Does $L^\infty$ convergence imply convergence almost surely?

Question

I'm not quite sure if the following proof is correct. Suppose we are working on the measure space $(\Omega, \mathcal{F}, P)$ and let $X$ denote the $L^{\infty}$ limit

Let $A = \{\omega \in \Omega: |X_n(\omega) - X(\omega)| \leq \|X_n - X\|_\infty \quad \forall n \in \mathbb{N}\}$

$A$ occurs almost surely since it is the intersection of countably many almost sure sets. Fix $\epsilon > 0 $ and $N$ such that $n \ge N$ implies $\| X_n - X\|_\infty < \epsilon $. Now for $\omega \in A$, we have for all $n \ge N$, $$|X_n(\omega) - X(\omega)| \leq \|X_n - X\|_\infty < \epsilon$$

so that $X_n(\omega) \rightarrow X(\omega)$ almost surely and, in fact, the convergence is uniform on $A$, i.e. almost sure convergence uniformly.

Answer 1

Yes, your proof is correct. There aren't any other subtleties.