Let $X_k$ be a decreasing and uniformly bounded sequence of nonnegative random variables which is completely determined by $\mathcal F_k$ (not necessarily $\sigma(X_1,...,X_k)$). Mathematically $ X_{k+1}\leq X_k$ a.s. and $0\leq X_k\leq M$ a.s. for some $M\in\mathbb R$. Let $\tau_n$ be a sequence of $\mathcal F_k$-stopping time such that $\tau_n\to \infty $ almost surely. Assume one has: \begin{align} X_{\tau_n} \stackrel{\text{ a.s. }}{\longrightarrow} 0 \text{ as } n\to\infty \ \ \ \end{align} I think that this does not imply $X_k\to 0$ a.s., or in probability. But if it would hold for all sequences of stopping time, then we would have the implication $X_k\to 0$. Also if we would have $\tau_n\leq K_n\in\mathbb R$ a.s. then we would have $X_k\to 0$. That is all I know...
Question. Can we actually say something about the convergence of $X_k$? Some type of convergence, maybe under additional conditions?
I'm also satisfied by citing some references on this. Thank you!
Consider the following elementary statement: if $x_n$ is a sequence of nonnegative real numbers which is decreasing and has a subsequence converging to $0$, then the entire sequence $x_n$ converges to $0$.
You have $X_{T_n} \to 0$ a.s. for some (fixed) sequence of stopping times $T_n$.
This implies that (almost surely) the random sequence $(X_n)$ has a subsequence which converges to $0$.
But the random sequence $(X_n)$ is almost surely nondecreasing. This implies that $X_n \to 0$ a.s. You don't even need $T_n$ to be stopping times, just that they tend to $\infty$ a.s.