Proposition 18.13.4AB19DE43A7F6

Proof. Put $Q=P_ h(F)$. The constant $\min _ XF$ is a competitor. The envelope construction of section 5.40, which preserves the exact lower current bound for the nonclosed form $h$, gives $Q\in \mathrm{PSH}(X,h)$ and

Since $Q$ is bounded, all its multiplier ideals are trivial. Thus $Q\sim _{\mathcal I}u$, making $Q$ itself a competitor for the ideal envelope below $F$. Every ideal competitor is an unrestricted competitor for $P_ h(F)$, so $P_{h,F}[u]_{\mathcal I}=Q$.

For $0<t<1$ set

Convexity gives $A_ t\in \mathrm{PSH}(X,h)$, and $w\leq 0$ implies $A_ t\leq (1-t)F-t\| F\| _\infty \leq F$. Boundedness of $Q$ and (18.13.3) with $b=1/t$ give $A_ t\preceq u$. Hence the ordinary envelope $O=P_{h,F}[u]$ has a competitor and satisfies $A_ t\leq O\leq Q$. Its regularized supremum is $h$-plurisubharmonic by the same envelope construction. Where $w$ is finite, $A_ t\to Q$ as $t\downarrow 0$. Thus $O=Q$ almost everywhere, and lemma 5.24.3 gives equality everywhere. For $F=0$, zero is an unrestricted competitor, so $P_ h(0)=0$. $\square$