Question
Let $(X_n)$ be a sequence of random variables taking values in $[0, \infty)$. Let $D=\{X_n=0\; \text{for some $n\geq 1$}\}$ and assume that
$$
P(D\mid X_{1}, \dotsc, X_n)\geq \delta(x)>0
\quad \text{a.s. on $\{X_n\leq x\}$}.\tag{0}$$
Use Lévy's Zero-One Law to conclude that $P(D\cup \{\lim_{n} X_n=\infty\})=1$.
My attempt Let $\mathcal{F}_n=\sigma(X_1, \dotsc, X_n)$ and $\mathcal{F}_{\infty}=\bigcup \mathcal{F}_n$. Since $D\in \mathcal{F}_{\infty}$, Lévy Zero-One Law implies that
$$
P(D\mid \mathcal{F}_n)\to I(D)\tag{1}
$$
where $I$ is the indicator function. For $m\geq 1$ let $A_m$ be the set on which $X_n\leq m$ eventually. By $(0)$ and $(1)$ it follows that $I(D)=1$ on the set $\bigcup_{m=1}^\infty A_m=\{\limsup X_n<\infty\}$ i.e.
$$
\bigcup_{m=1}^\infty A_m=\{\limsup X_n<\infty\}\subset D.
$$
Since $D\cup D^c\subset D\cup \{\limsup X_n=\infty\}$, it follows that $P(D\cup \{\limsup X_n=\infty\})=1$.
My problem
Assuming that everything above is correct, I have only been able to show that $P(D\cup \{\limsup X_n=\infty\})=1$. But ostensibly this is not enough since $P(D\cup \{\lim_{n} X_n=\infty\})\leq P(D\cup \{\limsup X_n=\infty\})$. Unless there is a typo in the question. But I got the question from Durrett.
Best Answer
Hint: Let $B_m$ be the event that $X_n\le m$ infinitely often, and replace $A_m$ with $B_m$ in your proof.