General-state MCMC ergodicity and mixing

For measurable $f:\mathsf{X}\to [0,1]$, put $A_t=\{x:f(x)>t\}$ for $t\in [0,1]$. The identity $f(x)={\int }_0^1{1}_{A_t}(x)dt$ and [F3] give $\int fd\mu -\int fd\nu ={\int }_0^1[\mu (A_t)-\nu (A_t)]dt$. Its absolute value is at most ${\int }_0^1\mathrm{\mid }\mu (A_t)-\nu (A_t)\mathrm{\mid }dt\le \parallel \mu -\nu {\parallel }_{\mathrm{T}\mathrm{V}}$ by [F1]. Conversely, each indicator $1_A$ is an admissible $f$ and attains the setwise difference for its set $A$. Taking suprema proves the dual formula.

For each $A\in \mathcal{X}$, the function $x\mapsto P(x,A)$ is measurable and takes values in $[0,1]$ by [F2]. Apply step 1.1 to that function to obtain $\mathrm{\mid }\mu P(A)-\nu P(A)\mathrm{\mid }\le \parallel \mu -\nu {\parallel }_{\mathrm{T}\mathrm{V}}$. Taking the supremum over $A$ proves kernel contraction.

For any $A\in \mathcal{X}$, $\mathrm{\mid }\mu (A)-\nu (A)\mathrm{\mid }=\mathrm{\mid }\mathbb{E}[1_A(U)-1_A(V)]\mathrm{\mid }\le \mathbb{E}\mathrm{\mid }{1}_A(U)-1_A(V)\mathrm{\mid }\le \mathbb{P}(E^c)$, because the two indicators agree on $E$. Taking the supremum over $A$ proves the stated bound.

Choose a disjoint measurable partition $(E_j{)}_{j\ge 1}$ with $\mu (E_j)<\mathrm{\infty }$ and positive total measure. Set $c_j=2^{-j}\mathrm{/}(1+\mu (E_j))$ and $c={\sum }_j{c}_j\mu (E_j)$. Then $0<c<\mathrm{\infty }$, and $\bar{\mu }(A)=c^{-1}{\sum }_j{c}_j\mu (A\cap E_j)$ is an equivalent probability measure. Its density $w=c_j\mathrm{/}c$ on $E_j$ is positive. For each fixed $x$, choose $K(x,\cdot )=f_x\bar{\mu }+S_x$ from [F3]. Let $({\mathcal{P}}_n)$ be increasing finite partitions generating $\mathcal{X}$, and set $f_n(x,y)={\sum }_{B\in {\mathcal{P}}_n:\bar{\mu }(B)>0}\frac{K(x,B)}{\bar{\mu }(B)}{1}_B(y),$ with value zero on zero-measure cells. For a positive-measure cell $B\in {\mathcal{P}}_n$, its average at level $n+1$ integrates over $B$ to the sum of $K(x,C)$ over the positive-measure children $C\subseteq B$, which is at most $K(x,B)$; zero-measure parents and all their children have value zero. Thus $(f_n(x,\cdot ))$ is a nonnegative $\bar{\mu }$-supermartingale. Also $\int f_n(x,\cdot )d\bar{\mu }\le K(x,\mathsf{X})=1$, so $-f_n(x,\cdot )$ has uniformly bounded positive-part expectations and [F4] gives a finite almost-sure limit for $f_n(x,\cdot )$.

Define $f_{\mathrm{\infty }}(x,y)={\mathrm{lim\hspace{0.17em}sup}}_n{f}_n(x,y)$, a jointly measurable extended nonnegative function; for each fixed $x$ it agrees almost everywhere with the finite limit from step 1.1. The conditional-expectation averages $g_n^x$ of the integrable density $f_x$ over ${\mathcal{P}}_n$ are defined by $g_n^x(y)=\bar{\mu }(B{)}^{-1}{\int }_B{f}_{x}d\bar{\mu }$ on positive-measure cells $B$, and by zero on zero-measure cells. Since $\int f_{x}d\bar{\mu }\le 1$, [F4] and the AC hypothesis [A1] give $g_n^x\to f_x$ almost everywhere: ${\mathcal{P}}_n$ generates $\mathcal{X}$. Cellwise $g_n^x\le f_n(x,\cdot )$ because ${\int }_B{f}_{x}d\bar{\mu }\le K(x,B)$, with both averages zero on null cells. Taking limits proves $f_x\le f_{\mathrm{\infty }}(x,\cdot )$ almost everywhere.

