Lemma 16.5.40.4C2A27A8F97A Let $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Then there is $\gamma \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ such that
4E22C768C562
16.5.41
4C2A27A8F97ALemma 16.5.40.4C2A27A8F97A Let $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Then there is $\gamma \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ such that
4E22C768C562
16.5.41
Proof. For each $j\geq 1$, let $\varphi _ j\coloneqq \varphi \lor (-j)$, $\psi _ j\coloneqq \psi \lor (-j)$. These functions are not necessarily $\theta $-psh. Let
By theorem 16.5.9, we can find $\gamma _ j\in \mathcal{E}(X,\theta ;\phi )$ such that
5B35DF529607
16.5.42
Take a constant $C>0$ so that $\psi -2C\leq \varphi \leq \psi +2C$. Let
Then $\eta \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ and we have the following inequalities:
Hence, $\gamma _ j\geq \eta $ by theorem 16.5.9. By theorem 16.5.9 again, $\gamma _ j$ is decreasing in $j$; let $\gamma =\inf _{j>0}\gamma _ j$. Then $\gamma \geq \eta $, hence $\gamma \in \mathcal{E}^{\infty }(X,\theta ;\phi )$.
Now observe that
as $j\to \infty $, by theorem 13.13.9. Finally observe that there is a constant $C'>0$ so that
since $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Therefore, for each $j>0$,
Hence, (16.5.41) follows from (16.5.42) and the dominated convergence theorem by letting $j\to \infty $. $\square$
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