Limit of smooth function (Sobolev space)

Question

If a sequence of smooth function $u_n$ converges to $u$ in $W^{k,p}$. Does that mean $u$ is continuous almost everywhere? Does $u_n$ converges to $u$ pointwise almost everywhere? I know converge in $W^{k,p}$ implies convergence in $L^p$, so there is a subsequence convnges to $u_n$ pointwise almost everywhere. I am not sure if that is still true for the original sequence.

Answer 1

Every measurable function is continuous a.e. by Lusin's theorem.

If $W^{k,p}$ embeds into the continuous functions, then you get convergence everywhere.

Otherwise, it does not hold. Indeed, for every $\varepsilon > 0$ you can find $v_\varepsilon \in W^{k,p}$ with $\|v_\varepsilon\|_{W^{k,p}} \le \varepsilon$ and $v_\varepsilon(0) = 1$. Then, you can construct a sequence $\{u_n\}$ by taking many shifted copies of $v_{1/m}$. If you do this carefully, $u_n(x)$ does not converge for all $x$.