Uniform ergodicity solves the Poisson equation $\pi$-a.e. in $L^2$

Let $X$ be a uniformly ergodic Markov chain with invariant distribution $\pi$, so that $\|P^n(x,\cdot) - \pi\| \le R\,t^n$ with $R$ independent of the starting point and $t < 1$, and let $f \in L^2(\pi)$. Then the Poisson equation

has a solution $\hat g \in L^2(\pi)$, with $P\hat g \in L^2(\pi)$, the equation holding for $\pi$-almost every $x$.

The solution is the geometric series $\hat g = \sum_{n \ge 0} P^n \bar f$ with $\bar f = f - E_\pi f$. Uniform ergodicity is exactly what makes it converge: the Dobrushin contraction coefficient of $P^n$ is bounded by $\sup_x \|P^n(x,\cdot) - \pi\| \le R t^n$, and the Dobrushin coefficient dominates the operator norm of $P^n$ on $L^p_0(\pi)$ for every $p$ by interpolation between $L^1$ and $L^\infty$. So $\|P^n\bar f\|_2$ decays geometrically, the series converges in $L^2$, and telescoping it gives the equation. That $P\hat g \in L^2$ is then automatic, a Markov kernel being an $L^2(\pi)$-contraction by Jensen and invariance.

Why the conclusion is almost-everywhere and not pointwise. An $L^2$ limit is only determined up to a $\pi$-null set, so no pointwise representative is canonical. More seriously, the left-hand side need not be meaningful at every point: $f \in L^2(\pi)$ does not give $f \in L^1(P(x,\cdot))$ for every $x$, because $P(x,\cdot)$ need not be absolutely continuous with respect to $\pi$; at such an $x$ the integral $\int \hat g \,\mathrm{d}P(x,\cdot)$ is not defined (and is $0$ by the usual Bochner convention), so a pointwise equation would force $\hat g(x) = f(x) - E_\pi f$ there for no reason. Repairing $\hat g$ on the exceptional set does not help, since it breaks the equation at every $x$ whose $P(x,\cdot)$ charges that set.

The weakening costs nothing downstream. In the martingale-approximation proof of the central limit theorem the equation is only ever evaluated along the trajectory, and every coordinate of the stationary chain has law $\pi$, so a $\pi$-a.e. identity holds almost surely at every coordinate simultaneously (a countable intersection of full-measure events).

This supersedes an earlier version of this statement that quantified over every $x$; that form is believed unprovable for the reason above.