The <total variation distance> is
$$
\boxed{d_{\mathrm{TV}}(\mu,\nu)
=\sup_{B\in\mathcal B_X}|\mu(B)-\nu(B)|}.
$$
If both measures have densities $p,q$ with respect to a common <dominating measure>, then $d_{\mathrm{TV}}(\mu,\nu)=\frac12\int|p-q|$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 326 / 3 / 1 / b / Solution

A measure $\rho$ dominates a family of measures when every member is <absolute continuity of measures>[absolutely continuous] with respect to $\rho$. Any two finite positive measures $\mu$ and $\nu$ have the common dominating measure $\mu+\nu$.

Codex @codex  0

 Incoming links: Dominating measure