The proof is by contradiction. Assume that there exists

\(z_0=(x_0,m_0)\) such that

\(h(z_0)\ne {\tilde{h}}(z_0)\). Let

Open image in new window be a Markov chain with transition kernel

*Q*. It can be easily checked that

\({\tilde{h}}\) is also a

*Q*-harmonic function, so that

\(h(Z_{n})\) and

\({\tilde{h}}(Z_{n})\) are bounded martingales that converge to

*H* and

\({\tilde{H}}\),

Open image in new window We have

Open image in new window , so that

Open image in new window . Assume, for instance, that

Open image in new window . The case

Open image in new window can be treated in the same way. Note first that there exist

\(a<b\) such that

Open image in new window . Let

\(A=\left\{ z\,:\;h(z)<a\right\} \),

\(B=\left\{ z\,:\;{\tilde{h}}(z)>b\right\} \). Since

\(a<b\), it follows that

Since

\(h(Z_n)\) converges to

*H* and

\(\tilde{h}(Z_n)\) converges to

\(\tilde{H}\), it follows that

Open image in new window , and since

Open image in new window , we have

Define

$$\begin{aligned} D_k = \bigcap _{n=k}^\infty \{Z_{n}\in A\cap B\} \;, \quad \quad D = \bigcup _{k\ge 0} D_k \;. \end{aligned}$$

Then (

11.A.3) implies that

Open image in new window . Define

Open image in new window . By the Markov property,

We first show that

Open image in new window . Indeed, since

\(D_k\) is increasing, we have for all

\(m\le k\),

Letting

*k* then

*m* tend to infinity yields

We obtain

Thus

Let

*C* be an accessible small set. By Lemma

9.3.3, there exist a probability measure

\(\nu \),

\(\epsilon \in \left( 0,1\right] \), and

Open image in new window such that

*C* is both an

\((m,\epsilon \nu )\)- and an

\((m+1,\epsilon \nu )\)-small set, i.e., for all

\(x\in C\),

$$\begin{aligned} P^m(x,\cdot ) \ge \epsilon \nu \;, \quad \quad P^{m+1}(x,\cdot ) \ge \epsilon \nu \;. \end{aligned}$$

(11.A.5)

Since

*C* is accessible, Proposition

10.2.2 implies that

Open image in new window . By (

11.A.4), it is also the case that

Open image in new window . Therefore, there exists

\(z_1=(x_1,n_1)\) such that

\(x_1\in C\) and

\(g(z_1)>1-(\epsilon /4)\nu (C)\), i.e.,

Define

$$\begin{aligned} C_0&= \left\{ x \in C\,:\;(x, n_1+m)\notin A\cap B\right\} \;, \\ C_1&= \left\{ x\in C\,:\;(x, n_1+m+1) \notin A\cap B\right\} \;. \end{aligned}$$

We have, using the first inequality in (

11.A.5),

This yields

\(\nu (C_0)<\nu (C)/4\). Similarly, using the second inequality in (

11.A.5), we obtain

Thus

\(\nu (C_1)<\nu (C)/4\), and altogether these two bounds yield

\(\nu (C_0\cup C_1)\le \nu (C)/2<\nu (C)\), and

*C* contains a point

*x* that does not belong to

\(C_0\cup C_1\), i.e.,

\((x, n_1+m)\in A\cap B\) and

\((x, n_1+m+1)\in A\cap B\). This contradicts (

11.A.2).

\(\Box \)