Uniform convergence on a measurable set implies convergence a.e.? Suppose for each $\epsilon$ there exists a measurable set $F$ such that $\mu(F^c) < \epsilon$ and $f_n$ converges to $f$ uniformly on $F$. Prove that $f_n$ converges to $f$ a.e.
I have been thinking about this question for a while and I am not quite sure how to proceed. My thoughts so far have been to try to think about the series of sets $F^c(\epsilon)$. Since the limit as $\epsilon$ approaches $0$ of $\mu(F^c(\epsilon))$ is $0$, I have been considering the limit as $\epsilon$ approaches $0$ of the $F^c(\epsilon)$ sets, to try to prove that all $f_n \rightarrow f$ for all $x$ not in that set. But, that feels like a dead end, since I can't seem to justify moving the limits inside the measure like I have, and I don't know how to proceed from there even if I could. Any thoughts?
Thanks.

This can be proved by contradiction. Suppose there is a measurable set $A$ and $\delta > 0$ such that $\mu(A) \ge \delta > 0$ and $f_n$ does not converge to $f$ pointwise on $A$, (such a set exists if $f_n$ does not converge a.e., then $f_n$ does not converge uniformly on any set intersecting $A$.)
Then, for all $\epsilon$, we have that $A \subset F^c(\epsilon)$, since it can't intersect $F(\epsilon)$ for any $\epsilon$. Hence, for $\epsilon < \delta$, we have that $A \subset F^c(\epsilon)$, but $\mu(A) \ge \delta > \epsilon > \mu(F^c(\epsilon)$, which contradicts monotonicity of the measure.
It can also be proved directly, but this seems fine.
