Total variation distance for probability laws

Let $\mu$ and $\nu$ be probability measures on the same measurable space $(E,\mathcal{E})$ (Probability measures and probability spaces). Their total variation distance is

$\parallel \mu -\nu {\parallel }_{\mathrm{T}\mathrm{V}}:={\sup }_{A\in \mathcal{E}}\mid \mu (A)-\nu (A)\mid \mathrm{.}$

For every event $A$ the difference $\mu (A)-\nu (A)$ is a real number in $[-1,1]$, because both measures have total mass one; hence the set being maximized is a nonempty subset of $[0,1]$ and the supremum lies in $[0,1]$. The empty event and the whole space give the values $0$ and $\mathrm{\mid }\mu (E)-\nu (E)\mathrm{\mid }=0$, so the distance is zero when $\mu =\nu$, and it is symmetric in $\mu$ and $\nu$. This convention carries no factor $1\mathrm{/}2$ in the supremum; it is therefore not the total variation norm of the signed measure $\mu -\nu$, which is twice the quantity above when that signed measure is considered on a measurable space where the decomposition is attained. On a countable state space with its power-set sigma-algebra the supremum equals the half-${\mathrm{\ell }}^1$ sum $\frac{1}{2}{\sum }_{x\in E}\mathrm{\mid }\mu (\{x\})-\nu (\{x\})\mathrm{\mid }$; that identity is proved in Half-${\mathrm{\ell }}^1$ formula for total variation on a countable space, not assumed here.