Harmonic function and Markov chain

Question

Let $X=(X_k)_{k \in \mathbb{N}}$ be a Markov chain with countable countable state space $S$ and transition matrix $P.$ Let $\mathcal{T}$ be the tail $\sigma$-field of $X:\mathcal{T}=\bigcap_{k \in \mathbb{N}}\sigma(X_k,X_{k+1},...).$

Suppose that for every initial distribution $\mu$ and every $F \in \mathcal{T},P_{\mu}(F) \in \{0,1\}.$ Consider Theorem $5.7.4$ from Durrett's book (picture). How can we prove using this theorem that in this case every bounded space-time harmonic function $h(x,k)$ ($x \in S,k \in \mathbb{N}$) is constant? Is there a reference for this ?

Answer 1

Edited following comment from O.P.

Claim Suppose that $h(x,k)$ is a bounded space-time harmonic function. Suppose that $P_\mu(F)\in\{0,1\}$ for every tail event $F$ and initial distribution $\mu$. Then there exists a real $L$ such that $h(x,k)=L$ for every state $(x,k)$ of the space-time chain.

Proof: For any initial distribution $\mu$ and integer $k \ge 0$, the limit $Y_k:=\lim_n h(X_n,n+k)$ exists almost surely, and is tail-measurable. (See footnote 1.) Since tail events have $\mu$ measure 0 or 1, there exists $L_k$ such that $P_\mu(Y_k= L_k)=1 \quad (*) \,.$ (See footnote 2.)

Now Choose $\mu$ that has full support, i.e., satisfies $\mu(x)>0$ for every $x$ in the state space. Let $(x,k)$ be a state of the space-time chain. Then $$h(x,k)=E_\mu[h(X_n,n+k)|X_0=x] \,.$$ By the bounded convergence theorem, the right-hand side converges to $L_k$, so $h(x,k)=L_k$. Finally, since $h(x,k)=E_x[h(X_1,k+1)]$, we conclude that $L_k=L_{k+1}$ for all $k$. $\qquad$ QED

This type of argument goes back to Blackwell [1].

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(Footnote 1) (In Durrett's notation, $Y_k=Z\circ \theta^k$ where $\theta$ is the left-shift on sequence space.)

(Footnote 2) This is standard, included for completeness: For every real $r$, we have $P_\mu(Y_k>r)\in \{0,1\}$. Define $$ L_k:= \sup\{ r>0 : P_\mu(Y_k>r)=1\}.$$ Then for every rational $r<L_k$, we have $P_\mu(Y_k>r)=1$, so taking a countable intersection over all such $r$ gives $P_\mu(Y_k \ge L_k)=1$. For every rational $q>L$, we have $P_\mu(Y_k>q)=0$, so taking a countable union over all such $q$ gives $P_\mu(Y_k> L_k)=0$. Thus, (*) holds.

[1] Blackwell, David. "On transient Markov processes with a countable number of states and stationary transition probabilities." The Annals of Mathematical Statistics (1955): 654-658.