left(\mu -\nu \right)\left(A\cup {D}^{+}\right)+\left(\mu -\nu \right)\left(A\cup {D}^{-}\right)\right)\\ & =|\mu -\nu |\left(A\right).\end{array}$

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