A geometrically ergodic chain has a small set of positive invariant measure

Let $P$ be a Markov transition kernel on a state space $\mathsf{X}$ whose $\sigma$-field is countably generated, let $\pi$ be an invariant probability measure, and suppose the chain is Harris ergodic and geometrically ergodic: there are a function $M \ge 0$ and a constant $t < 1$ with $\|P^n(x,\cdot) - \pi\| \le M(x)\,t^n$ for every $x$ and every $n \ge 1$. Then the chain possesses a measurable small set of positive invariant measure: there is a measurable $C \subseteq \mathsf{X}$ with $\pi(C) > 0$ for which one can find an integer $n_0 \ge 1$, a constant $\varepsilon > 0$ and a probability measure $Q$ satisfying the minorization $P^{n_0}(x, A) \ge \varepsilon\, Q(A)$ simultaneously for every $x \in C$ and every measurable $A$.

This is the measure-theoretic half of Meyn and Tweedie's Theorem 15.0.1 (i) $\Rightarrow$ (iii), isolated from its analytic half. In the classical treatment the small set is produced by Theorem 5.2.2 from $\psi$-irreducibility, by Nummelin splitting together with a rectangle-extraction argument. Under geometric ergodicity that machinery is unnecessary, and the proof given here avoids it. Write $f_n(x, y)$ for the Radon-Nikodym density of the absolutely continuous part of $P^n(x,\cdot)$ with respect to $\pi$, which is jointly measurable precisely because the $\sigma$-field is countably generated. The elementary observation driving the argument is that if $\pi(A) \le \mu(A) + u$ for every measurable $A$, then $\pi\{y : (\mathrm{d}\mu/\mathrm{d}\pi)(y) < 1/2\} \le 2u$; applied with $\mu = P^n(x,\cdot)$ and $u = M(x)t^n$, this says that the set of states where the $n$-step density is small is itself small in measure, not merely of positive complement, and it is this quantitative strengthening that removes the need for rectangle extraction. Cutting the state space along a level set of $x \mapsto \pi\{y : f_n(x,y) < 1/2\}$ produces a set $C$ of positive $\pi$-measure on which that bound is uniform; one application of Tonelli's theorem on $\pi \times \pi$, followed by Markov's inequality, yields a set $D$ with $\pi(D) \ge 1/2$ whose points are reached with density at least $1/2$ from all but a quarter of $C$. Composing the two half-steps gives $P^{2n}(x, A) \ge \tfrac{1}{8}\,\pi(C)\,\pi(D \cap A)$ for every $x \in C$, which is the required minorization with $n_0 = 2n$, $\varepsilon = \tfrac{1}{8}\pi(C)\pi(D)$ and $Q$ the normalized restriction of $\pi$ to $D$.

Two remarks on the scope of the statement. First, the conclusion is not free. The identity kernel on $\mathbb{R}$ leaves the uniform law on $[0,1]$ invariant, yet under it every small set is a singleton or empty and therefore null, so no measurable small set of positive measure exists: some ergodicity hypothesis is genuinely doing work. Second, countable generation of the $\sigma$-field is load-bearing rather than decorative, and enters twice — as the hypothesis of the kernel Radon-Nikodym theorem that makes the densities $f_n$ jointly measurable, and as the assumption that excludes the countable/co-countable state spaces on which the accepted refutation of the sibling problem MarkovChainCLT.geometricallyErgodic_integrable_rate is built, and on which the present statement is in fact false. It should be recorded that the proof consumes only the geometric total-variation bound: the invariance of $\pi$ carried by HarrisErgodic is never used, and the hypothesis is retained solely so that the statement matches the binder list of the parent problem MarkovChainCLT.geoDriftCondition_of_geometricallyErgodic.