# Almost sure convergence and bounded supremum

Suppose $$\lim_{n→\infty} X_n=X$$ a.s. and $$|X|<\infty$$ a.s. Let $$Y=\sup_n|X_n|$$. Show that $$Y<\infty$$ a.s.

I stumbled across this problem while reading through Jacod and Protter's Probability Essentials, which assumes for the chapter in which the problem is stated that all random variables are real-valued.

I found this thread, with an answer, but it was not clear to me. So, I came up with my own attempt and I would appreciate any feedback regarding the attempt, whether it works, or, if it does not work, where the problem lies. Here is my attempt:

Without loss of generality, assume $$X=0$$ a.s. Otherwise, take $$Z_n = X_n - X$$, then $$Z_n \rightarrow 0$$ almost surely and $$\sup_n |Z_n| < \infty$$ if and only if $$\sup_n|X_n|<\infty$$ as $$|X|<\infty$$ by hypothesis.

Now, let $$N= \{\omega: \lim_{n \rightarrow \infty} X_n(\omega) \ne 0\}$$. Then, $$\lim_{n\rightarrow \infty}X_n = 0$$ everywhere on $$N^C$$. But this implies that there exists $$M\in \mathbb N$$ such that for all $$n>M$$, we have $$|X_n|<\epsilon$$. So, \begin{aligned} \sup_n |X_n| &= \max\left\{\max_{n\le M} |X_n|, \sup_{n>M} |X_n|\right\} \\& \le \max\left\{\max_{n\le M}|X_n|, \epsilon\right\} \\&< \infty \end{aligned} on $$N^C$$. As $$X_n \rightarrow 0$$ a.s., $$P(N^C) = 1$$ and we are done.

Your argument is OK, but he question has nothing to do with probability theory. For each $$\omega$$ such that $$X_n(\omega) \to X(\omega)$$ we have $$\sup_n |X_n(\omega)|<\infty$$ because if a sequence of real numbers is convergent then it is bounded. Hence your proof is really unnecessary.