# Variation of Markov Chain Convergence Theorem

Assume the chain $\{X_n\}_{n\in\mathbb{N}}$ on the statespace $(S,\mathcal{F})$ (we may assume it is countable) is aperiodic, irreducible and positive recurrent. We denote with $\pi$ its (unique) stationary distribution. Is it true that for any $x\in S$ $$\underset{n\rightarrow\infty}{\text{lim}}~\underset{\text{A}\in\mathcal{F}^{\mathbb{Z}_+}}{\text{sup}}~|P_x((X_{n},X_{n+1},\dots)\in\text{A})-P_\pi(X\in A)| =0?$$

I would assume this holds true due to the similarity of the statement to the usual convergence theorem $$\underset{n\rightarrow\infty}{\text{lim}}~||P_i(X_n\in\cdot)-\pi(\cdot)||_V=0$$ and the feeling that the Markov chain is reproducing itself. But I can't figure out how to prove it. I tried to come up with coupling which seems to be the way to deal with those Variance norms. Thanks a lot for any suggestion.

You are asking about the (one half of) the total variation distance between the measure $\mathbf P_\pi$ and the shifted by $n$ measure $\mathbf P_x$. The latter measure is Markov with the initial distribution $\delta_x P^n$ (here $P$ is the transition operator of the Markov chain). Now, for Markov measures determined by two different initial distributions of the same chain their total variation distance is the same as the total variation distance between their initial distributions.