Total variation bounds differences of integrals of $[0,1]$-valued functions

Let $\mu, \nu$ be finite measures on a measurable space $\mathsf{X}$ and let $f : \mathsf{X} \to \mathbb{R}$ be measurable with $0 \le f \le 1$ everywhere. Then

What it upgrades. The total variation distance is defined as a supremum over measurable sets — equivalently, over indicator functions. This theorem says the same bound holds for every measurable function with values in $[0,1]$. The set version is the special case $f = \mathbf{1}_A$; the function version is what is actually needed, and it does not follow formally from the definition.

Why it is indispensable for Markov chains. Every quantitative mixing estimate has the same shape: one has a rate $\|P^n(x,\cdot) - \pi\| \le C$ and wants to conclude that some functional of the chain started at $x$ differs from the same functional under stationarity by at most $C$. The functional in question is almost never an indicator of a set in $\mathsf{X}$ — it is $x \mapsto K(x, B)$ for a kernel $K$ and a set $B$ in a different space, typically path space. Concretely, for a Markov kernel $K$ and measurable $B$,

since $x \mapsto K(x,B)$ takes values in $[0,1]$. That is the data-processing inequality: applying a Markov kernel cannot increase total variation distance. It is the step that transports a rate of convergence on the state space to a bound on path-space events, and hence the step that turns an ergodicity hypothesis into a mixing-coefficient bound.

Proof. The definition of $\|\mu-\nu\|$ as a supremum over sets gives no direct handle on integrals; the bridge is the unsigned Hahn decomposition. Choose a measurable $s$ with $\nu \le \mu$ on subsets of $s$ and $\mu \le \nu$ on subsets of $s^{c}$; equivalently, as measures, $\nu|_s \le \mu|_s$ and $\mu|_{s^c} \le \nu|_{s^c}$. Split each integral over $s$ and $s^c$. On $s^c$ the difference $\int_{s^c} f\,\mathrm{d}\mu - \int_{s^c} f\,\mathrm{d}\nu$ is $\le 0$ because $f \ge 0$ and $\mu \le \nu$ there. On $s$, applying the same monotonicity to $1 - f \ge 0$ gives

Adding the two gives $\int f\,\mathrm{d}\mu - \int f\,\mathrm{d}\nu \le \mu(s) - \nu(s) \le \|\mu-\nu\|$. The reverse inequality is the same argument with the roles of $\mu$ and $\nu$ exchanged and $s$ replaced by $s^c$. Note the two applications of "$1-f$ rather than $f$" are what make the bound $\mu(s)-\nu(s)$ appear rather than the useless $\int_s f \,\mathrm{d}\mu$.