I think this may be a trivial question. But I could not find a proper reference due to my lack of knowledge. Suppose I have a Markov chain $X=(X_n)_{n\geq 1}$ on the uncountable (but compact!) state space $\mathcal{S}$. It is known that $X$ admits a stationary distribution $\pi$.
Let $\mathbb{P}_{\pi}$ denote the canonical law of $X$ on the path space $\mathcal{S}^{\mathbb{N}}$ and assume that the probability preserving dynamical system $(\mathcal{S}^{\mathbb{N}},\mathbb{P}_{\pi},\theta)$ is also ergodic where $\theta$ is the left-shift on $\mathcal{S}^{\mathbb{N}}$. By an application of pointwise Ergodic theorem and bounded convergence theorem, it is known that for $f\in C_b(\mathcal{S})$, we have \begin{equation} \tag{1} \lim\limits_{n\to\infty}\frac{1}{n}\sum_{l=1}^{n}\mathbb{E}_x[f(X_l)]=\int_{\mathcal{S}}f(y)\,\rm{d}\pi(y),\quad\text{ for }\ \pi \ \text{ a.e }\ x\in\mathcal{S}. \end{equation}
Suppose we also know that $\pi$ reversible. By this I mean if $P(x,dy)$ is the probability transition kernel and $f,g\in C_B(\mathcal{S})$ then, $$ \int_{\mathcal{S}} f(x)\int_{\mathcal{S}}g(y)P(x,dy)\,d\pi(x) = \int_{\mathcal{S}} g(x)\int_{\mathcal{S}}f(y)P(x,dy)\,d\pi(x).$$ My question is that if reversibility of $\pi$ is enough to get the strong convergence out of (1), i.e., does it hold, $$ \lim\limits_{n\to\infty}\mathbb{E}_x[f(X_n)] =\int_{\mathcal{S}}f\,\rm{d}\pi$$ for $\pi$ a.e $x$. I found the following reference (Theorem 2) https://www.ams.org/journals/bull/1962-68-02/S0002-9904-1962-10737-X/S0002-9904-1962-10737-X.pdf
which seems to suggest that at least $\lim\limits_{n\to\infty}\mathbb{E}_x[f(X_{2n})]$ exists for $\pi$ a.e $x$. I am not much familiar with the ergodic theory terminology.
Update: $\lim\limits_{n\to\infty}\mathbb{E}_x[f(X_{2n})]$ exists because, if $T\,:\,L^2(\pi)\to L^2(\pi)$ is the Markov operator, $$ Tf(x) := \int_{\mathcal{S}} f(y)\,P(x,dy), $$ then $T = T^*$ as $\pi$ is reversible. Also $T^nf(x) = \mathbb{E}_x[f(X_n)]$ for $f\in C_b(\mathcal{S})$. Thus by the Theorem 2, $\mathbb{E}_x[f(X_{2n})] = (T^n) (T^*)^nf(x)$ converges for $\pi$ a.e $x$.
As mentioned in the comment, I am ok with assuming that the chain is aperiodic.
Any help is very much appreciated!
From the convergence of $E_x f(X_{2n})$ you can deduce convergence of $E_x f(X_{2n+1})$ (by conditioning on $X_1$.) Convergence of $E_x [f(X_{n})+f(X_{n+1})]$ also follows.
Without further assumptions, the convergence of $E_x f(X_n)$ may fail even on finite spaces, e.g., consider simple random walk on a bipartite graph (or a single edge) if $f$ takes value $1$ on one side and $0$ on the other.