Lemma 10.9.13D94E16CFCA6C

Proof. Let $\mathcal F_ A$ be the family of competitors for $T_ A$. It contains the constant $0$, so $T_ A\geq 0$.

We first show that if $\psi _ j\in \mathrm{PSH}(X,\omega )$ and $\sup _ X\psi _ j=0$ for $j\geq 1$, then $\Psi =\sum _{j\geq 1}2^{-j}\psi _ j$ belongs to $\mathrm{PSH}(X,\omega )$. By lemma 10.9.3(4), there is a constant $M$ with $\| \psi _ j\| _{L^1(X)}\leq M$ for all $j$. The partial sums $\Psi _ N=\sum _{j=1}^ N2^{-j}\psi _ j$ decrease with $N$, since $\psi _ j\leq 0$, and $\mathrm{dd}^{\mathrm{c}}\Psi _ N\geq -\sum _ j2^{-j}\omega \geq -\omega $. On a coordinate ball with a function $q$ as in (10.9.2), the functions $\Psi _ N+q$ are plurisubharmonic and decrease, so by theorem 5.10.1 their limit $\Psi +q$ is plurisubharmonic or identically $-\infty $ on the ball. By monotone convergence $\Psi _ N\to \Psi $ in $L^1(X)$ and $\int _ X|\Psi |\, \omega ^ n\leq \sum _ j2^{-j}M\leq M$, so $\Psi $ is finite almost everywhere and $\Psi +q$ is not identically $-\infty $ on any ball. Hence $\Psi $ is quasi-plurisubharmonic, and $\Psi \in \mathrm{PSH}(X,\omega )$ because the inequality $\omega +\mathrm{dd}^{\mathrm{c}}\Psi _ N\geq 0$ passes to the limit.

Suppose now that $\sup \{ \sup _ Xu:u\in \mathcal F_ A\} =\infty $. Choose $u_ j\in \mathcal F_ A$ with $b_ j=\sup _ Xu_ j\geq 2^ j$ for $j\geq 1$, and put $\psi _ j=u_ j-b_ j$. Then $\psi _ j\in \mathrm{PSH}(X,\omega )$, $\sup _ X\psi _ j=0$ and $\psi _ j\leq -2^ j$ on $A$, so $\Psi =\sum _ j2^{-j}\psi _ j$ belongs to $\mathrm{PSH}(X,\omega )$. Moreover $\Psi =-\infty $ on $A$, because every term $2^{-j}\psi _ j$ is at most $-1$ there and all terms are nonpositive. Thus $A$ is globally pluripolar.

Otherwise $T_ A\leq C$ for a constant $C$, and by lemma 10.9.3(1) the regularization $T_ A^*$ belongs to $\mathrm{PSH}(X,\omega )$ and satisfies $0\leq T_ A^*\leq C$. Thus at least one of the alternatives holds. They exclude each other: if $A\subset \{ \Phi =-\infty \} $ for a quasi-plurisubharmonic $\Phi $, choose $t>0$ with $t\Phi \in \mathrm{PSH}(X,\omega )$; then $t\Phi +c\in \mathcal F_ A$, hence $T_ A\geq t\Phi +c$, for every real $c$, so $T_ A^*$ is not bounded.

A quasi-plurisubharmonic function is locally integrable (lemma 6.14.4(1)), so its pole set has Lebesgue measure zero in every chart and contains no nonempty open set. Therefore a set with nonempty interior is not globally pluripolar.

Finally let $A_ j\subset \{ \varphi _ j=-\infty \} $, $j\geq 1$, with $\varphi _ j$ quasi-plurisubharmonic. After multiplying by small positive constants and subtracting constants, we may assume that $\varphi _ j\in \mathrm{PSH}(X,\omega )$ and $\sup _ X\varphi _ j=0$. By the claim proved at the start, $\sum _ j2^{-j}\varphi _ j$ belongs to $\mathrm{PSH}(X,\omega )$; it equals $-\infty $ on every $A_ j$, because one term is $-\infty $ there and all terms are nonpositive. (A finite union is the case in which $A_ j=\varnothing $ for all large $j$.) $\square$