Almost surely either $X_n=0$ for some $n$ or $\lim_{n\to\infty}X_n=\infty$ How I came to this:
Let $\{X_n\}_{n\in\mathbb{N}}$ be a sequence of non negative random variables, $\mathcal{F}_n=\sigma(X_l,l\leq n)$ the sequence of corresponding sigma-algebras and define $Z=\{X_n=0 \text{ for some } n\geq1\}$.
Assume there is a sequence of numbers $\varepsilon_k>0$ such that $P(Z|\mathcal{F}_n)\geq\varepsilon_k$ almost surely on $\{X_n\leq k\}$ for all $k,n\in\mathbb{N}$.
Show that almost surely either $X_n=0$ for some $n$ or $\lim_{n\to\infty}$ $X_n=\infty$.
My Problem:
I don not really understand how I can get information about $X_n$ or $\lim_{n\to\infty}$ $X_n$ and how one proves "almost surely either ... or" in a clever way.
Weak ideas so far:
One can consider $M_n=P(Z|\mathcal{F}_n)=E(\mathbb{1}_Z|\mathcal{F}_n)$ as a martingale (which seems reasonable since we get lots of helpful theory). Since $\mathbb{1}_Z$ is in $L_1$ we can show that $M_n$ is uniformly integrable. Therefore $M_n\rightarrow M_\infty$ a.s. and in $L^1$ with $L_1\ni M_\infty=E(\mathbb{1}_Z|\mathcal{F}_\infty)$ but I can't really see how I can go further from here.
Any suggestions are really appreciated. Thanks a lot! 
 A: Let $\Omega'$ be the set of probability one of the $\omega$'s such that for all $n,k$, 
$$\tag{*}
P(Z|\mathcal{F}_n)(\omega)\mathbf 1\{X_n\leq k\}(\omega)\geqslant\varepsilon_k\mathbf 1\{X_n\leq k\}(\omega)
$$
and $P(Z|\mathcal{F}_n)(\omega) \to P(Z|\mathcal{F}_\infty)(\omega)$ (that this set has probability one follows from the assumption and the martingale convergence theorem applied to the martingale $\left(P(Z|\mathcal{F}_n)\right)_{n\geqslant 1}$.
Let $\omega\in \Omega'$ be such that $\lim_{n\to\infty}X_n(\omega)=+\infty$ does not hold. This means that there exists an integer $k_0$ and an increasing sequence of integers $(n_l)_{l\geqslant 1}$ such that for all $l$, the inequalities $0\leqslant X_{n_l}(\omega)\leqslant k_0$ hold. By (*) applied with $k:=k_0$, we get that for all integer $n$, 
$$ 
P(Z|\mathcal{F}_n)(\omega) \geqslant\varepsilon_{k_0} 
$$ 
and letting $n$ going to infinity and accouting that $Z$ is $\mathcal F_\infty$-measurable, we get that $\mathbf 1_Z(\omega)\geqslant \varepsilon_{k_0} \gt 0$ hence there exists an integer $n$ such that $X_n(\omega)=0$.
