Theorem 20.3.35.5CC5BF99E84F Choose $\Omega _0$ and $v$ as in proposition 20.3.24 and set $\Omega =2\Omega _0$. Then $T=\mathcal T_\Omega (u;Y)$ belongs to $\mathrm{PSH}(Y,\Omega )$ and satisfies all the divisorial inequalities in (20.3.25). Moreover,
\[ P_\Omega [T]_{\mathcal I}=T,\qquad P_\Omega [T]=T, \qquad P_{\Omega /2}(T)\not\equiv -\infty . \]
Writing $r=\dim Y$, its extended measure satisfies
\[ 0<\int _ Y\mathrm{MA}^{\mathrm{ext}}_\Omega (T)<+\infty , \qquad \mathrm{MA}^{\mathrm{ext}}_\Omega (T) \leq \mathbf1_{\{ T=0\} }\Omega ^ r. \]
Proof.
Since $\theta |_ Y+\epsilon _ m\omega |_ Y\leq \Omega $ for all sufficiently large $m$, the restricted approximants belong to $\mathrm{PSH}(Y,\Omega )$. Theorem 20.3.27, applied with $\delta _ m=0$, gives $T$, its defining inequalities and both fixed-point identities.
Normalize $v$ so that $\sup _ Yv=0$. Since $v\in \mathrm{PSH}(Y,\Omega /2)$ is a trace competitor, it lies below $T$ and is a competitor for $P_{\Omega /2}(T)$. The contact estimate for envelope fixed points, proposition 16.28.7, applied on $Y$, gives the bound for the measure and its finiteness, and lemma 16.28.69 gives strict positivity.
$\square$
Pome
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