Proof.
1, 2, 3, 4 are immediate consequences of the corresponding properties proved in 18.1.23.
5 Using 3 and 4, it suffices to establish the following: Suppose that
5EAAF326E66A
18.1.29
\begin{equation} T=\sum _ i c_ i[E_ i] \end{equation}
is a closed positive $(1,1)$-current on $X$, where $\{ E_ i\} $ is a countable collection of prime divisors on $X$ and $c_ i>0$. Then
A35A4555844E
18.1.30
\begin{equation} \operatorname{vol}\left(T,T_2,\ldots ,T_ n \right)=0 \end{equation}
for any closed positive $(1,1)$-currents $T_2,\ldots ,T_ n$ on $X$.
Step 1. We first assume that (18.1.29) is a finite sum. Fix a Kähler form $\omega $ on $X$.
In this case, thanks to 3 and 4 again, we may assume that $T=[E]$ for some prime divisor $E$ on $X$. Write $[E]=\theta _{\varphi }$. Then $\varphi $ has analytic singularities thanks to proposition 5.20.4. Therefore, $P_{\theta +\epsilon \omega }[\varphi ]_{\mathcal{I}}\sim \varphi $ for any $\epsilon >0$ due to proposition 16.6.22. Write $T_ i=\theta _ i+\mathrm{dd}^{\mathrm{c}}\varphi _ i$ for $i=2,\ldots ,n$, and choose $C>0$ so that $C\omega +\theta _ i$ are Kähler forms for each $i=2,\ldots ,n$. Then
\[ \begin{aligned} & \operatorname{vol}(T,T_2,\ldots ,T_ n)\\ =& \lim _{\epsilon \to 0+} \int _ X \left([E]+\epsilon \omega \right)\wedge \left( \theta _2+\epsilon \omega +\mathrm{dd}^{\mathrm{c}}P_{\theta _2+\epsilon \omega }[\varphi _2]_{\mathcal{I}}\right)\wedge \cdots \\ & \wedge \left( \theta _ n+\epsilon \omega +\mathrm{dd}^{\mathrm{c}}P_{\theta _ n+\epsilon \omega }[\varphi _ n]_{\mathcal{I}}\right)\\ =& \lim _{\epsilon \to 0+} \epsilon \int _ X \omega \wedge \left( \theta _2+\epsilon \omega +\mathrm{dd}^{\mathrm{c}}P_{\theta _2+\epsilon \omega }[\varphi _2]_{\mathcal{I}}\right)\wedge \cdots \\ & \wedge \left( \theta _ n+\epsilon \omega +\mathrm{dd}^{\mathrm{c}}P_{\theta _ n+\epsilon \omega }[\varphi _ n]_{\mathcal{I}}\right)\\ \leq & \lim _{\epsilon \to 0+} \epsilon \int _ X \omega \wedge \left( \theta _2+C\omega \right)\wedge \cdots \wedge \left( \theta _ n+C\omega \right)\\ =& 0, \end{aligned} \]
where the inequality step follows from the monotonicity theorem theorem 13.13.11 and the fact that $P_{\theta _ i+\epsilon \omega }[\varphi _ i]_{\mathcal{I}}\leq 0$ for all $i=2,\ldots ,n$.
Step 2. We prove the case where $\{ E_ i\} $ is infinite. We may assume that $i$ runs over $\mathbb {Z}_{>0}$.
Write $T_ i=\theta _{i}+\mathrm{dd}^{\mathrm{c}}\varphi _ i$ for $i=2,\ldots ,n$ as before. Fix a Kähler form $\omega $ on $X$ so that $\theta _ i+\omega $ is a Kähler form for each $i=2,\ldots ,n$. Fix $\epsilon >0$.
We can find $N_0>0$ so that for any $N\geq N_0$, the class of
\[ \epsilon \omega + \sum _{i=N+1}^{\infty }c_ i[E_ i] \]
is Kähler. Take a Kähler form $\omega _ N$ in this class. Then the currents
\[ \sum _{i=1}^ N c_ i [E_ i]+\omega _ N,\quad \sum _{i=1}^{\infty } c_ i [E_ i]+\epsilon \omega \]
all lie in the same cohomology class. Moreover,
\[ \sum _{i=1}^{\infty } c_ i [E_ i]+\epsilon \omega \preceq _{\mathcal{I}}\sum _{i=1}^ N c_ i [E_ i]+\omega _ N. \]
Therefore, by the monotonicity in 5, followed by the finite divisorial case of Step 1,
\[ \begin{aligned} \operatorname{vol}\left(T,T_2,\ldots ,T_ n\right)\leq & \operatorname{vol}\left(T+\epsilon \omega ,T_2,\ldots ,T_ n\right)\\ \leq & \operatorname{vol}\left(\sum _{i=1}^ N c_ i [E_ i]+\omega _ N,T_2,\ldots ,T_ n\right)\\ =& \operatorname{vol}\left(\omega _ N,T_2,\ldots ,T_ n\right)\\ \leq & \int _ X \omega _ N\wedge (\theta _2+\omega )\wedge \cdots \wedge (\theta _ n+\omega ). \end{aligned} \]
Letting $N\to \infty $, and using $\{ \omega _ N\} \to \{ \epsilon \omega \} $, we get
\[ \operatorname{vol}\left(T,T_2,\ldots ,T_ n\right)\leq \epsilon \int _ X \omega \wedge (\theta _2+\omega )\wedge \cdots \wedge (\theta _ n+\omega ). \]
Since $\epsilon >0$ is arbitrary, we conclude (18.1.30).
6 Write $T=\theta _{\varphi }$. In more concrete terms, we need to show that
\[ \lim _{\epsilon \to 0+}\int _ X (\theta +\epsilon \omega +\mathrm{dd}^{\mathrm{c}}P_{\theta +\epsilon \omega }[\varphi ]_{\mathcal{I}})^ n=\int _ X (\theta +\mathrm{dd}^{\mathrm{c}}P_{\theta }[\varphi ]_{\mathcal{I}})^ n. \]
We may replace $\varphi $ by $P_{\theta }[\varphi ]_{\mathcal{I}}$ and assume that $\varphi $ is $\mathcal{I}$-model in $\mathrm{PSH}(X,\theta )$. Then we claim that
\[ \varphi =\inf _{\epsilon >0}P_{\theta +\epsilon \omega }[\varphi ]_{\mathcal{I}}. \]
From this, our assertion follows from proposition 16.4.4.
The $\leq $ direction is clear. For the converse, since $\varphi $ is $\mathcal{I}$-model, it suffices to show that for each prime divisor $E$ over $X$, we have
\[ \nu (\varphi ,E)\leq \nu \left(\inf _{\epsilon >0}P_{\theta +\epsilon \omega }[\varphi ]_{\mathcal{I}},E\right). \]
We simply compute
\[ \nu \left(\inf _{\epsilon >0}P_{\theta +\epsilon \omega }[\varphi ]_{\mathcal{I}},E\right)\geq \sup _{\epsilon >0}\nu \left(P_{\theta +\epsilon \omega }[\varphi ]_{\mathcal{I}},E\right)=\nu (\varphi ,E). \]
$\square$
Pome
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