Harris geometric ergodicity admits an integrable total-variation envelope

Apply [F2] to $M_0$ to obtain a countably generated $\mathcal{G}\subseteq \mathcal{X}$, the restricted kernel $P_{\mathcal{G}}$, and invariant probability ${\pi }_{\mathcal{G}}=\pi {\mathrm{\mid }}_{\mathcal{G}}$. If $B\in \mathcal{G}$ has ${\pi }_{\mathcal{G}}(B)>0$, [F1] implies $P^n(x,B)>0$ for all sufficiently large $n$, for each $x$, because $M_0(x)$ is finite. Thus $P_{\mathcal{G}}$ is ${\pi }_{\mathcal{G}}$-irreducible by [F7]. Applying [F3] on this countably generated space gives an $m\ge 1$, a $D\in \mathcal{G}$ with $\pi (D)>0$, a probability ${\nu }_0$ on $\mathcal{G}$ with ${\nu }_0(D)>0$, and ${ϵ}_0>0$ such that $P_{\mathcal{G}}^m(x,\cdot )\ge {ϵ}_0{\nu }_0$ on $D$. No small-set assertion on the original sigma-field is being made.

Since $M_0$ is finite everywhere, $D\cap \{M_0\le K\}$ increases to $D$ as the positive integer $K$ increases. Choose $K$ so $C=D\cap \{M_0\le K\}$ has both $\pi (C)>0$ and ${\nu }_0(C)>0$. Choose a finite integer $L\ge K$ and put $a={\nu }_0(M_0\le L)>0$, $\nu (B)={\nu }_0(B\cap \{M_0\le L\})\mathrm{/}a$. Choose $0<ϵ<\min (1,{ϵ}_{0}a)$. For $Q=P_{\mathcal{G}}^m$ this gives $Q(x,\cdot )\ge ϵ\nu$ on $C$, $\nu (M_0)\le L$, and $M_0\le K$ on $C$.

For $n\ge 1$, the restricted-iteration identity in [F2] and [F1] give $\mathrm{\mid }\nu Q^n(C)-\pi (C)\mathrm{\mid }\le \int M_0(x)r_0^{mn}\nu (dx)\le L(r_0^m{)}^n\mathrm{.}$ For $n=0$ the absolute difference is at most $1$. Therefore [F4] applies with constant $\max (1,L)$ and rate $r_0^m$. It supplies $s>1$ and a measurable $H(x)={\mathbb{E}}_x[s^{\sigma }]$, where $\sigma =\inf \{j\ge 0:X_{jm}\in C\}$, such that $\pi (H)<\mathrm{\infty }$. Indeed the skeleton path restricted to $\mathcal{G}$ under the full canonical $P$-chain has the $Q$-chain law: its finite-dimensional cylinder probabilities agree by the iterated-kernel identity in [F2]. Thus the same $H$ is the entrance-time moment for the original chain of [F5].

Choose $\max (r_0,s^{-1\mathrm{/}m})<r_1<1$, and set $S=m\sigma$ and $K_1=\max (1,K)$. Fix $A\in \mathcal{X}$ and $n\ge 1$. On $\{S\le n\}$, decompose into the finitely many events $\{S=j\}$ and use [F6] to replace the conditional terminal indicator by $P^{n-j}(X_j,A)$. Since $X_j\in C$, [F1] bounds its difference from $\pi (A)$ by $K_1{r}_0^{n-j}$; when $j=n$, use $\mathrm{\mid }{1}_A(X_n)-\pi (A)\mathrm{\mid }\le 1\le K_1$. On $\{S>n\}$ use the bound $1$. Subtracting $\pi (A)$ on these two events gives $\mathrm{\mid }{P}^n(x,A)-\pi (A)\mathrm{\mid }\le {\mathbb{P}}_x(S>n)+K_1{\mathbb{E}}_x[r_0^{n-S};S\le n]\le (1+K_1)r_1^{n}H(x)\mathrm{.}$ For the final inequality, $r_1^{-S}\le s^{\sigma }$, and on $S\le n$ one has $r_0^{n-S}\le r_1^n{r}_1^{-S}$; on $S>n$, $1\le r_1^n{r}_1^{-S}$. This argument is for each original measurable event $A$, so taking its supremum controls the full original total variation and does not require measurability of a total-variation supremum.

The measurable set $N=\{H=\mathrm{\infty }\}$ is $\pi$-null by step 3.1. Define $M_1(x)=(1+K_1)H(x)$ for $x\mathrm{\notin }N$ and $M_1(x)=M_0(x)$ for $x\in N$. It is measurable and finite everywhere, and $\pi (M_1)<\mathrm{\infty }$. Off $N$, step 4.1 gives the desired bound with rate $r_1$; on $N$, [F1] gives it because $r_0<r_1$. Taking $M=M_1$ and $r=r_1$ proves the stronger conclusion from invariance and the everywhere geometric bound alone: the restricted irreducibility used in step 1.1 was derived there directly. The original additional Harris, irreducibility and aperiodicity hypotheses therefore give the same conclusion as a special case. AC was used precisely through [F3] and the path-space and Markov suppliers [F4], [F5] and [F6].