Aperiodicity makes minorization times eventually consecutive

For $n,k\ge 1$, let $E_{n,k}=\{x:P^n(x,\cdot )\ge k^{-1}\nu (\cdot )\}$. Fix a countable generating sequence $(G_j)$. The union of the finite Boolean algebras generated by $G_1,\dots ,G_j$ is a countable algebra ${\mathcal{A}}_0$ generating $\mathcal{X}$. The inequality defining $E_{n,k}$ is equivalent to its restrictions to $B\in {\mathcal{A}}_0$: if those hold, approximate any measurable $A$ under the finite measure $P^n(x,\cdot )+k^{-1}\nu$ by an algebra set; the difference of the two sides changes by at most that symmetric-difference mass, which tends to zero. Thus $E_{n,k}={\bigcap }_{B\in {\mathcal{A}}_0}\{x:P^n(x,B)\ge k^{-1}\nu (B)\}$ is measurable by [F2] and [F4]. The statewise hypothesis makes their union the whole space. Since $\nu$ is a probability, one such set has positive $\nu$-mass. Integrating its inequality gives $\nu P^n\ge k^{-1}\nu (E_{n,k})\nu$. Thus the set $T=\{n\ge 1:\nu P^n\ge c_n\nu \text{ for some }{c}_n>0\}$ is nonempty.

If $r,s\in T$, then $(\nu P^r)P^s\ge c_r\nu P^s\ge c_r{c}_s\nu$, so $T$ is additive. Let $d=\gcd (T)$.

Suppose $d>1$. For $1\le i\le d$, define $X_i=\{x:\mathrm{\exists }\mathrm{\ell }\ge 1,\mathrm{\ell }d-i\ge 1,\delta >0,P^{\mathrm{\ell }d-i}(x,\cdot )\ge \delta \nu (\cdot )\}\mathrm{.}$ Every positive witness time $n$ has exactly one representation $n=\mathrm{\ell }d-i$ with $1\le i\le d$ and $\mathrm{\ell }d-i\ge 1$, so the statewise minorization covers the space by these sets; rational thresholds and the countable testing algebra make them measurable. Put $O={\bigcup }_{i\mathrm{\ne }j}(X_i\cap X_j)$. For every $m\ge 0$, $\nu P^m(O)=0$: otherwise a countable subdivision gives a positive $\nu P^m$-mass subset of an overlap on which both phase exponents and both positive constants are fixed. Integrating the two minorization inequalities gives two times in $T$ with different residues modulo $d$, contradicting that $d$ divides every member of $T$.

Let $S=O\cup \{x:\mathrm{\exists }m\ge 0,P^m(x,O)>0\}$ and $X_i^{\mathrm{\prime }}=X_i\setminus S$. Since $\nu P^m(O)=0$ for every $m$, $\nu (S)=0$. The $X_i^{\mathrm{\prime }}$ are disjoint and cover $S^c$. For $x\in X_i^{\mathrm{\prime }}$, $P(x,S)=0$: positive mass on the countable union defining $S$ would, by integration of a positive hitting probability, imply $P^{m+1}(x,O)>0$ for some $m\ge 0$, contradicting $x\mathrm{\notin }S$. Thus $P(x,{\bigcup }_j{X}_j^{\mathrm{\prime }})=1$. If $P(x,X_j^{\mathrm{\prime }})>0$ for $j\mathrm{\ne }i+1(\mathrm{m}\mathrm{o}\mathrm{d}d)$, countably partition $X_j^{\mathrm{\prime }}$ by its witnessing exponent and rational minorization constant; one part has positive $P(x,\cdot )$-mass. Integrating the common later minorization after that one-step transition puts $x$ in $X_{j-1}$, contradicting disjointness of the $X_i^{\mathrm{\prime }}$. Thus $P(x,X_{i+1})=1$ for $x\in X_i^{\mathrm{\prime }}$. Since $\nu (S)=0$, the disjoint classes satisfy $\nu ({\bigcup }_i{X}_i^{\mathrm{\prime }})=1$. Absolute continuity gives $\pi ({\bigcup }_i{X}_i^{\mathrm{\prime }})>0$, so some $X_i^{\mathrm{\prime }}$ has positive $\pi$-mass. Invariance propagates positive mass from each class to its successor, so all $X_i^{\mathrm{\prime }}$ have positive $\pi$-mass. This is a cyclic decomposition, contrary to [F1]. Hence $d=1$.

Choose finitely many $t_1,\dots ,t_q\in T$ with gcd one. Their nonnegative combinations modulo $t_1$ form the subgroup generated by their residues; gcd one makes this all residues. Choose one represented $u_j\in T$ for each residue $j$. Every sufficiently large integer in residue $j$ is $u_j+\mathrm{\ell }{t}_1$ and belongs to $T$ by additivity. The defining positive constant for each such $n$ is ${ϵ}_n$.