Proposition 16.23.13499B9FD713C

Proof. Subtracting the same constant from $u$ and $v$ changes neither the hypothesis nor the conclusion, so we may assume that $u,v\leq 0$. Put $m=\int _ X\langle \theta _{\max \{ u,v\} }^ n\rangle $ and $\phi =P_\theta [\max \{ u,v\} ]$. By monotonicity $\int _ X\langle \theta _ u^ n\rangle \leq m$ and $\int _ X\langle \theta _ v^ n\rangle \leq m$, so the strict inequality forces both masses to be positive. Hence $m>0$, and by theorem 16.22.11, $\phi $ is a model potential of mass $m$. Moreover $u,v\leq \max \{ u,v\} \leq \phi $, because the nonpositive potential $\max \{ u,v\} $ is a competitor for $\phi $.

Since $\mu _ u$ and $\mu _ v$ do not charge pole sets, $\mu _ u(\{ u>\phi -k\} )\to \int _ X\langle \theta _ u^ n\rangle $ and $\mu _ v(\{ v>\phi -k\} )\to \int _ X\langle \theta _ v^ n\rangle $ as $k\to \infty $. Choose $k>0$ so large that

For $C>k$ put

Since $\phi -C$ is a common minorant, all three potentials have the singularity type of $\phi $ and mass $m$. As $C\to \infty $, $w_ C$ decreases to $P_\theta (u,v)$ by proposition 16.11.6. Suppose that $P_\theta (u,v)=-\infty $, so that $w_ C$ decreases to $-\infty $ identically. Then $\sup _ Xw_ C\to -\infty $ by corollary 10.3.6, and lemma 16.22.2 gives $w_ C\leq \phi -k$ for all large $C$. On $\{ w_ C=u_ C\} $ we then have $u_ C\leq \phi -k$, and similarly for $v_ C$, so corollary 16.13.5 gives

For the equality we used that, since $C>k$, $u_ C=u$ on the plurifine open set $\{ u>\phi -k\} $, so $\mu _{u_ C}=\mu _ u$ there, and similarly for $v$. This contradiction shows that $P_\theta (u,v)\ne -\infty $. $\square$