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Let $(X_n)_{n\in\mathbb N_0}$ be a time-homogeneous Markov chain on a probability space $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\pi$, invariant measure $\mu$ and initial distribution $\nu$. Let $f\in\mathcal L^1(\mu)$ and $$A_{b,\:n}f:=\frac1n\sum_{i=b}^{b+n-1}f(X_i).$$

Assuming that the total variation distance $|\mu-\nu\pi^n|$ tends to $0$ as $n\to\infty$, it is claimed here in Theorem 3.18 that $$A_{b,\:n}f\xrightarrow{n\to\infty}\int f\:{\rm d}\mu.$$ In the proof, the author is basically reducing the problem to the case $\nu=\mu$ (i.e. the chain is started in stationarity). However, even in that case, we should need that $\operatorname P_\nu:=\nu\pi$ (composition of transition kernels) is ergodic with respect to the shift $$\tau:\mathbb R^{\mathbb N_0}\to\mathbb R^{\mathbb N_0}\;,\;\;\;(x_n)_{n\in\mathbb N_0}\mapsto(x_{n+1})_{n\in\mathbb N_0}.$$ What am I missing?

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  • $\begingroup$ Could you write explicitly what $\nu \pi$ is? I don't see why it's a measure on $\mathbb R^{\mathbb N_0}$. $\endgroup$ – Roberto Rastapopoulos Mar 2 at 17:03
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    $\begingroup$ You could also add that the convergence seeked is in the almost sure sense. For the constant Markov chain ($X_{n+1} = X_{n})$, all measures are invariant and the statement obviously does not hold, so I agree that an additional ergodicity condition should be assumed. $\endgroup$ – Roberto Rastapopoulos Mar 2 at 17:08
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[There was a mistake in my initial reply; I've corrected it now.]

I think you're right:

If $\mu$ is only invariant and not ergodic,$^\ast$ if we take the simple case $\nu=\mu$, the ergodic averages $A_{b,n}f$ converge almost surely to the random variable $\mathbb{E}_\mu[f|\mathcal{I}](X_k)$ (which may alternatively be written as $\mathbb{E}_\mathrm{P}[f(X_k)|X_k^{-1}\mathcal{I}]$), where $k$ may be any natural number and $\mathcal{I}$ is the $\sigma$-algebra of measurable sets $A$ satisfying $\pi(x,A)=1$ for $\mu$-almost all $x \in A$. This limiting random variable is only equal (mod null sets) to the constant $\mu(f)$ if $\mu$ is ergodic.

So I think an additional condition is meant to be included in the theorem, namely that $\mu$ is ergodic.$^\ast$ Indeed, Corollary 2.15 -- which is probably the "stationary case" of the CLT referred to several times in Section 3.4.3 (including the statement of Theorem 3.18) -- explicitly assumes ergodicity.

[By the way, $\mathrm{P}_\nu$ is not $\nu\pi$ -- which would just be the same as $\nu$ if $\nu=\mu$ -- nor is it any other kind of "composition" of $\nu$ with $\pi$. It is a measure on the sequence space, just as you have written, defined as the law of a homogeneous Markov process with transition kernel $\pi$ and initial distribution $\nu$.]

$^\ast$Ergodicity of $\mu$ (with respect to $\pi$) is defined as meaning that the equivalent conditions in Theorem 2.1 are satisfied.

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