Let $\Omega \subset \mathbb{R}^N$ be a smooth domain with bounded complement (e.g. $\Omega = \mathbb{R}^N \setminus \overline{B(0;1)}$), and $(u_n)$ be a bounded sequence in $H^1_0(\Omega) = W_0^{1,2}(\Omega)$.
Since $H^1_0(\Omega)$ is a reflexive Banach space, weak compactness implies the existence of a subsequence $(u_{n_k})$ and a function $u_0 \in H_0^1(\Omega)$ such that $u_{n_k} \rightharpoonup u_0$ weakly in $H^1_0(\Omega)$. May I assume that (going if necessary to a subsequence, which I still denote by $(u_{n_k})$) that $u_{n_k} \rightarrow u_0$ almost everywhere on $\Omega$?
According to the Rellich theorem, for $\omega \Subset \Omega$ (compactly embedded), the embedding $$H_0^1(\omega) \subset L^p(\omega)$$ is compact for all $1 \leq p < 2^*$. Hence, we may assume that $u_n \rightarrow u_0$ in $L^p(\omega)$, with $1 \leq p < 2^*$. Thus, by some theorem, there exists a subsequence (still denoted $(u_n)$) such that $u_n \rightarrow u_0$ almost everywhere on $\Omega$. Can we (as proposed here) exhaust $\Omega$ by balls of increasing radius and conclude that $u_n \rightarrow u_0$ almost everywhere on $\Omega$?
Thanks in advance.
Let's consider the following:
Fix the sequence $u_n\to u$ weakly in $H_0^1(\Omega)$ as you suggest in your post. Define $\Omega_k:=\Omega\cap B(0,k)$. Then each $\Omega_k$ is compact and hence we could consider to use Rellich theorem.
But before we do that, let me point out that the boundary $\partial \Omega_k$ may not be smooth, actually it can even be pretty bad, even if $\partial\Omega$ is smooth... So the Rellich theorem won't work here directly.
But now, let me wave my hand and assume that $\partial \Omega_k$ is good enough so that we could use Rellich on it. Take $k=1$, then by Rellich, from the sequence $u_n$ we could extract a future subsequence, call it $u_{n_{(1)}}$, such that $u_{n_{(1)}}\to u$ at least in $L^1(\Omega_1)$ strongly. Hence you know that $u_{n_{(1)}}\to u$ a.e. (subject to a subsequence of course) in $\Omega_1$. Continuous on this method, by diagonalization argument, you could obtain a subsequence, still denote as $u_n$, such that $u_n\to u$ weakly and $u_n\to u$ a.e. on $\Omega$. But be careful that you can not conclude $u_n\to u$ strongly in $L^1(\Omega)$, you could only say that $u_n\to u$ in $L^1_{\text{loc}}(\Omega)$
Having said that, now let's consider what if $\partial \Omega_k$ is not smooth, or, not a extension domain. Here is one thing can probably rescue us. On Leoni's book, exercise 11.17, he gives the following statement:
The space $W^{1,p}(\Omega)$ is compact embedded in $L^q(\Omega)$ when $q<p$ and $\Omega$ is finite.
In our argument, of course $\Omega_k$ is finite for each $k$ and hence we at least can have $H_0^1(\Omega_k)$ is compact embedded in $L^1(\Omega_K)$ and hence we are good.
However, here I must warn you that I only see this statement in Leoni's book but no-where-else. (On Adam's book, it has the inverse statement, i.e., compact embedding implies that $\Omega$ is finite.) I also receive concern from my friend that this exercise is wrong but they are not sure... Anyhow, at least I have a prove of this exercise by myself but maybe you could work out a proof yourself too...