Proof.
Step 1. We first reduce to the case where $\theta _1=\cdots =\theta _ n$, $\varphi _1=\cdots =\varphi _ n$ and $f_1=\cdots =f_ n$.
Assume for the moment that this special case is known. Let us handle the general case.
For $\lambda =(\lambda _1,\dots ,\lambda _ n)\in (\mathbb {R}_{>0})^ n$, we set
\[ \theta _\lambda \coloneqq \sum _{j=1}^ n\lambda _ j\theta _ j,\quad \varphi _\lambda \coloneqq \sum _{j=1}^ n\lambda _ j\varphi _ j,\quad f_\lambda \coloneqq \sum _{j=1}^ n\lambda _ j f_ j. \]
Then $\varphi _\lambda \in \mathrm{PSH}(X,\theta _\lambda )$, $f_\lambda \in C^{1,\overline{1}}(X)$, $\varphi _\lambda \leq f_\lambda $, and
\[ \bigcap _{j=1}^ n\{ \varphi _ j=f_ j\} \subseteq \{ \varphi _\lambda =f_\lambda \} . \]
So
\[ \mathds {1}_{\bigcap _ j\{ \varphi _ j=f_ j\} }(\theta _\lambda )^ n_{\varphi _\lambda } = \mathds {1}_{\bigcap _ j\{ \varphi _ j=f_ j\} }(\theta _\lambda )^ n_{f_\lambda }. \]
Both sides are polynomials in $\lambda _1,\dots ,\lambda _ n$; comparing the coefficient of $\lambda _1\cdots \lambda _ n$ gives (13.13.58).
From now on, we assume
\[ \theta =\theta _1=\dots =\theta _ n,\quad \varphi =\varphi _1=\cdots =\varphi _ n\in \mathrm{PSH}(X,\theta ), \]
and
\[ f=f_1=\cdots =f_ n\in C^{1,\overline{1}}(X) \]
with
\[ \varphi \leq f. \]
We wish to prove that
1FE4032CE3EE
13.13.59
\begin{equation} \mathds {1}_{\left\{ \varphi =f\right\} }\, \theta _{\varphi }^ n=\mathds {1}_{\left\{ \varphi =f\right\} }\, \theta _{f}^ n. \end{equation}
Step 2. We first handle the special case where $f\in \mathrm{PSH}(X,\theta )$.
Step 2.1. We assume that $\theta $ is a Kähler form, denoted by $\omega $ for clarity.
Pick a decreasing sequence $(\phi _ j)_ j$ in $C^{\infty }(X)$ with pointwise limit $\varphi $ and set
\[ \psi _ j\coloneqq P_{\omega }\left(\min \{ \phi _ j,f\} \right). \]
Then $\psi _ j\in \mathrm{PSH}(X,\omega )$, and
\[ \varphi \leq \psi _ j\leq \min \{ \phi _ j,f\} \leq f, \]
and hence $\psi _ j\searrow \varphi $. Applying 13.13.51 with $f_1=\phi _ j$ and $f_2=f$ gives
88AC8C43CD82
13.13.60
\begin{equation} \omega ^ n_{\psi _ j}\geq \mathds {1}_{\{ \psi _ j=f\} }\, \omega ^ n_ f \geq \mathds {1}_{\{ \varphi =f\} }\, \omega ^ n_ f. \end{equation}
Fix $C>0$ with $\min _ X f>-C$ and $\epsilon >0$. Let
\[ h_ j^{C,\epsilon }\coloneqq \frac{(\psi _ j+C)\lor 0}{\left((\psi _ j+C)\lor 0 \right)+\epsilon },\qquad h^{C,\epsilon }\coloneqq \frac{(\varphi +C)\lor 0}{\left((\varphi +C)\lor 0\right)+\epsilon }. \]
These functions are both quasi-continuous with values in $[0,1]$. Observe that $h_ j^{C,\epsilon }$ converges in capacity to $h^{C,\epsilon }$ as $j\to \infty $.
