Small-set coupling yields almost-everywhere total-variation convergence

By [F5], $\pi (C)>0$. Integrating the small-set inequality over $C$ and using invariance gives $\pi (A)=\pi P^m(A)\ge ϵ\pi (C)\nu (A)$ for all $A$, hence $\nu \ll \pi$.

Fix $x$. Irreducibility gives $k_x$ with $P^{k_x}(x,C)>0$, so $P^{k_x+m}(x,\cdot )\ge ϵP^{k_x}(x,C)\nu$. By [F6], $\nu P^t(C)\ge c_t\nu (C)>0$ for every sufficiently large $t$. Thus $P^n(x,C)>0$ for every sufficiently large $n$, and in particular $P^{mq}(x,C)>0$ for every sufficiently large integer $q$. This holds for every $x$.

Let $\bar{Q}=(P^m)\otimes (P^m)$. For measurable rectangles $A\times B$, $({\pi }^2\bar{Q})(A\times B)=(\pi P^m)(A)(\pi P^m)(B)=\pi (A)\pi (B)$. By [F11], this proves that ${\pi }^2$ is invariant for $\bar{Q}$. By step 2.1, every starting pair has positive probability of hitting $C^2$ under $\bar{Q}$. Apply the stationary positive-hitting lemma [F5] to $\bar{Q}$, invariant law ${\pi }^2$, and target $C^2$: the stationary pair hits $C^2$ almost surely from ${\pi }^2$-almost every pair.

Start the pair with law ${\pi }^2$ and run independent $\bar{Q}$ transitions before each split attempt. By step 3.1 and [F5], the first hit of $C^2$ occurs almost surely. Its location law is absolutely continuous with respect to ${\pi }^2$: for any ${\pi }^2$-null set $N$, the event that the first hit lies in $N$ is contained in ${\bigcup }_{k\ge 0}\{Z_k\in N\}$, which has probability zero by stationarity. More generally, if the pair at the start of a waiting period has any subprobability law $\lambda \ll {\pi }^2$, then $\lambda {\bar{Q}}^k\ll {\pi }^2$ for every $k$, because invariance gives ${\bar{Q}}^k(z,N)=0$ for ${\pi }^2$-almost every $z$; hence its next first-hit location is also absolutely continuous and the hit occurs almost surely by [F5]. At a visit $(x,y)\in C^2$ with $0<ϵ<1$, use one common Bernoulli($ϵ$) coin: success draws a shared $\nu$ value and couples immediately, while failure draws independently from $R_x=(P^m(x,\cdot )-ϵ\nu )\mathrm{/}(1-ϵ)$ and $R_y$. Each coordinate retains its $P^m$ marginal. For every ${\pi }^2$-null $N$, stationarity gives $\bar{Q}((x,y),N)=0$ for ${\pi }^2$-almost every $(x,y)$, while the product expansion gives $\bar{Q}((x,y),\cdot )\ge (1-ϵ{)}^2{R}_x\otimes R_y$; thus the failed endpoint law is absolutely continuous whenever the attempt-state law is. It therefore reaches $C^2$ almost surely again, with an absolutely continuous first-hit law. If $ϵ=1$, the first split succeeds and no failure branch occurs.

Each visit succeeds with probability $ϵ$, so $j$ successive failures have probability $(1-ϵ{)}^j\to 0$. The coupling succeeds almost surely for ${\pi }^2$-almost every initial pair. Fubini gives that for $\pi$-almost every $x$, a chain from $x$ couples almost surely to an independent stationary chain. The coupling inequality gives convergence along times $mn$. For $n=mk+s$, $0\le s<m$, contraction by $P^s$ and invariance of $\pi$ gives convergence for every $n$.