A stationary sticky chain with finite variance and no root-n CLT

Let $\mathsf{X}=\{0,1\}$, $P(0,\{0\})=P(1,\{1\})=1$, and let $\pi (\{0\})=\pi (\{1\})=1\mathrm{/}2$. The identity kernel leaves $\pi$ invariant, so the chain initialized with $X_0\sim \pi$ is stationary by [F2]. Set $f(0)=-1$ and $f(1)=1$. Then $\pi (f)=0$ and $\pi (f^2)=1$.

The transition is deterministic, hence $X_k=X_0$ for every $k$ and $S_n:={\sum }_{k=0}^{n-1}f(X_k)=nf(X_0)$. Thus $S_n\mathrm{/}\sqrt{n}=\sqrt{n}f(X_0)$ takes values $\pm \sqrt{n}$ with equal probabilities. For each fixed $t>0$, once $n>t^2$ its distribution function at $t$ is $1\mathrm{/}2$, while any $N(0,{\sigma }^2)$ has distribution function strictly greater than $1\mathrm{/}2$ at $t$ when ${\sigma }^2>0$ and equal to $1$ when ${\sigma }^2=0$. Since $t$ is a continuity point of every such normal law, [F4] rules out convergence to any normal distribution.

Choose $\varphi =\pi$ and $A=\{1\}$. Then $\varphi (A)=1\mathrm{/}2>0$, but $P^n(0,A)=0$ for every $n\ge 1$ because the chain stays at $0$. By [F5] the chain is not $\varphi$-irreducible and is not Harris recurrent relative to $\varphi$. This counterexample shows only that finite variance without an ergodicity assumption does not guarantee a root-$n$ CLT.