For a cell $B$ at partition level $N$, every $n\ge N$ has ${\int }_B{f}_n(x,y)\bar{\mu }(dy)\le K(x,B)$, since the positive-measure level-$n$ cells inside $B$ contribute only a sub-sum of the kernel masses of a partition of $B$. Fatou therefore gives ${\int }_B{f}_{\mathrm{\infty }}(x,y)\bar{\mu }(dy)\le K(x,B)$. By [F8], the left side defines a measure; Fatou at the whole space gives its total mass at most one, so both sides are finite measures. Their sum is a finite measure, and [F6] approximates any measurable set in symmetric difference under that sum; the inequality extends from the generating partition algebra to every measurable set. Choose a singular support $N_x$ with $\bar{\mu }(N_x)=0$ and $S_x(N_x^c)=0$. Since the left measure has density $f_{\mathrm{\infty }}$ with respect to $\bar{\mu }$, it assigns zero mass to $N_x$. Applying the inequality to $A\cap N_x^c$ therefore gives ${\int }_A{f}_{\mathrm{\infty }}d\bar{\mu }\le {\int }_A{f}_{x}d\bar{\mu }$. Together with step 2.1, this proves $f_{\mathrm{\infty }}=f_x$ almost everywhere.

Since $\bar{\mu }(dy)=w(y)\mu (dy)$, set $p(x,y)=f_{\mathrm{\infty }}(x,y)w(y)$ and $S(x,A)=K(x,A)-{\int }_{A}p(x,y)\mu (dy)$. Step 3.1 shows sectionwise that $S(x,\cdot )=S_x$, a nonnegative subprobability measure singular to $\mu$. The integral is measurable in $x$ by [F5]; hence $S(\cdot ,A)$ is measurable for every $A$, and its sections are countably additive. Thus $S$ is a subprobability kernel and the decomposition holds.

Put $\eta ={\varphi }^{\otimes 3}(E)>0$. The product measure is finite because $\varphi$ is finite. Fix a generating sequence $(G_j{)}_{j\ge 1}$ for $\mathcal{X}$. Let ${\mathcal{P}}_n$ consist of the nonempty sets ${\bigcap }_{j=1}^n{H}_j$, where each $H_j$ is $G_j$ or $G_j^c$. These are finite partitions and ${\mathcal{P}}_{n+1}$ refines ${\mathcal{P}}_n$. Their finite-union algebras increase and contain every $G_j$. The union of the corresponding tensor-cell algebras on ${\mathsf{X}}^3$ is therefore an algebra generating ${\mathcal{X}}^{\otimes 3}$: it contains every coordinate lift of a $G_j$, which generate the product sigma-algebra, and all its cells are product-measurable. Apply [F2] to obtain a member $H$ of that algebra with ${\varphi }^{\otimes 3}(E\mathrm{△}H)<\eta \mathrm{/}6$. Then ${\varphi }^{\otimes 3}(H)>5\eta \mathrm{/}6$ and ${\varphi }^{\otimes 3}(H\setminus E)<\eta \mathrm{/}6<{\varphi }^{\otimes 3}(H)\mathrm{/}4$, so $E$ occupies more than three quarters of $H$. The set $H$ is a finite union of tensor cells at a common partition level. If every positive-measure cell had occupancy at most three quarters, finite additivity would give the same upper bound on $H$, a contradiction. Thus some cell $U\times V\times W$ has positive product measure and ${\varphi }^{\otimes 3}(E\cap (U\times V\times W))>\frac{3}{4}\varphi (U)\varphi (V)\varphi (W)>0$. Only one finite-algebra approximant and one cell of its finite partition are selected; no conditional-expectation representative or choice axiom is used.

