Lemma 5.9.4598B886D700E

Proof. Let $A$ be the set of points of $X$ having an open neighborhood on which $v\equiv -\infty $, and $B$ the set of points having an open neighborhood on which $v$ is plurisubharmonic. Both sets are open, and they cover $X$ by hypothesis. They are disjoint: if $x\in A\cap B$, with open neighborhoods $W_1$ on which $v\equiv -\infty $ and $W_2$ on which $v$ is plurisubharmonic, then the restriction of $v$ to the nonempty open set $W_1\cap W_2$ is plurisubharmonic and identically $-\infty $, which is impossible because plurisubharmonic functions are locally integrable. Since $X$ is connected, either $X=A$, and then $v\equiv -\infty $, or $X=B$, and then $v$ is plurisubharmonic near every point, hence on $X$, because plurisubharmonicity is a local property. $\square$