Lemma 16.23.4F4294ED919CC

Proof. Put $\lambda _ j=\mathbf1_{\{ q_ j=0\} }\mu _{q_ j}$. Since $q_ j\leq V_\theta \leq 0$, the set $\{ q_ j=0\} $ is contained in $\{ q_ j=V_\theta \} $, and lemma 16.13.1 gives $\lambda _ j\leq \mu _{V_\theta }$. Let $A$ be the proper analytic subset of lemma 16.13.2: potentials with minimal singularities are locally bounded on $X\setminus A$, and $\mu _{V_\theta }(A)=0$.

For $t>0$ put $b_ j=\max \{ 1+tq_ j,0\} $ and $b=\max \{ 1+tq,0\} $, with value zero at the poles. Fix a coordinate ball $D\Subset X\setminus A$ with a smooth potential $\rho $ of $\theta $ near $\overline D$ (lemma 5.4.2) and a number $L>1/t$, and put $\widetilde q_ j=\max \{ q_ j,V_\theta -L\} $ and $\widetilde q=\max \{ q,V_\theta -L\} $. On $D$ the functions $\widetilde q_ j$ are uniformly bounded and decrease to $\widetilde q$, so they converge in capacity (proposition 7.5.6). Since $V_\theta -L<-1/t$, the weight $b_ j$ vanishes on $\{ q_ j\leq V_\theta -L\} $, and on the complementary plurifine open set $\widetilde q_ j=q_ j$; hence $b_ j\mu _{q_ j}=b_ j\mu _{\widetilde q_ j}$ by plurifine locality, and likewise $b\mu _ q=b\mu _{\widetilde q}$. Moreover $b_ j=\max \{ 1+t\widetilde q_ j,0\} $ and $b=\max \{ 1+t\widetilde q,0\} $ on $D$, so the weights are bounded and quasi-continuous and converge in capacity to $b$. The weighted form of theorem 7.5.8, applied to the bounded plurisubharmonic functions $\rho +\widetilde q_ j$, therefore gives

Let $\delta >0$. Since $\mu _{V_\theta }(A)=0$, lemma 16.12.1(1) gives a compact set $K\subset X\setminus A$ with $\mu _{V_\theta }(X\setminus K)<\delta $; choose a continuous function $0\leq \zeta \leq 1$ with compact support in $X\setminus A$ and $\zeta =1$ on $K$ (proposition 5.2.6). Since $b_ j=1$ on $\{ q_ j=0\} $ and $\lambda _ j\leq \mu _{V_\theta }$,

Cover the support of $\zeta $ by finitely many balls $D_ l$ as above, and write $\zeta =\sum _ l\zeta _ l$ with continuous functions $\zeta _ l\geq 0$ with compact support in $D_ l$ (for instance $\zeta _ l=\zeta \psi _ l/\sum _ k\psi _ k$, where $\psi _ l\geq 0$ is continuous with compact support in $D_ l$ and $\sum _ k\psi _ k>0$ on the support of $\zeta $). The weak convergence on each $D_ l$ gives $\int _ X\zeta b_ j\, \mu _{q_ j}\to \int _ X\zeta b\, \mu _ q\leq \int _ Xb\, \mu _ q$, and letting then $\delta \downarrow 0$ gives $\varlimsup _ j\lambda _ j(X)\leq \int _ Xb\, \mu _ q$. Finally, as $t\to \infty $ the weights $b$, which depend on $t$, decrease to $\mathbf1_{\{ q=0\} }$, because $q\leq 0$; dominated convergence for the finite measure $\mu _ q$ proves the lemma. $\square$