I am seeking a clarification to this question on almost convergence in Sobolev spaces. The essence of the answer there was to prove that if a sequence $(u_n) \rightharpoonup u$ in $H_0^1({\mathbb{R}}^N)$, then upto subsequences, $u_k(x) \rightharpoonup u(x)$ a.e on $\mathbb{R}^N$. Since the embedding theorems cannot be directly applied for $\mathbb{R}^N$, the approach was to consider open balls of increasing radii and somehow prove the statement. This method was also given here.
I'm not sure if this method or the reasoning behind it works. If we define $\Omega_k = B(0,k)$, we can find, for each $k$, a subsequence $\{u_n^{(k)}\}_n$ that converges a.e to $u(x)$ on $\Omega_k$, but it is unclear how one would take the limit.
More precisely, for any $\varepsilon >0 $, let $N_k$ be an integer such that $n > N_k \implies |u_n^{(k)}(x) - u(x)| < \varepsilon, \text{ a.e on } \Omega_k$. To prove the statement, we need to find a subsequence $(u_n)$ such that there always exists an $N$ with the property $n > N \implies |u_n(x) - u(x)| < \varepsilon$. I don't see how one can prove that, since $N$ can "escape to infinity". Any help is much appreciated.
I believe you can use a "diagonal argument".
First, you extract a subsequence $\{ u_n^{(1)} \}$ from $\{ u_n \}$ such that $u_n^{(1)}$ converges pointwise inside $\Omega_1$.
Then, from this subsequence, you extract $\{ u_n^{(2)} \}$ which converges pointwise inside $\Omega_2$.
Iteratively, you defined the subsequence $\{ u_n^{(k+1)} \}$ as being extracted from $\{ u_n^{(k)} \}$ and converging pointwise inside $\Omega_{k+1}$.
Eventually, you define, say, $v_n := u_n^{(n)}$ (hence the "diagonal" name for the argument).
Now take any $x \in \mathbb{R}^N$. For $k$ large enough $x \in \Omega_k$. And for $n \geq k$ large enough, $v_n$ is an element of a subsequence of $\{ u^{(k)} \}$, so $v_n(x) \to u(x)$.