Let $X_n$ be a Markov chain. Use Levy's zero-one law to show that if $$P(\cup_{m=n+1}^{\infty}\{X_m\in B_m\}|X_n)\geq\delta>0 \text{ on }\{X_n\in A_n\}\text{, then}$$ $$P(\{X_n\in A_n\text{ i.o.}\} - \{X_n\in B_n\text{ i.o.}\})=0$$
My attempt: $$\text{Let }C_n = \cup_{m=n+1}^{\infty}\{X_m\in B_m\}\implies \cap C_n=\{X_n\in B_n \text{ i.o.}\}$$ $$P(\cup_{m=n+1}^{\infty}\{X_m\in B_m\}|X_n) = E(1_{C_n}|X_n)=E(1_{C_n}|\mathcal{F}_n)\rightarrow E(1_{\cap C_n}|\mathcal{F}_{\infty}) = 1_{\cap C_n} \text{ by 0-1 law}$$
where $\mathcal{F}_n = \sigma(X_1,..,X_n)$ and the second equality holds because of Markov property. We know that on $$\{X_n\in A_n\}, E(1_{C_n}|X_n)\geq \delta>0 \text{ and thus on } \{X_n\in A_n \text{ i.o.}\}, E(1_{C_n}|X_n)\geq \delta>0 \text{ i.o.}$$ and since the limit of $E(1_{C_n}|X_n) = \cap C_n = \{X_n\in B_n \text{ i.o.}\}$, I guess $$\{X_n\in A_n \text{ i.o.}\}\subset \{X_n\in B_n \text{ i.o.}\}$$
May I know how that leads to the answer, I am missing something small. Thanks.
Since the convergence is hold a.s. You actually have shown $$1_{\{X_n\in A_n \text{ i.o.}\}}\leq 1_{\{X_n\in B_n \text{ i.o.}\}} \text{ a.s.}$$ Note that two sets $U$ and $V$ with $U\subset V$ imply that $U-V=\emptyset$. Hence $$\{X_n\in A_n \text{ i.o.}\}- \{X_n\in B_n \text{ i.o.}\}$$ is a null set and then has probability $0$.