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Let $X$ be a Markov chain with countable state space $S$ and transition kernel $P$, and let $h \colon S \to [0,1]$ be a sub-harmonic or super-harmonic function. Assume that for all $\varepsilon >0$ there exists $\delta >0$ such that

$$ \forall x \in S : \ \varepsilon \leq h(x) \leq 1 - \varepsilon \Rightarrow \mathbb{P}^x [ | h(X_1 ) - h(x) | \geq \delta ] \geq \delta .$$

Then I would like to show that $\lim_{k \to \infty} h(X_k) \in \{ 0, 1 \}$ almost surely. I think I can use martingale convergence to show that the limit exists and then somehow show that it cannot take values in $(0,1)$.

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$\newcommand\ep\varepsilon\newcommand\de\delta$You were almost there.

Indeed, without loss of generality, $h$ is sub-harmonic. So, $(Y_n)$ is a bounded submartingale, where $Y_n:=h(X_n)$. So, $Y_n\to Y$ (as $n\to\infty$) almost surely for some random variable $Y$ with values in $[0,1]$.

Take any real $\ep>0$. Then, by your displayed condition, for some real $\de>0$ and all $n$ such that $P(Y_n\in[\ep,1-\ep])>0$ we have $$P(|Y_{n+1}-Y_n|\ge\de\,|\,Y_n\in[\ep,1-\ep])\ge\de$$ and hence $$P(Y_n\in[\ep,1-\ep])\le\frac1\de\,P(|Y_{n+1}-Y_n|\ge\de)\to0,$$ since $Y_n\to Y$. So, $P(Y\in(\ep,1-\ep))=0$, for each real $\ep>0$. So, $P(Y\in\{0,1\})=1$. That is, $P(\lim_n h(X_n)\in\{0,1\})=1$. $\quad\Box$

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