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.
Best Answer
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.