Proposition 13.9.5FEA17FAE6BAA

Proof. Let $\varepsilon >0$. The function $w=\max \{ u,v-\varepsilon \} $ is $\theta $-plurisubharmonic. The sets $\{ u<v-\varepsilon \} $ and $\{ u>v-\varepsilon \} $ are disjoint and plurifine open. Indeed, near each point we have $u-v=(u+h)-(v+h)$ for a smooth local potential $h$ of $\theta $, and $u+h$, $v+h$ are plurisubharmonic, so by proposition 5.47.25 the function $u-v$ is plurifinely continuous off the common pole set of $u$ and $v$, which is disjoint from both sets. Moreover $w$ equals $v-\varepsilon $ on the first set and $u$ on the second. By plurifine locality (proposition 13.2.6),

The left side is at most $\operatorname {vol}(\{ \theta \} )$ by proposition 13.9.2. The complement of $\{ u>v-\varepsilon \} $ consists of $\{ u\leq v-\varepsilon \} $, which is contained in $\{ u<v\} $ up to the common pole set of $u$ and $v$, and that pluripolar set carries no mass. Hence

As $\varepsilon \downarrow 0$, the sets $\{ u<v-\varepsilon \} $ increase to $\{ u<v\} $, which proves the inequality. If $u$ has full mass, the last two terms cancel. If $u\geq v-O(1)$, then $w-u$ is bounded, so $\mu _ w(X)=\mu _ u(X)$ by corollary 13.8.8, and the same computation with $\mu _ u(X)$ in place of the volume gives the inequality without these terms. $\square$