Since $E\subseteq A{\times }_{\text{\hspace{0.17em}⁣}y}B$, Tonelli gives ${\int }_{U\times V}{1}_A(x,y)d\varphi (x)d\varphi (y)>\frac{3}{4}\varphi (U)\varphi (V),$ and likewise ${\int }_{V\times W}{1}_B(y,z)d\varphi (y)d\varphi (z)>\frac{3}{4}\varphi (V)\varphi (W)$. Consequently the sets $C=\{x\in U:\varphi (A_1(x)\cap V)>\frac{3}{4}\varphi (V)\}$ and $D=\{z\in W:\varphi (B_2(z)\cap V)>\frac{3}{4}\varphi (V)\}$ have positive measure; otherwise the corresponding integral could not exceed three quarters of the rectangle measure.

For $x\in C$ and $z\in D$, both sections occupy more than three quarters of $V$. The complement of their intersection in $V$ is the union of two sets, each of measure less than $\varphi (V)\mathrm{/}4$. Thus $\varphi (A_1(x)\cap B_2(z))\ge \varphi (V)\mathrm{/}2>0$. Set $\gamma =\varphi (V)\mathrm{/}2$.

Run the two coordinates independently until both lie in $D$. At each such visit, use the minorization to toss a common Bernoulli($ϵ$) split: on success draw one shared $\nu$ state; on failure draw independently from $R_x,R_y$. Let $T=\inf \{k+1:\text{the split at time }k\text{ succeeds}\}$, with $\inf \mathrm{\varnothing }=\mathrm{\infty }$. Record success with a measurable binary flag and evolve synchronously thereafter, without testing equality before success. The shared transition is the pushforward of $Q(X_k,\cdot )$ under $z\mapsto (z,z)$; this map is measurable on every measurable space. This construction has the required $Q$-marginals by [F1], [F2], and [F5].

Let $N_k$ count failed split transitions completed by time $k$, with $N_0=0$, and set $Z_k=a^{-k}{B}^{-N_k}h(X_k,Y_k)1_{\{T>k\}}$. If $ϵ=1$, a visit to $D^2$ couples on the next transition, so the uncoupled contribution there is zero. If $0<ϵ<1$, off $D^2$ the conditional expectation of the next $h$ is at most $ah$; on $D^2$ the uncoupled transition has probability $1-ϵ$ and expected $h$ equal to $R_x\otimes R_y(h)$. The definition of $B$ bounds its contribution after the extra factor $B^{-1}$ by $ah(x,y)$, since $h\ge 1$. Thus $(Z_k)$ is a nonnegative supermartingale in both cases.

For any integer $j\ge 1$, the probability of at least $j$ failed attempts is at most $(1-ϵ{)}^j$. On the event $T>k$ with at most $j-1$ failures, $h\ge 1$ and $B^{N_k}\le B^{j-1}$, so the supermartingale bound gives $\mathrm{Pr}(T>k)\le (1-ϵ{)}^j+a^k{B}^{j-1}h(x,y)\mathrm{.}$

Choose $\theta >0$ so $a{B}^{\theta }<1$ and set $j=\max (1,\lfloor \theta k\rfloor )$. Both terms in step 3.1 are bounded by a constant times $r^k$ for some $r<1$. Apply the coupling inequality [F3] to obtain the asserted geometric bound after increasing $M$ to cover the finite initial values of $k$.

If $C=\mathrm{\varnothing }$, then $PV\le \lambda V$ everywhere; iterating gives $1\le P^{N}V(x)\le {\lambda }^{N}V(x)$ for every state $x$ and $N$, impossible for sufficiently large $N$ because $V(x)<\mathrm{\infty }$. Thus the hypotheses force $C\mathrm{\ne }\mathrm{\varnothing }$. Let ${\tau }_C=\inf \{n\ge 0:X_n\in C\}$. The killed kernel $K(x,A)=1_{C^c}(x)P(x,A)$ satisfies $KV\le \lambda V$ off $C$. Since $V\ge 1$, induction gives $K^{N}1(x)\le K^{N}V(x)\le {\lambda }^{N}V(x)$. Hence for $x\in \{V\le d\}$, ${\mathrm{Pr}}_x({\tau }_C\le N)\ge 1-d{\lambda }^N$; for $x\in C$, ${\tau }_C=0$.

For each $t\le N$, restrict paths to the disjoint event $\{{\tau }_C=t\}$ and apply [F5] followed by the small-set minorization. This gives $P^{t+m}(x,A)\ge ϵ{\mathrm{Pr}}_x({\tau }_C=t)\nu (A)\mathrm{.}$ Summing over $t$ yields ${\sum }_{t=0}^N{P}^{t+m}(x,A)\ge ϵ(1-d{\lambda }^N)\nu (A)$ uniformly on $D$.

