Lemma 16.3.32.BCC1B8FD1BED Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )$. Assume that (16.3.30) holds and
\[ \int _ X \theta _{\varphi \land \psi }^ n>0. \]
Suppose that $\gamma ,\eta \in \mathrm{PSH}(X,\theta )$ and satisfy
\[ \gamma \preceq \varphi ,\quad \eta \preceq \psi ,\quad \int _ X \theta _{\gamma }^ n=\int _ X \theta _{\varphi }^ n,\quad \int _ X \theta _{\eta }^ n=\int _ X \theta _{\psi }^ n. \]
Then $\gamma \land \eta \not\equiv -\infty $ and
725175FB4F3C
16.3.33
\begin{equation} \int _ X \theta _{\gamma \land \eta }^ n=\int _ X \theta _{\varphi \land \psi }^ n. \end{equation}
Proof.
Choose constants $A,B>0$ such that
\[ \gamma -A\leq \varphi ,\quad \eta -B\leq \psi . \]
Since
\[ (\gamma -A)\land (\eta -B)\sim \gamma \land \eta , \]
the non-pluripolar masses of these two rooftop potentials are equal by the monotonicity theorem theorem 13.13.11. Hence it suffices to prove the lemma after replacing $\gamma ,\eta $ by $\gamma -A,\eta -B$. We can therefore assume that
\[ \gamma \leq \varphi ,\quad \eta \leq \psi . \]
Step 1. We first show that $\gamma \land \psi \not\equiv -\infty $ and
500FC88C1939
16.3.34
\begin{equation} \int _ X \theta _{\gamma \land \psi }^ n=\int _ X \theta _{\varphi \land \psi }^ n. \end{equation}
For any $a>1$, we define
\[ \gamma _ a\coloneqq P_{\theta }\Bigl ( a\gamma +(1-a)\varphi \Bigr ). \]
When $a=1$, we simply write $\gamma _1=\gamma $.
Thanks to corollary 13.13.42, for any $a\geq 1$, we have $\gamma _ a\leq \varphi $ and
\[ \int _ X \theta _{\gamma _ a}^ n=\int _ X \theta _{\varphi }^ n. \]
By our assumption, for any $a\geq 1$, we have
\[ \int _ X \theta _{\gamma _ a}^ n+\int _ X \theta _{\varphi \land \psi }^ n> \int _ X \theta _{\varphi }^ n,\quad \gamma _ a\preceq \varphi ,\quad \varphi \land \psi \preceq \varphi . \]
So proposition 16.3.25 gives
\[ \gamma _ a\land \varphi \land \psi =\gamma _ a\land \psi \in \mathrm{PSH}(X,\theta ). \]
In particular, for $a=1$, we find $\gamma \land \psi \not\equiv -\infty $. It remains to verify (16.3.34).
For $a>1$, we have
\[ \gamma \geq a^{-1}\gamma _ a+\left(1-a^{-1}\right)\varphi , \]
thanks to the definition of $\gamma _ a$ (cf. (13.13.34)). Hence
\[ \gamma \land \psi \geq a^{-1}\left( \gamma _ a\land \psi \right)+\left(1-a^{-1}\right)\left(\varphi \land \psi \right). \]
Therefore, by the monotonicity theorem theorem 13.13.11,
\[ \int _ X \theta _{\gamma \land \psi }^ n\geq \left(1-a^{-1}\right)^ n \int _ X \theta _{\varphi \land \psi }^ n. \]
Letting $a\to \infty $ and taking theorem 13.13.11 into account, (16.3.34) follows.
Step 2. We complete the proof.
For any $a>1$, we let
\[ \eta _ a\coloneqq P_{\theta }\Bigl ( a\eta +(1-a)\psi \Bigr ). \]
When $a=1$, we set $\eta _1=\eta $. Thanks to corollary 13.13.42, for each $a\geq 1$, we have $\eta _ a\leq \psi $ and
\[ \int _ X \theta _{\eta _ a}^ n=\int _ X \theta _{\psi }^ n. \]
From Step 1, we have
\[ \int _ X \theta _{\gamma \land \psi }^ n+\int _ X \theta _{\eta _ a}^ n=\int _ X \theta _{\varphi \land \psi }^ n+\int _ X \theta _{\psi }^ n>\int _ X \theta _{\psi }^ n. \]
Applying proposition 16.3.25 again, we find
\[ \gamma \land \psi \land \eta _ a =\gamma \land \eta _ a\in \mathrm{PSH}(X,\theta ). \]
When $a=1$, we find $\gamma \land \eta \in \mathrm{PSH}(X,\theta )$. It remains to prove (16.3.33).
By definition of $\eta _ a$, we have
\[ \eta \geq a^{-1}\eta _ a+\left(1-a^{-1}\right)\psi . \]
Therefore,
\[ \gamma \land \eta \geq a^{-1} (\gamma \land \eta _ a)+\left(1-a^{-1}\right) \left( \gamma \land \psi \right). \]
By the monotonicity theorem theorem 13.13.11,
\[ \int _ X \theta _{\gamma \land \eta }^ n\geq \left(1-a^{-1}\right)^ n\int _ X \theta _{\gamma \land \psi }^ n. \]
Letting $a\to \infty $ and applying theorem 13.13.11 again, we arrive at (16.3.33).
$\square$
Pome
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