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Proposition 17.13.38 (Hisamoto)3746B0074A79

Proposition 17.13.38 (Hisamoto).3746B0074A79 Let $\ell \in \mathcal{R}(X,\theta ;\phi )$. Given any $\tau <\ell ^*_{\max }$, we have

16A5FF346A3B 17.13.39
\begin{equation} \int _ X \left(\theta +\mathrm{dd}^{\mathrm{c}}\ell ^*_{\tau }\right)^ n=\int _{\{ \dot{\ell }_0\geq \tau \} } \theta _{\phi }^ n. \end{equation}

Proof. Fix $\tau <\ell ^*_{\max }$. We first observe that it suffices to show that

0893138962D7 17.13.40
\begin{equation} \int _{\left\{ \dot{\ell }_0\geq \tau \right\} } \theta _{\phi }^ n=\int _{\left\{ \ell ^*_{\tau }=\phi \right\} } \theta _{\phi }^ n. \end{equation}

Assuming (17.13.40) for the moment. By theorem 17.13.21, $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$. In particular, $\ell ^*_\tau $ is a model potential and

\[ \ell ^*_\tau \leq \ell ^*_{-\infty }=\phi . \]

Since $\phi $ is model as well, proposition 16.3.41 gives

\[ \theta ^ n_{\ell ^*_{\tau }}=\mathds {1}_{\left\{ \ell ^*_{\tau }=0\right\} }\, \theta ^ n,\quad \theta ^ n_{\phi }=\mathds {1}_{\{ \phi =0\} }\, \theta ^ n. \]

Since $\ell ^*_{\tau }\leq \phi \leq 0$, we have

\[ \mathds {1}_{\left\{ \ell ^*_{\tau }=\phi \right\} }\, \theta _{\phi }^ n =\mathds {1}_{\left\{ \ell ^*_{\tau }=\phi =0\right\} }\, \theta ^ n =\mathds {1}_{\left\{ \ell ^*_{\tau }=0\right\} }\, \theta ^ n =\theta ^ n_{\ell ^*_{\tau }}. \]

Therefore, (17.13.39) follows from (17.13.40).

In order to prove (17.13.40), it suffices to show that

7F6505CF081C 17.13.41
\begin{equation} \left\{ \dot{\ell }_0\geq \tau \right\} =\left\{ \ell ^*_{\tau }=\phi \right\} \quad \textup{outside a pluripolar set}. \end{equation}

Take $x\in X$ so that $(\ell _ t(x))_{t\geq 0}$ is finite and right-differentiable at $t=0$. Note that this condition holds quasi-everywhere. Suppose that $\ell ^*_{\tau }(x)=\phi (x)$. Then

\[ \dot{\ell }_0(x)=\inf _{t>0} \frac{\ell _ t(x)-\phi (x)}{t}\geq \inf _{t>0} \frac{\ell ^*_{\tau }(x)+t\tau -\phi (x)}{t}=\tau . \]

Therefore, the $\supseteq $ direction in (17.13.41) follows. Conversely, suppose that $\dot{\ell }_0(x)\geq \tau $, then

\[ \ell ^*_{\tau }(x)=\inf _{t>0} \left(\ell _ t(x)-t\tau \right)\geq \inf _{t>0} \left(\phi (x)+t\dot{\ell }_0(x)-t\tau \right)\geq \phi (x). \]

Together with $\ell ^*_\tau \leq \phi $, this gives $\ell ^*_\tau (x)=\phi (x)$. Therefore, the $\subseteq $ direction in (17.13.41) follows as well. $\square$

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