Theorem 16.8.2368E00C8C52A1

Proof. By our assumption and (16.3.8), we have $\varphi \preceq P_{\theta }[\varphi ]\leq P_{\theta }[\psi ]$. Due to proposition 16.3.10, the interval (16.8.24) remains unchanged after replacing $\psi $ by $P_{\theta }[\psi ]$. Fix $a$ in this interval. So by 13.13.30 and the discussions afterwards,

We first check that $\eta $ is a genuine $\theta $-psh function. Choose a closed non-pluripolar set $K\subseteq X$ on which $\varphi $ and $\psi $ are bounded. Then the obstacle $a\varphi +(1-a)\psi $ is bounded above on $K$. Every candidate in the definition of $\eta $ is therefore uniformly bounded above on $X$ by 5.17.4, so $\eta \not\equiv \infty $ and hence $\eta \in \mathrm{PSH}(X,\theta )\cup \{ -\infty \} $. On the other hand, since $\psi \preceq P_{\theta }[\psi ]$, we can find $C>0$ such that $\psi \leq P_{\theta }[\psi ]+C$. Then

quasi-everywhere, hence $\eta \geq \eta '-(a-1)C$. Thus $\eta '\preceq \eta $, and the monotonicity theorem theorem 13.13.11 gives $\eta \in \mathrm{PSH}(X,\theta )_{>0}$.

By (13.13.20),

quasi-everywhere. Equivalently, (16.8.25) holds quasi-everywhere. Since both sides of (16.8.25) are $\theta $-psh, proposition 5.14.6 upgrades the inequality to an everywhere inequality.

It remains to prove the assertion about the $P$-order. Using proposition 16.4.2, we get

By the assumption $\varphi \preceq _ P\psi $, we also have $P_{\theta }[\varphi ]\leq P_{\theta }[\psi ]$. Therefore,

quasi-everywhere and hence everywhere by proposition 5.14.6. It follows that $P_{\theta }[\eta ]\leq P_{\theta }[\varphi ]$, namely $\eta \preceq _ P\varphi $.

Finally assume that $\varphi \sim _ P\psi $. Then $\int _ X\theta _{P_{\theta }[\psi ]}^ n=\int _ X\theta _{\varphi }^ n$ by proposition 16.3.10. Applying corollary 13.13.42 gives

Since $\eta '\preceq \eta $, the monotonicity theorem theorem 13.13.11 gives

The reverse inequality follows from $\eta \preceq _ P\varphi $ and proposition 16.8.22. Hence $\eta \sim _ P\varphi $ by proposition 16.8.7. $\square$