Theorem 5.10.1FE144A69076E

Proof. The limit $u$ is upper semicontinuous, since $\{ u<c\} =\bigcup _ i\{ u_ i<c\} $ is open for every real $c$. Fix a connected component $X_0$ of $X$; it has a countable basis.

Step 1: a decreasing sequence with the same limit. Let $i_0\in I$. We show that there are indices $i_0\leq i_1\leq i_2\leq \cdots $ such that $u_{i_ l}\downarrow u$ pointwise on $X_0$. Replacing $I$ by the cofinal directed set $\{ i\in I:i\geq i_0\} $, which does not change the limit, we only have to find indices $i_1\leq i_2\leq \cdots $. For each rational $c$, the open sets $\{ u_ i<c\} \cap X_0$ cover $\{ u<c\} \cap X_0$, and countably many of them suffice: for each basis element of $X_0$ contained in some set $\{ u_ i<c\} $, choose one such index $i$; every point of $\{ u<c\} \cap X_0$ lies in such a basis element. Let $k_1,k_2,\ldots $ be an enumeration, with repetitions allowed, of the indices chosen for all rational $c$. Since $I$ is directed, we can choose inductively $i_1\geq k_1$ and $i_{l+1}\geq i_ l,k_{l+1}$. The sequence $(u_{i_ l})$ is decreasing and bounded below by $u$. If $u(x)<c$ with $c$ rational and $x\in X_0$, then $u_{k_ l}(x)<c$ for some $l$, hence $u_{i_ l}(x)\leq u_{k_ l}(x)<c$. Therefore $u_{i_ l}\downarrow u$ on $X_0$.

Step 2: the limit. Let $V\subset X_0$ be a connected coordinate neighborhood, identified with a domain in $\mathbb {C}^ n$. On each connected component of each affine complex-line section of $V$, the restrictions of the $u_{i_ l}$ form a decreasing sequence of subharmonic functions, whose limit is subharmonic by theorem 3.4.9. Since $u$ is upper semicontinuous, it is either plurisubharmonic on $V$ or identically $-\infty $ on $V$, by definition 5.8.1. As $X_0$ is connected, lemma 5.9.4 shows that either $u$ is plurisubharmonic on $X_0$ or $u\equiv -\infty $ on $X_0$.

Step 3: convergence. Suppose first that $u$ is plurisubharmonic on $X_0$. Fix $i_0\in I$, take the sequence of Step 1, and let $K$ be a compact subset of a coordinate neighborhood in $X_0$. Then $0\leq u_{i_ l}-u\leq u_{i_0}-u$ almost everywhere, namely wherever $u$ is finite, and $u_{i_0}-u$ is integrable on $K$. By dominated convergence, $\| u_{i_ l}-u\| _{L^1(K)}\to 0$. Given $\varepsilon >0$, choose $l$ with $\| u_{i_ l}-u\| _{L^1(K)}<\varepsilon $; for every $i\geq i_ l$ we have $0\leq u_ i-u\leq u_{i_ l}-u$ almost everywhere, hence $\| u_ i-u\| _{L^1(K)}<\varepsilon $. Thus the net converges to $u$ in $L^1(K)$, and covering an arbitrary compact subset of $X_0$ by finitely many such sets $K$ gives convergence in $L^1_{\mathrm{loc}}(X_0)$.

Suppose now that $u\equiv -\infty $ on $X_0$, and let $K\subset X_0$ be compact and $A>0$. The open sets $\{ u_ i<-A\} $ cover $K$, so finitely many of them, say those with indices $j_1,\ldots ,j_ p$, already cover $K$. If $i\geq j_1,\ldots ,j_ p$, then $u_ i\leq u_{j_ q}$ for each $q$, and hence $u_ i<-A$ on $K$. This is locally uniform convergence to $-\infty $. $\square$