almost everywhere convergence

Question

Let $E\subseteq\mathbb{R}^l$ be s.t. $E$ is Lebesgue measurable and $m(E)>0$. Let for all $k\in\mathbb{N}$, $f_k:E\rightarrow \mathbb{R}$ be measurable functions. If for all $\epsilon>0$ we can find $F\subseteq E$ s.t. $F$ is closed and $m(F)\leq \epsilon$ AND $f_k\rightarrow f$ uniformly on $E-F$, is it true that $f_k\rightarrow f$ uniformly a.e. on $E$?

Answer 1

No. Take $f_n(x) = x^n$ on $[0,1]$. Then $f_n \to 0$ on any closed set that does not contain $1$.

However, $f_n$ does not converge uniformly to zero on the complement of any null set.

To see this, note that $m \{ x | f_n(x) > \frac{1}{2} \} = 1-\frac{1}{\sqrt[n]{2}} > 0$ for all $n$, so there are always 'lots' of points (from a measure zero perspective) that satisfy $f_n(x) > \frac{1}{2}$.