# Application of Lévy Zero-One Law

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.

Hint: Let $$B_m$$ be the event that $$X_n\le m$$ infinitely often, and replace $$A_m$$ with $$B_m$$ in your proof.