Total variation controls integrals of any bounded function, with constant $2M$

Let $\mu,\nu$ be probability measures on $\mathsf X$ and let $g$ be measurable with $|g| \le M$ everywhere. Then

The general test-function form. The total variation distance is defined by testing against sets; the companion result for $[0,1]$-valued functions extends that to indicators' convex hull. This statement removes the range restriction entirely: any bounded measurable $g$ works, at the price of the factor $2M$. The constant is sharp — for $g = M\cdot(\mathbf 1_A - \mathbf 1_{A^c})$ the two sides agree when $A$ is a Hahn set.

Why the general form is the one needed for limit theorems. Convergence in distribution is characterized by convergence of $\int g\,\mathrm{d}\mu_n$ for bounded continuous $g$ — functions with no reason to be indicators or to take values in $[0,1]$. So transferring a weak limit from one sequence of measures to another that is close in total variation requires exactly this inequality: if $\|\mu_n - \nu_n\| \to 0$ then $\int g\,\mathrm{d}\mu_n$ and $\int g\,\mathrm{d}\nu_n$ have the same limit for every bounded continuous $g$, hence $\mu_n$ and $\nu_n$ have the same weak limit.

For Markov chains this is the last analytic ingredient in the passage from the stationary central limit theorem to the statement for an arbitrary initial distribution. A uniformly ergodic chain satisfies $\|\mathbb P_{\lambda P^m} - \mathbb P_\pi\| \le Rt^m$ on path space, and the test function is $g\bigl(\sqrt n(\bar f_n - \mathbb E_\pi f)\bigr)$ for a bounded continuous $g$ — bounded, but taking both signs and values outside $[0,1]$.

Proof. If $M = 0$ then $g \equiv 0$ and both sides vanish. Otherwise rescale: $h = (g+M)/(2M)$ is measurable with values in $[0,1]$, so the $[0,1]$-valued bound applies to $h$. Since $\mu$ and $\nu$ are probability measures, $\int h\,\mathrm{d}\rho = \bigl(\int g\,\mathrm{d}\rho + M\bigr)/(2M)$ for $\rho \in \{\mu,\nu\}$, and the additive constants cancel in the difference:

Multiplying the $[0,1]$ bound by $2M$ gives the claim. That $\mu$ and $\nu$ are probability measures is used exactly here — with different total masses the constants would not cancel.