Proof.
Since $\nu (V_\theta , Y)=0$, we have $Y\not\subseteq \operatorname{nK}(\alpha +\epsilon \{ \omega \} )$ for all $\epsilon >0$. Therefore, by (20.2.31) and (20.2.33), only the last equality of (20.2.35) needs to be argued.
Thanks to 16.9.50, we have
\[ V_{\theta +\epsilon \omega }\xrightarrow {d_ S}V_{\theta } \]
as $\epsilon \to 0+$.
Therefore, using proposition 20.2.22, we find
\[ \operatorname{Tr}_ Y\left( V_{\theta +\epsilon \omega } \right) \xrightarrow {d_ S} \operatorname{Tr}_ Y\left(V_{\theta }\right). \]
Set
\[ M\coloneqq \lim _{\epsilon _0\to 0+}\int _ Y \left(\theta |_ Y+\epsilon _0\omega |_ Y\right)^ m_{\operatorname{Tr}_ Y^{\theta +\epsilon _0\omega }(V_{\theta })}. \]
By example 20.2.17, if $\operatorname{Tr}_ Y^{\theta }(V_{\theta })$ is defined, then
\[ M=\int _ Y (\theta |_ Y)^ m_{\operatorname{Tr}_ Y^{\theta }(V_{\theta })}, \]
whereas if $\operatorname{Tr}_ Y^{\theta }(V_{\theta })$ is not defined, then $M=0$. Thus $M$ is the last term of (20.2.35), with the convention made in the statement.
Thanks to theorem 16.9.30, for any $\epsilon _0>0$, we have
\[ \begin{aligned} \lim _{\epsilon \to 0+}\int _ Y \left(\theta |_ Y + \epsilon _0 \omega |_ Y\right)^ m_{\operatorname{Tr}_ Y^{\theta + \epsilon _0 \omega }(V_{\theta + \epsilon \omega })} & = \int _ Y (\theta |_ Y + \epsilon _0 \omega |_ Y)^ m_{\operatorname{Tr}_ Y^{\theta + \epsilon _0 \omega }(V_{\theta })}. \end{aligned} \]
By the definition of $M$, this implies the existence of $(\epsilon _ j)_{j>0}$ such that $0<\epsilon _ j<1/j$ and
6EEF71F49B54
20.2.36
\begin{equation} \lim _{j\to \infty }\int _ Y \left(\theta |_ Y + j^{-1}\omega |_ Y\right)^ m_{\operatorname{Tr}_ Y^{\theta + j^{-1} \omega }(V_{\theta + \epsilon _ j \omega }) } = M. \end{equation}
Moreover, for each $j>0$,
E74D68200A33
20.2.37
\begin{equation} \begin{aligned} \int _ Y \left(\theta |_ Y + j^{-1} \omega |_ Y\right)^ m_{\operatorname{Tr}_ Y^{\theta + j^{-1} \omega }(V_{\theta + \epsilon _ j \omega }) } \geq & \int _ Y \left(\theta |_ Y + \epsilon _ j \omega |_ Y\right)^ m_{\operatorname{Tr}_ Y^{\theta + \epsilon _ j\omega }(V_{\theta + \epsilon _ j \omega }) } \\ \geq & M. \end{aligned} \end{equation}
Indeed, the first inequality follows from the multi-linearity of the non-pluripolar products proposition 13.11.4 6. For the second one, if $M=0$ there is nothing to prove; if $M>0$, then example 20.2.17 shows that $\operatorname{Tr}_ Y^\theta (V_\theta )$ is defined, and the inequality follows from the multi-linearity of the non-pluripolar products proposition 13.11.4 6 and the monotonicity theorem proposition 16.8.22. Putting (20.2.36) and (20.2.37) together, it follows that
\[ \lim _{j\to \infty }\int _ Y \left(\theta |_ Y + \epsilon _ j \omega |_ Y\right)^ m_{\operatorname{Tr}_ Y^{\theta + \epsilon _ j \omega }(V_{\theta + \epsilon _ j \omega }) } = M. \]
Finally, since $\int _ Y \left(\theta |_ Y + \epsilon \omega |_ Y + \mathrm{dd}^{\mathrm{c}}\operatorname{Tr}_ Y^{\theta + \epsilon \omega }(V_{\theta + \epsilon \omega }) \right)^ m$ depends monotonically on $\epsilon >0$, we conclude that
\[ \lim _{\epsilon \to 0+}\int _ Y \left(\theta |_ Y + \epsilon \omega |_ Y + \mathrm{dd}^{\mathrm{c}}\operatorname{Tr}_ Y^{\theta + \epsilon \omega }(V_{\theta + \epsilon \omega }) \right)^ m=M. \]
This completes the proof.
$\square$
Pome
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