Lemma 16.13.115A418789279

Proof. For $\varepsilon >0$ put $w_\varepsilon =\max \{ u,v-\varepsilon \} $. As $\varepsilon \downarrow 0$, $w_\varepsilon $ increases to $\max \{ u,v\} =v$, so $\mu _{w_\varepsilon }\to \mu _ v$ weakly along every sequence $\varepsilon \downarrow 0$, by corollary 13.8.9. On the plurifine open set $\{ u>v-\varepsilon \} $ we have $w_\varepsilon =u$, hence $\mathbf1_{\{ u>v-\varepsilon \} }\mu _{w_\varepsilon }=\mathbf1_{\{ u>v-\varepsilon \} }\mu _ u$. This set contains every point of $\{ u=v\} $ at which $u$ is finite, and $\mu _ u$ does not charge the pole set of $u$. Therefore

The right side is a fixed finite measure. Integrating a continuous function $\chi \geq 0$ against both sides and letting $\varepsilon \downarrow 0$, we obtain $\int _ X\chi \, \mu _ v\geq \int _ X\chi \mathbf1_{\{ u=v\} }\, \mu _ u$, so $\mu _ v\geq \mathbf1_{\{ u=v\} }\mu _ u$ by lemma 16.12.1. Restricting both sides to $\{ u=v\} $ proves the lemma. $\square$