I am working through exercise 5.2.11 in Durrett's Probability: Theory and Examples (Version 5). The question is as follows:
Let $V_A = \inf \{n \ge 0 : X_n \in A\}$ and $g(x) = \mathbf{E}_x V_A$. Suppose that $S−A$ is finite and for each $x \in S−A$, $P_x(V_A < \infty)>0$. (i) Show that $(∗)$ g(x)=1+$\sum_{y} p(x,y)g(y)$ for $x \not \in A$ (ii) Show that if $g$ satisfies (∗), $g(X_{n \wedge V_A}) + n \wedge V_A$ is a martingale. (iii) Use this to conclude that $g(x) = \mathbf{E}_x V_A$ is the only solution of (∗) that is 0 on $A$.
I think that I have shown i), which seems to hold for any set $A \subset S$. We know that $\mathbf{E}_x [V_A-1 | X_1 = y] = \mathbf{E}_y[V_A]$ and then we can consider $\mathbf{E}_x[V_A-1]$ and use the law of total expectation, averaging over the first step to yield: $g(x) -1 = \mathbf{E}_x[V_A-1]= \sum_y p(x,y) \mathbf{E}_x[V_A-1 | X_1 = y] =\sum_y p(x,y) \mathbf{E}_y[V_A] =\sum_y p(x,y)g(y)$.
However, I am struggling to prove (ii) and (iii). For (ii), I need to show that $\mathbf{E} [g(X_{(n+1) \wedge V_A}) + (n+1) \wedge V_A | X_n] = g(X_{n \wedge V_A}) + n \wedge V_A$ but I'm stuck on proving this and still struggle to intuitively understand what this equation means in terms of probabilistic events. I know that for each $n$, $X_n$ takes values in $S$, which is countable. So I tried writing the first term of the conditional expectation out as: $\mathbf{E}[g(X_{(n+1) \wedge V_A}) | X_n] = \sum_y (g(X_{(n+1) \wedge V_A})| X_n=y) P(X_n=y)$. However, I'm not sure how to simplify this given that the only thing that I can assume about $g$ is that it satisfies $(*)$. I'd appreciate any help