On a countably generated space, $x \mapsto \|Q(x,\cdot)-\nu\|$ is measurable

Let $Q$ be a Markov kernel from $\mathsf X$ to a countably generated space $\mathsf Y$, and let $\nu$ be a probability measure on $\mathsf Y$. Then

is measurable.

Why this is not automatic. The total variation distance is defined as a supremum over all measurable sets — an uncountable index set — and an uncountable supremum of measurable functions need not be measurable. So nothing about the definition makes $x \mapsto \|Q(x,\cdot)-\nu\|$ measurable, even though each individual $x \mapsto Q(x,A)$ is.

Why it is needed. Every quantitative ergodicity statement about Markov chains produces bounds of the form $\|P^n(x,\cdot)-\pi\| \le \dots$, and turning such a bound into a statement about the chain's mixing coefficients requires integrating it over the starting state:

That integral is meaningless without this measurability. It is what makes the passage possible from Harris ergodicity — convergence $\|P^n(x,\cdot)-\pi\| \to 0$ from every starting point, but at no uniform rate — to $\alpha(n) \to 0$: the integrand is dominated by $1$, so dominated convergence applies, whereas a supremum over $x$ would simply fail to converge. This is exactly where the countably-generated hypothesis on the state space earns its place in the classical statement.

Proof. On a countably generated space there is a countable ring $\mathcal C$ of measurable sets containing $\mathsf Y$ and generating the $\sigma$-algebra. Since the total variation is attained on such a ring, for each $x$

— the inequality $\le$ is the approximation theorem, and $\ge$ holds because members of $\mathcal C$ are themselves measurable sets. Enumerating $\mathcal C$ as $\{e_0, e_1, \dots\}$ (possible: it is countable and nonempty) turns the right-hand side into $\sup_{n\in\mathbb N}\bigl|Q(x,e_n)-\nu(e_n)\bigr|$.

Each term is measurable in $x$, being $|\,\cdot\,|$ of the difference of the measurable function $x \mapsto Q(x,e_n)$ and a constant. The family is uniformly bounded by $2$, since both measures are probability measures, so the countable supremum is measurable.

Both suprema are taken in $\mathbb R$ rather than $[0,\infty]$, so the boundedness is not decoration: it is what makes the suprema real numbers at all, and it is used three times — for the defining set of the total variation, for the ring supremum, and for the countable supremum.