Simple equality of total variation distance

Question

I am trying to understand the following equality involving probability measures in Wikipedia:

$$\|\mu-\nu\|=|\mu-\nu|(X)=2 \sup \{|\mu(A)-\nu(A)|: A \in \Sigma\}$$

where the total variation norm $\|\cdot\|$ and the total variation of a measure $|\cdot|$ are defined in the article. 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.

(1) Proposition 4.2 in Markov Chains and Mixing Times, 2017

you can adapt it by using integrals, no? – kodlu Mar 08 '22 at 14:58 — kodlu, Mar 08 '22 at 14:58
Not quite. See my answer – Martin Geller Mar 08 '22 at 23:51 — Martin Geller, Mar 08 '22 at 23:51

score 1 · Answer 1 · answered Mar 08 '22 at 18:10

First we need to show $\mu-\nu(\cdot)$ 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 ($D^+\sqcup D^-$) of $X$ under $\mu-\nu$ s.t. $\mu-\nu$ is positive on $D^+$. It is then easy to prove that for all $A \in \Sigma$, $\mu(A)-\nu(A) \le \mu(B)-\nu(B)$ and $\nu(B)-\mu(A)$ $\le \nu(B^c)-\mu(B^c) = \mu(B)-\nu(B) $. Then you go on to say that

$$\sup_{A\in\Sigma} |\mu(A)-\nu(A)| \le \mu(B)-\nu(B)$$

and, since $B\in\Sigma$,

$$\sup_{A\in\Sigma} |\mu(A)-\nu(A)| \ge \mu(B)-\nu(B).$$

From this, it is possible to conclude

$$\begin{aligned}\sup_{A\in\Sigma} |\mu(A)-\nu(A)| &= \mu(B)-\nu(B) \\ &=\frac{1}{2}(\mu(B)-\nu(B) + \nu(B^c)-\mu(B^c) ) \\ &=\frac{1}{2}(\mu(A\cup D^+)-\nu(A\cup D^+) + \nu(A\cup D^-)-\mu(A\cup D^-) ) \\ &=\frac{1}{2}( (\mu-\nu)(A\cup D^+) + (\mu-\nu)(A\cup D^-) ) \\ &=|\mu-\nu|(A).\end{aligned}$$

Simple equality of total variation distance

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