Invariant mass and almost-sure hitting of accessible sets

For the additional conclusion, let $f(x)={\sum }_{n\ge 1}{P}^n(x,A)$. By [F1], $f(x)>0$ everywhere. Tonelli and stationarity give $\int fd\pi ={\sum }_{n\ge 1}\pi (A)$. If $\pi (A)=0$, this integral is zero, contradicting positivity of $f$ under a probability measure. Hence $\pi (A)>0$; [F1] also supplies the positive-hitting premise for the first assertion.

It remains to prove the first assertion under its positive-hitting premise. Set $q_0=1_A$ and recursively $q_{n+1}=1_A+1_{A^c}P{q}_n$, where $P{q}_n(x)=\int q_n(y)P(x,dy)$. Then $q_n(x)={\mathrm{Pr}}_x({\tau }_A\le n)$; induction and [F7] show each $q_n$ is measurable, and $q_n\uparrow q:={\mathrm{Pr}}_x({\tau }_A<\mathrm{\infty })$, so $q$ is measurable. If the bad set $\{q<1\}$ has positive mass, a countable level-set union gives some ${\delta }_1>0$ for which $B_1=\{x:q(x)\le 1-{\delta }_1\}$ has positive $\pi$-measure. For each $x\in B_1$, $q(x)>0$ and $x\mathrm{\notin }A$, so some ${\mathrm{\ell }}_0\ge 1$ has $P^{{\mathrm{\ell }}_0}(x,A)>0$. Countable unions over times and rational thresholds then give a common ${\mathrm{\ell }}_0\ge 1$, ${\delta }_2>0$, and $B_2\subseteq B_1$ with $\pi (B_2)>0$ and $P^{{\mathrm{\ell }}_0}(x,A)\ge {\delta }_2$ on $B_2$.

Let $\eta =\mathrm{\#}\{k\ge 1:X_{k{\mathrm{\ell }}_0}\in B_2\}$. At each such visit, the conditional probability of avoiding $A$ at the next ${\mathrm{\ell }}_0$-step time is at most $1-{\delta }_2$. Induction using [F4] gives ${\mathrm{Pr}}_x({\tau }_A=\mathrm{\infty },\eta \ge r)\le (1-{\delta }_2{)}^r$ for each finite $r$, and hence ${\mathrm{Pr}}_x({\tau }_A=\mathrm{\infty },\eta =\mathrm{\infty })=0$. Thus for $x\in B_2$, ${\mathrm{Pr}}_x({\tau }_A=\mathrm{\infty },\eta <\mathrm{\infty })\ge {\delta }_1$. To select a measurable positive-mass set with a common finite last-visit bound, for each $\mathrm{\ell }\ge 1$ forbid $A$ at every time and also forbid $B_2$ at times $k{\mathrm{\ell }}_0$ with $k\ge \mathrm{\ell }$. For a finite horizon $N$, let $F_n^{(\mathrm{\ell })}=A\cup B_2$ if $n=k{\mathrm{\ell }}_0$ for some $k\ge \mathrm{\ell }$, and $F_n^{(\mathrm{\ell })}=A$ otherwise. Set $g_{N,N}=1_{(F_N^{(\mathrm{\ell })}{)}^c}$ and recursively set $g_{N,n}=1_{(F_n^{(\mathrm{\ell })}{)}^c}P{g}_{N,n+1}$ for $n=N-1,\dots ,0$. Then $g_{\mathrm{\ell },N}=g_{N,0}$ is the finite-horizon event probability and is measurable by [F7]. As $N\to \mathrm{\infty }$, these events decrease to $\{{\tau }_A=\mathrm{\infty },\sup \{k\ge 1:X_{k{\mathrm{\ell }}_0}\in B_2\}<\mathrm{\ell }\}$, so continuity from above makes their limit $g_{\mathrm{\ell }}(x)$ measurable. The events increase with $\mathrm{\ell }$ and their union is $\{{\tau }_A=\mathrm{\infty },\eta <\mathrm{\infty }\}$, so $g_{\mathrm{\ell }}(x)\uparrow$ a value at least ${\delta }_1$ on $B_2$. A countable rational threshold selection therefore gives $\mathrm{\ell }\ge 1$, $\delta >0$, and $B\subseteq B_2$ with $\pi (B)>0$ and ${\mathrm{Pr}}_x\text{\hspace{0.17em}⁣}({\tau }_A=\mathrm{\infty },\sup \{k\ge 1:X_{k{\mathrm{\ell }}_0}\in B_2\}<\mathrm{\ell })\ge \delta (x\in B)\mathrm{.}$

Put $L=\mathrm{\ell }{\mathrm{\ell }}_0$ and $S=\sup \{k\ge 1:X_{kL}\in B\}$, with $S=-\mathrm{\infty }$ if no such visit occurs. For $1\le r\le j$, consider the event that $X_{rL}\in B$, there is no later $B$-visit at a positive multiple of $L$, and no hit of $A$ at or after $rL$. It is a subset of $\{S=r,X_{jL}\mathrm{\notin }A\}$. Stationarity and the Markov property at time $rL$ therefore give ${\mathrm{Pr}}_{\pi }(S=r,X_{jL}\mathrm{\notin }A)\ge {\int }_B\pi (dy){\mathrm{Pr}}_y({\tau }_A=\mathrm{\infty },S=-\mathrm{\infty })\ge \pi (B)\delta \mathrm{.}$ Indeed the event in step 2.1 rules out every $B_2$-visit at skeleton indices $k\ge \mathrm{\ell }$, hence every later $B$-visit at a positive multiple of $L=\mathrm{\ell }{\mathrm{\ell }}_0$, as well as every future hit of $A$. The events $S=r$ are disjoint, so stationarity yields $\pi (A^c)\ge j\pi (B)\delta$. Taking $j>1\mathrm{/}(\pi (B)\delta )$ contradicts $\pi (A^c)\le 1$. Therefore the bad set is $\pi$-null.