Corollary 13.13.42.992A84CBAC5D Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )_{>0}$, $\varphi \preceq \psi $. Assume that $\int _ X \theta _{\varphi }^ n=\int _ X \theta _{\psi }^ n$. Then for any $\epsilon \in (0,1)$, there is $\eta \in \mathrm{PSH}(X,\theta )$ such that
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$\int _ X \theta _{\eta }^ n=\int _ X \theta _{\varphi }^ n$;
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$\epsilon \eta +(1-\epsilon )\psi \leq \varphi $.
In fact, we can take
\[ \eta =P_{\theta }\left(\epsilon ^{-1}\varphi +(1-\epsilon ^{-1})\psi \right). \]
Proof.
Fix $\epsilon \in (0,1)$. We define $\eta $ as in the statement of the corollary. By 13.13.30, $\eta \in \mathrm{PSH}(X,\theta )_{>0}$. Observe that 2 is satisfied by (13.13.34).
Thanks to (13.13.41), for each $a'>\epsilon ^{-1}$, we have
\[ \int _ X \theta _{\eta }^ n>\left(\frac{a'-\epsilon ^{-1}}{a'}\right)^ n\int _ X \theta _{\varphi }^ n. \]
Letting $a'\to \infty $, we conclude that
\[ \int _ X \theta _{\eta }^ n\geq \int _ X \theta _{\varphi }^ n. \]
On the other hand, since $\eta \preceq \psi $, using the monotonicity theorem theorem 13.13.11 we find that
\[ \int _ X \theta _{\eta }^ n\leq \int _ X \theta _{\psi }^ n= \int _ X \theta _{\varphi }^ n. \]
Hence,
\[ \int _ X \theta _{\eta }^ n= \int _ X \theta _{\varphi }^ n. \]
Therefore, 1 holds as well.
$\square$
Pome
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