# Simple equality of total variation distance

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? Commented Mar 8, 2022 at 14:58
• Not quite. See my answer Commented Mar 8, 2022 at 23:51

## 1 Answer

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}