Choose $N$ with $d{\lambda }^N\le 1\mathrm{/}2$ and assign weight $1\mathrm{/}(N+1)$ to times $m,\dots ,m+N$. The resulting probability mixture dominates $ϵ\nu \mathrm{/}[2(N+1)]$, so it satisfies the petite definition with positive multiplier $\alpha =ϵ\mathrm{/}[2(N+1)]$.

If $C=\mathrm{\varnothing }$, then $PV\le \lambda V$ everywhere. For each integer $N\ge 1$, $V_N=V\wedge N$ satisfies $P{V}_N-V_N\le -(1-\lambda )V$ on $\{V\le N\}$ and $P{V}_N-V_N\le 0$ on $\{V>N\}$. Invariance for the bounded function $V_N$ gives integral zero, so $\pi (\{V\le N\})=0$. Since $V$ is finite-valued, these sets increase to the whole space, contradicting that $\pi$ is a probability. Thus no invariant probability exists in the empty-set case, and the implication is vacuous. Assume $C\mathrm{\ne }\mathrm{\varnothing }$ below.

Set $V_N=V\wedge N$. Since $V_N\le V$ and $V_N\le N$, $P{V}_N\le \min (PV,N)$. On $C$, $P{V}_N-V_N\le \lambda {\sup }_{C}V+b$. On $C^c\cap \{V\le N\}$ it is at most $-(1-\lambda )V$, and on $C^c\cap \{V>N\}$ it is at most zero.

Invariance for bounded $V_N$ gives $\int (P{V}_N-V_N)d\pi =0$. Using the three pointwise bounds from step 1.2, $(1-\lambda ){\int }_{C^c\cap \{V\le N\}}Vd\pi \le (\lambda {\sup }_{C}V+b)\pi (C)\le \lambda {\sup }_{C}V+b\mathrm{.}$

The domains $C^c\cap \{V\le N\}$ increase to $C^c$; monotone convergence bounds ${\int }_{C^c}Vd\pi$ by the right side divided by $1-\lambda$. The integral over $C$ is at most ${\sup }_{C}V$. Hence $\pi V<\mathrm{\infty }$.

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.

Apply [F3] to each $P^n$, obtaining preliminary densities $p_n^0$ and singular residuals $S_n$. Inductively set $p_1=p_1^0$ and, for $n>1$, $p_n(x,y)=\max (p_n^0(x,y),{\sup }_{1\le k<n}\int P^{n-k}(x,dz)p_k(z,y))\mathrm{.}$ For each $k<n$, the lifted kernel $((x,y),A)\mapsto P^{n-k}(x,A)$ is a probability kernel by [F2]. Apply [F4] to this kernel and the nonnegative product-measurable integrand $((x,y),z)\mapsto p_k(z,y)$; finite maxima preserve measurability. Thus the right side is jointly measurable. For each fixed $x$, composition of kernels gives $P^n(x,A)\ge {\int }_A\int P^{n-k}(x,dz)p_k(z,y)\varphi (dy)$. Since $S_n(x,\cdot )$ is singular, this domination forces each candidate density to be at most $p_n^0(x,\cdot )$ $\varphi$-almost everywhere. Thus $p_n=p_n^0$ $\varphi$-a.e. for each $x$, so it is still the density part and has both displayed convolution lower bounds pointwise by construction.

Fix $x$. For each $n$, choose a $\varphi$-null support $N_n$ of $S_n(x,\cdot )$ and let $F={\bigcap }_{n\ge 1}{N}_n^c$. Then $\varphi (F^c)=0$ and every residual assigns zero mass to $F$. If $B\subseteq F$ has $\varphi (B)>0$, [F1] gives an $m$ with $P^m(x,B)>0$; because the residual gives $B$ zero mass, ${\int }_B{p}_m(x,y)\varphi (dy)>0$.

Therefore every positive-$\varphi$ subset of $F$ intersects $\{{\sum }_n{p}_n(x,\cdot )>0\}$ in positive measure. Its complement in $F$ must be $\varphi$-null; otherwise take that complement as $B$ in step 1.2, which would make every integral zero. Hence the density sum is positive $\varphi$-almost everywhere.