Proposition 12.10.2462BDDC322780

Proof. Choose $\alpha <\beta <1$ and $\varepsilon >0$ with $\varepsilon \beta C\leq \delta $. The number

is finite, and $v_\delta =-\varepsilon (1-\psi )^\beta -K\leq u$. The function $t\mapsto -\varepsilon (1-t)^\beta $ is increasing and convex on $t\leq 0$, with derivative at most $\varepsilon \beta $. Applying the chain inequality for quasi-plurisubharmonic functions to $\max (\psi ,-j)$ and passing to the decreasing limit as $j\to \infty $, we obtain $\mathrm{dd}^{\mathrm{c}}v_\delta \geq -\varepsilon \beta C\omega \geq -\delta \omega $. The limit is not identically $-\infty $ because $\psi $ is finite almost everywhere; at the poles of $\psi $ it takes the value $-\infty $, where the comparison with $u$ still holds. Thus $v_\delta \in \mathrm{PSH}(X,\delta \omega )$ is a competitor for $P_{\delta \omega }(u)$ for each $\delta $, and the Lelong estimate preceding theorem 12.10.23 proves the last assertion. $\square$