Lemma 16.22.952ED2F2B0A63

Proof. Put $m=\int _ X\langle \theta _{V_\theta }^ n\rangle $. Since $u\preceq V_\theta $ and $\int _ X\langle \theta _ u^ n\rangle >0$, we can choose $a>1$ so close to one that $a^ n\bigl (m-\int _ X\langle \theta _ u^ n\rangle \bigr )<m$. Theorem 16.22.4 gives $\eta \in \mathrm{PSH}(X,\theta )$ with $a^{-1}\eta +(1-a^{-1})V_\theta \leq u$. Since the class is big, it contains a closed current $T\geq \delta _0\omega $ for some $\delta _0>0$, and by theorem 10.2.15 we can write $T=\theta +\mathrm{dd}^{\mathrm{c}}\psi $ with $\psi $ quasi-plurisubharmonic; after subtracting a constant, $\psi \in \mathrm{PSH}(X,\theta )$ satisfies $\psi \leq 0$ and $\theta _\psi \geq \delta _0\omega $, and $\psi \leq V_\theta $ because $\psi $ is a competitor for $V_\theta $. Therefore