I'll elaborate on my comment.
To make the notation more explicit, I'll write $N_K = \min \{ n : X_n \le -K \}$, and abbreviate $\{\omega : N_K(\omega) = \infty\}$ as $\{N_K = \infty\}$. Since $X_{n \wedge N_K} \ge -K-M$ for all $n$, $X_{n \wedge N_K}$ is a martingale which is bounded below, thus it converges a.s. But on the set $\{N_K = \infty\}$, $X_n = X_{n \wedge N_K}$ for all $n$, so $X_n$ converges a.s. on $\{N_K = \infty\}$.
$X_n$ need not converge on the set $\{N_K < \infty\}$. If $X_n$ is simple random walk, then $P(N_K < \infty) = 1$, and $X_n$ converges with probability 0.
(This is why the theorem read $P(C \cup D) = 1$ instead of $P(C) = 1$.)
To complete the proof, since $K$ was arbitrary, we have that $X_n$ converges a.s. on $\bigcup_K \{N_K = \infty\}$. This latter set is precisely $\{\liminf X_n > -\infty\}$. Replacing $X_n$ by $-X_n$ we can also show that $X_n$ converges a.s. on $\{\limsup X_n < \infty\}$. Together, these are equivalent to the statement $P(C \cup D) = 1$.
I think that the question is same as this.
I answered there as:
Here, you can prove that $$\{\lim Z_n / \mu^n > 0 \} = \{Z_n > 0 \mbox{ for all } n\} a.s.$$
With this note, let $N = \inf \{ n : Z_n = 0\}$ be a stopping time. Since $\rho^{Z_n}$ is martingale, $\rho^{Z_{n \wedge N}}$ is martingale, and thus $$\rho^x = \mathbb{E}[\rho^{Z_{0 \wedge N}}] = \mathbb{E}[\rho^{Z_{n \wedge N}}].$$ Since $\rho < 1$, we can apply the dominated convergence theorem, to get $$\rho^x = \rho^0 \mathbb{P}(N < \infty) + \rho^{\infty}\mathbb{P}(N = \infty) = \mathbb{P}(N < \infty)$$ where the first equality from the first notification that I gave above.
EDIT:
Above almost sure equality may hold under the condition that $\mathbb{P}(\lim Z_{n}/\mu^n = 0) < 1$ and I'm not sure whether it holds or not.
First, we may assume that $p_0 > 0$ since otherwise it becomes trivial.
Now, the following lemma is helpful:
Lemma let $X_n$'s be random variables taking values in $[0, \infty)$, and $D = \{ X_n = 0 \mbox{ for some } n \ge 1\}$. Assume that $\mathbb{P}(D | X_1, \cdots, X_n) \ge \delta(x) > 0$ a.s. on $\{X_n \le x\}$, $$\mathbb{P}(D \cup \{\lim X_n = \infty\}) = 1$$
Proof On $\{\liminf X_n \le M \}$, $X_n \le < M + 1$ infinitely often so $$\mathbb{P}(D | X_1, \cdots, X_n) \ge \delta(M+1) > 0$$ i.o.
By Levy's 0-1 law, LHS converges to $1_D$ so we have $\{\liminf X_n \le M \} \subseteq D$. Taking $M \to \infty$, $\{\liminf X_n < \infty \} \subseteq D$ a.s. and the result follows.
Now, from $p_0 > 0$, $\mathbb{P}(Z_{n+1} = 0 | Z_1, \cdots, Z_n) \ge p_0 ^k$ on $\{Z_n \le k\}$. Thus by the lemma, $$\mathbb{P}(\{Z_n = 0 \mbox{ for some } n\} \cup \{\lim Z_n = \infty\}) = 1$$ which allows us to do the calculation in the original answer.
Best Answer
We note that
$\begin{eqnarray} \mathbb{E}(\frac{Z_{n+1}}{\mu^{n+1}}|\mathcal{F}_n) &=& \frac{1}{\mu^{n+1}}\mathbb{E}(\xi_1^{n+1}+...+\xi_{Z_n}^{n+1}|\mathcal{F}_n)\\ &=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbb{E}((\xi_1^{n+1}+...+\xi_{k}^{n+1})\mathbf{1}_{\{Z_n = k\}}|\mathcal{F}_n)\\ &=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbf{1}_{\{Z_n = k\}}\mathbb{E}((\xi_1^{n+1}+...+\xi_{k}^{n+1})|\mathcal{F}_n); \ \text{because } \{Z_n = k\} \text{ is } \mathcal{F}_n\text{-measurable}\\ &=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbf{1}_{\{Z_n = k\}}(\mathbb{E}(\xi_1^{n+1})+...+\mathbb{E}(\xi_{k}^{n+1})); \ \text{because } \xi_j^{n+1} \text{ is independent of } \mathcal{F}_n\\ &=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbf{1}_{\{Z_n = k\}}k\mu = \frac{1}{\mu^n}\sum_{k=1}^{\infty}k\mathbf{1}_{\{Z_n = k\}} = \frac{1}{\mu^n}Z_n \end{eqnarray}$ And we have proved that $\frac{Z_n}{\mu^n}$ is a martingale.