Convergence on every compact set implies convergence almost everywhere Suppose I have a sequence of functions {$u_n$} that converges to $v$ uniformly on every compact subset of $\mathbb{R}^n$. Suppose further that {$u_n$} converges to $u$ in $L^1$ for every compact subset of $\mathbb{R}^n$ aswell, which means that i can extract a sub sequence such that $u_{n_k}\rightarrow u $ a.e. Does it then follow that $u = v$ a.e.?
if so, why? is it because the "points at infinity" have zero meausure? (I don't really know any measure theory so forgive me if that is silly).
note that $u$ is not in $L^1(\mathbb{R}^n)$.
Thanks!
 A: We indeed have $u = v$ a.e.
This can be seen as follows. For $k\in \mathbb N,$ denote by $B_k \subseteq \mathbb R^n$ the closed Euclidean ball with center $0$ and radius $k.$ Note that $B_k$ is compact and $\mathbb R^n = \bigcup_{k\in \mathbb N}B_k.$ That's a generally useful fact - $\mathbb R^n$ can be $exhausted$ by an ascending sequence of compact sets. Now, pick a subsequence $u^{(1)}_n$ of $u_n$ such that $u^{(1)}_n \rightarrow u$ pointwise on $B_1$ except on a measure zero set $Z_1 \subseteq B_1.$ Next, pick a subsequence $u^{(2)}_n$ of $u^{(1)}_n$ such that $u^{(2)}_n \rightarrow u$ pointwise on $B_2$ except on a meausre zero set $Z_2 \subseteq B_2.$ This can be done since $u^{(1)}_n$ converges to $u$ in $L^1$ on $B_2.$ Continuing in this manner, we get a sequence of sequences $u^{(m)}_n$ such that $u^{(m+1)}_n$ is a subsequence of $u^{(m)}_n$ and $u^{(m)}_n \rightarrow u$ pointwise on $B_m$ except on a measure zero set $Z_m \subseteq B_m.$ Define the "diagonal" sequence
$$
w_n := u^{(n)}_n.
$$
Then, for every $k\in\mathbb N,$ the "tail" sequence $(w_n)_{n \geq k}$ is a subsequence of $(u^{(k)}_n)_{n\in\mathbb N}.$ This implies $w_n \rightarrow u$ on $\bigcup_{k \in \mathbb N}B_k = \mathbb R^n$ except on the measure zero set $\bigcup_{k\in \mathbb N}Z_k.$
Since, by assumption, we also have $w_n \rightarrow v$ pointwise on $\mathbb R^n,$ we get $u=v$ on $\mathbb R^n$ except on the measure zero set $\bigcup_{k\in \mathbb N}Z_k,$ as desired.
A: Note that the balls $B[0,R]$ are compact. By assumption,
$$u_{n_k} \to u \qquad \text{and} \qquad u_n \to v$$
almost everywhere on $B[0,R]$. Since pointwise limits are (almost everyhwere) unique, this implies that
$$N_R := \{x \in B[0,R]; u(x) \neq v(x)\}$$
has measure zero. Since
$$\{x \in \mathbb{R}^n; u(x) \neq v(x)\} = \bigcup_{n \in \mathbb{N}} \{x \in B[0,n]; u(x) \neq v(x)\} = \bigcup_{n \in \mathbb{N}} N_n$$
we get that the set $\{x; u(x) \neq v(x)\}$ has measure zero (it is a countable union of null sets).
