Proposition 5.40.6F031A16FF76F

Proof. Put $m=\inf _ K\psi $ and $M=\sup _ K\psi $. Since $\theta $ is big, $\mathrm{PSH}(X,\theta )$ is nonempty, and $V_\theta \leq 0$; hence $V_\theta +m$ belongs to the family defining $R$. Applying lemma 5.40.1 with the constant obstacle $M$ gives a common upper bound $C$ for the larger family defining $Q$. The upper-envelope argument at the beginning of section 5.40 therefore shows that $R^*$ and $Q^*$ belong to $\mathrm{PSH}(X,\theta )$.

Now fix a competitor $u$ for $Q$ and define a bounded function on $X$ by

Then $u\leq f$ quasi-everywhere. By lemma 5.40.5, there are $u_ t\in \mathrm{PSH}(X,\theta )$ satisfying $u_ t\leq f$ everywhere and $u_ t\to u$ almost everywhere. Each $u_ t$ is a competitor for $R$, so $R^*\geq u$ almost everywhere, and lemma 5.24.3 gives this inequality everywhere. Taking the supremum over $u$ and regularizing yields $R^*\geq Q^*$; the inclusion of the competitor families gives the reverse inequality.

Write $F=R^*=Q^*$. Since $F\leq C$, the potential $F-C$ is a competitor for $V_\theta $. Thus

which proves minimal singularities and, when $\theta \geq 0$, boundedness.

Finally, if $E\subseteq K$ is pluripolar, the family constrained on $K\setminus E$ lies between the families defining $R$ and $Q$, so its regularized supremum also equals $F$. Note that this argument uses the upper bound obtained from the compact set $K$, even when $K\setminus E$ is not compact. $\square$