Q) Let $V_C = \text{inf}\{n\geq 0:X_n\in C\}$ and let $h(x)=P_x(V_A<V_B)$. Suppose $A\cap B = \emptyset$, $S-(A\cup B)$ is finite and $P_x(V_{A\cup B}<\infty)>0$ for all $x\in S-(A\cup B)$. Use the fact that $h(X_{\{n\wedge V_{A\cup B}\}})$ is a martingale and $$P_x(V_{A\cup B}>kN)\leq (1-\epsilon)^k \text{ (given the hypotheses of the question})$$ to show that $h(x) = P_x(V_A<V_B)$ is the only solution of
$$h(x) = \sum_y p(x,y)h(y)\text{ for }x\notin A\cup B\tag{*}$$
that is $1$ on $A$ and $0$ on $B$.
Ans) $h(X_{\{n\wedge V_{A\cup B}\}})$ is a martingale $\iff$ $h(x)=Eh(X_{\{n\wedge V_{A\cup B}\}})$ and since
$$P_x(V_{A\cup B}>kN)\leq (1-\epsilon)^k\implies\sum_k P_x(V_{A\cup B}>kN) = EV_{A\cup B}<\infty\implies V_{A\cup B}<\infty$$
But how do I know that any $h$ that satisfies $(*)$ is bounded or dominated so that $$h(x)=E_xh(X_{\{n\wedge V_{A\cup B}\}}) \rightarrow E_xh(X_{\{V_{A\cup B}\}}) = P_x(V_A<V_B).h(X_{V_A})+P_x(V_B<V_A).h(X_{V_B})$$ and I can claim that $h(x) = P_x(V_A<V_B)$ is the only solution of $(*)$ that is $1$ on $A$ and $0$ on $B$? Thanks.