Proposition 16.3.42.1F5A8A58FB4D Let $\varphi \in \mathrm{PSH}(X,\theta )$ and $f\in C^{1,\overline{1}}(X)$ with $\varphi \leq f$. Then
\[ \begin{aligned} \mathds {1}_{\left\{ \varphi =P_\theta (f)\right\} }\, \theta ^ n_{P_\theta (f)} =& \mathds {1}_{\left\{ \varphi =P_\theta (f)\right\} }\, \theta ^ n_\varphi , \\ \mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)\right\} }\, \theta ^ n_{P_\theta [\varphi ](f)} =& \mathds {1}_{\{ \varphi =P_\theta [\varphi ](f)\} }\, \theta ^ n_\varphi . \end{aligned} \]
Proof.
By corollary 13.13.61 and theorem 13.13.57, we have
\[ \mathds {1}_{\{ \varphi =P_\theta (f)\} }\, \theta ^ n_{P_\theta (f)}=\mathds {1}_{\{ \varphi =P_\theta (f)=f\} }\, \theta ^ n_{f}=\mathds {1}_{\{ \varphi =P_\theta (f)=f\} }\, \theta ^ n_{\varphi }\leq \mathds {1}_{\{ \varphi =P_\theta (f)\} }\, \theta ^ n_{\varphi }. \]
The reverse inequality follows from 13.13.14. The first equality follows.
As for the second, we compute using proposition 16.3.41 and theorem 13.13.57 that
\[ \begin{split} \mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)\right\} }\, \theta ^ n_{P_\theta [\varphi ](f)}=\mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)=f\right\} }\, \theta ^ n_{f}\\ =\mathds {1}_{\left\{ \varphi =P_\theta [\varphi ](f)=f\right\} }\, \theta _{\varphi }^ n\leq \mathds {1}_{\{ \varphi =P_\theta [\varphi ](f)\} }\, \theta ^ n_\varphi . \end{split} \]
Since $\varphi \leq P_{\theta }[\varphi ](f)$, 13.13.14 gives the opposite inequality. The second equality follows.
$\square$
Pome
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