Stationary Markov-chain total-variation convergence implies strong mixing

Fix $n\ge 1$, $j\ge 0$, $A\in {\mathcal{F}}_j$, and $B\in \sigma (X_{j+n},X_{j+n+1},\dots )$. By [F2], the indicator of this future-path event has conditional expectation $h(X_{j+n})$ for a measurable $0\le h\le 1$. Applying [F3] and the tower property gives $\mathbb{E}[{\mathbf{1}}_B\mid {\mathcal{F}}_j]=P^{n}h(X_j)$. Stationarity gives $\mathbb{P}(B)=\pi (h)$.

For probability measures $\mu ,\nu$ and measurable $0\le h\le 1$, the dyadic simple approximation $h_m=2^{-m}{\sum }_{r=1}^{2^m}{\mathbf{1}}_{\{h\ge r{2}^{-m}\}}$ differs from $h$ by at most $2^{-m}$. Each indicator integral difference is bounded by [F4], so the integrals of $h_m$ differ by at most $\mathrm{\parallel }\mu -\nu {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$. Letting $m\to \mathrm{\infty }$ gives $\mathrm{\mid }\mu (h)-\nu (h)\mathrm{\mid }\le \mathrm{\parallel }\mu -\nu {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$.

For fixed $n$ and $x$, ${\mu }_{n,x}:=P^n(x,\cdot )+\pi$ is a finite measure. Let $\mathcal{D}$ be the sets approximable in ${\mu }_{n,x}$-symmetric-difference measure by members of $\mathcal{C}$. It contains $\mathcal{C}$ and is closed under complements because $\mathcal{C}$ is an algebra. For $A_i\in \mathcal{D}$ and $\epsilon >0$, [F6] gives $N\ge 1$ with ${\mu }_{n,x}(({\bigcup }_i{A}_i)\setminus ({\bigcup }_{i=1}^N{A}_i))<\epsilon \mathrm{/}2$. Choose $C_i\in \mathcal{C}$ with ${\mu }_{n,x}(A_i\mathrm{△}{C}_i)<\epsilon \mathrm{/}(2N)$ for $i\le N$. The finite union $C={\bigcup }_{i=1}^N{C}_i$ belongs to $\mathcal{C}$ and ${\mu }_{n,x}(({\bigcup }_i{A}_i)\mathrm{△}C)<\epsilon$ by finite subadditivity. Thus $\mathcal{D}$ is a sigma-algebra containing the generator $\mathcal{C}$, so $\mathcal{D}=\mathcal{X}$. Approximating each measurable $A$ in this way shows $d_n(x):={\sup }_{C\in \mathcal{C}}\mathrm{\mid }{P}^n(x,C)-\pi (C)\mathrm{\mid }=\mathrm{\parallel }{P}^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$: the discrepancy for $A$ differs from that for an approximating $C$ by at most $P^n(x,A\mathrm{△}C)+\pi (A\mathrm{△}C)$. Since $\mathcal{C}$ is countable and each evaluation is measurable by [F8], $d_n$ is measurable; also $0\le d_n\le 1$.

Using [F5] and step 1.1, $\mathrm{\mid }\mathbb{P}(A\cap B)-\mathbb{P}(A)\mathbb{P}(B)\mathrm{\mid }\le \mathbb{E}\mathrm{\mid }{P}^{n}h(X_j)-\pi (h)\mathrm{\mid }\le \int d_n(x)\pi (dx)$. The last inequality follows from step 1.2 applied pointwise to $P^n(x,\cdot )$ and $\pi$, and then from step 1.3. By hypothesis $d_n(x)\to 0$ for $\pi$-almost every $x$; since $d_n\le 1$, [F7] makes the integral tend to zero. This bound is independent of $j,A,B$, so taking the supremum in [F1] proves ${\alpha }_X(n)\to 0$.