Harmonic function and Markov chain

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 ?

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].



(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 $$r, we have $$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.

• @john: You are right, and I modified the detailed argument above accordingly. The argument you suggest in your comment is also fine. Feb 19 at 4:36
• We do not need to take a particular $\mu$ ? (Since $L_k$ depends on it and $h(x,k)$ should be equal to a constant that doesn't depend on the choice of $\mu$)
– john
Feb 19 at 6:25
• Apriori $L_k$ depends on $\mu$. In the argument above, we took any $\mu$ of full support and then (under the given hypothesis on tail events) we get $L_k=h(x,k)$ for all $x$. So aposteriori $L_k$ does not depend on $\mu$ (which must be the case.) Is there a particular point in the argument you find unclear or unconvincing? Feb 19 at 16:40
• Why $\mu$ (with full support) exists ?
– john
Feb 20 at 6:06
• Enumerate the state space $\{x_j\}_{j \ge 1}$ and define $\mu(x_j)=2^{-j}$ for all $j \ge 1$. Feb 20 at 15:57