Theorem 3.4.962E4A7EF2B4F

Proof. The limit $u$ is upper semicontinuous, since $\{ u<a\} =\bigcup _ j\{ u_ j<a\} $ is open for every real $a$. Let $x\in \Omega $ and $\overline{B(x,r)}\subset \Omega $. By theorem 3.4.5, $u_ j(x)\leq M_{u_ j}(x,r)$ for every $j$. If $M$ is a real upper bound for $u_1$ on $\partial B(x,r)$, the nonnegative functions $M-u_ j$ increase to $M-u$ on this sphere, so the monotone convergence theorem gives $M_{u_ j}(x,r)\downarrow M_ u(x,r)$. Hence $u(x)\leq u_ j(x)\leq M_{u_ j}(x,r)$ for all $j$, and in the limit $u(x)\leq M_ u(x,r)$. Thus $u$ is subharmonic.

Let $\Omega '$ be a connected component of $\Omega $. By proposition 3.4.8, $u$ is either identically $-\infty $ or locally integrable on $\Omega '$. In the first case, fix a compact set $K\subset \Omega '$ and a real number $A$. The open sets $\{ u_ j<-A\} $ increase with $j$ and cover $\Omega '$, because $u_ j(x)\downarrow -\infty $ for every $x\in \Omega '$. Finitely many of them, and hence one of them, cover $K$; thus $u_ j<-A$ on $K$ for all large $j$. In the second case, $u_1\geq u$ is not identically $-\infty $ on $\Omega '$ and is therefore locally integrable there as well. Then $0\leq u_ j-u\leq u_1-u$ almost everywhere on $\Omega '$, and $u_ j-u\to 0$ wherever $u$ is finite, so dominated convergence gives $\int _ K(u_ j-u)\, \mathrm d x\to 0$ for every compact $K\subset \Omega '$. $\square$