Lemma 17.2.752C5A516B8C8

Proof. Put $\lambda _ u=\langle \theta _ u\wedge T_1\wedge \cdots \wedge T_{n-1}\rangle $ and define $\lambda _ v$ in the same way, so that $\mathrm{dd}^{\mathrm{c}}(u-v)\wedge R=\lambda _ u-\lambda _ v$ by (17.2.1). Since $u-v$ is bounded, corollary 13.8.8, applied after writing $T_ i=\theta _ i+\mathrm{dd}^{\mathrm{c}}\eta _ i$ with smooth $\theta _ i$ and $\eta _ i\in \mathrm{PSH}(X,\theta _ i)$ (theorem 10.2.15), gives $\lambda _ u(X)=\lambda _ v(X)$. Hence the integral does not change if a constant is added to $u$, and we may assume $u\leq v$. We must show that $\int _ X(v-u)\lambda _ v\leq \int _ X(v-u)\lambda _ u$.

Let $t\geq 0$ and $\varepsilon >0$, and put $q=\max \{ u,v-t-\varepsilon \} $. Then $v-t-\varepsilon \leq q\leq v$, so $\lambda _ q(X)=\lambda _ v(X)=\lambda _ u(X)$ by the same corollary. The sets $O_1=\{ u<v-t-\varepsilon \} $ and $O_2=\{ u>v-t-\varepsilon \} $ are disjoint and plurifine open, by proposition 5.47.25 applied in charts after adding the same smooth local potential of $\theta $ to $u$ and $v$; neither of them meets the common pole set of $u$ and $v$. Moreover $q=v-t-\varepsilon $ on $O_1$ and $q=u$ on $O_2$, so plurifine locality (proposition 13.2.6) gives $\lambda _ q=\lambda _ v$ on $O_1$ and $\lambda _ q=\lambda _ u$ on $O_2$, and hence

Since pole sets carry no mass, it follows that

As $\varepsilon \downarrow 0$ the sets $\{ u<v-t-\varepsilon \} $ increase to $\{ u<v-t\} $, and therefore

On the common finite locus the bounded function $v-u\geq 0$ equals $\int _0^\infty \mathbf1_{\{ u<v-t\} }\, \mathrm dt$. Integrating the last inequality in $t$ and using Tonelli's theorem, we obtain $\int _ X(v-u)\lambda _ v\leq \int _ X(v-u)\lambda _ u$, as required. $\square$