Now fix $g\in C^0(X)$, $g\geq 0$. Note that $h_ j^{C,\epsilon }=0$ on the set $\{ \psi _ j\leq -C\} $ and $h^{C,\epsilon }=0$ on the set $\{ \varphi \leq -C\} $. Using locality of the non-pluripolar product proposition 13.11.4 1, we have
\[ h_ j^{C,\epsilon }g \, \omega _{\psi _ j}^ n=h_ j^{C,\epsilon }g \, \omega _{\psi _ j\lor (-C)}^ n,\quad h^{C,\epsilon }g \, \omega _{\varphi }^ n=h^{C,\epsilon }g \, \omega _{\varphi \lor (-C)}^ n. \]
Therefore,
\[ \begin{aligned} \int _ X g\, \omega _{\varphi }^ n\geq & \int _ X h^{C,\epsilon }g \, \omega _{\varphi }^ n & \\ =& \int _ X h^{C,\epsilon }g \, \omega _{\varphi \lor (-C)}^ n & \\ =& \lim _{j\to \infty } \int _ X h_ j^{C,\epsilon }g \, \omega _{\psi _ j\lor (-C)}^ n & \textup{by theorem \href{../../tag/272876B00528/}{13.12.8}}\\ =& \lim _{j\to \infty } \int _ X h_ j^{C,\epsilon }g \, \omega _{\psi _ j}^ n & \\ \geq & \lim _{j\to \infty } \int _{\{ \varphi =f\} } h_ j^{C,\epsilon }g \, \omega _{f}^ n & \textup{by }(\href{../../tag/88AC8C43CD82/}{13.13.60})\\ =& \int _{\{ \varphi =f\} }h^{C,\epsilon }g \, \omega _{f}^ n, \end{aligned} \]
where the last equality follows from the dominated convergence theorem. By our choice of $C$, we have $h^{C,\epsilon }>0$ on $\{ \varphi =f\} $, and hence when $\epsilon \to 0+$, $h^{C,\epsilon }\to 1$ everywhere on this set. Letting $\epsilon \to 0+$, we find
\[ \int _ X g\, \omega _{\varphi }^ n\geq \int _{\{ \varphi =f\} }g \, \omega _{f}^ n \]
by the dominated convergence theorem.
By the Riesz representation theorem, we conclude that
\[ \omega _{\varphi }^ n\geq \mathds {1}_{\{ \varphi =f\} }\, \omega _{f}^ n. \]
Then by 13.13.14, the reverse inequality also holds, and (13.13.59) follows in this case.
Step 2.2. We no longer assume that $\theta $ is a Kähler form.
Choose a Kähler form $\omega $ on $X$ so that $\theta +\omega $ is a Kähler form and $f\in \mathrm{PSH}(X,\theta +\omega )$. Fix $g\in C^0(X)$ and consider, for $t\geq 0$,
\[ Q(t)\coloneqq \int _{\{ \varphi =f\} }g\, (\theta +t\omega +\mathrm{dd}^{\mathrm{c}}\varphi )^ n - \int _{\{ \varphi =f\} }g\, (\theta +t\omega +\mathrm{dd}^{\mathrm{c}}f)^ n. \]
By multilinearity of the non-pluripolar product, $Q$ is a polynomial in $t$. Step 2.1 applied to $\theta +t\omega $ for $t>1$ gives $Q(t)\equiv 0$ for $t>1$, hence $Q\equiv 0$, and in particular $Q(0)=0$. Since $g$ is arbitrary, (13.13.59) holds.
Step 3. We handle the general case. We no longer assume that $f\in \mathrm{PSH}(X,\theta )$.
Choose a Kähler form $\omega $ on $X$ so that $\theta +\omega $ is a Kähler form. Then
\[ \varphi \leq P_{\theta +\omega }(f)\leq f. \]
By 13.13.51 applied to $f_1=f_2=f$, we find that $f$ and $P_{\theta +\omega }(f)$ agree to the second order on the contact set $\{ P_{\theta +\omega }(f)=f \} $, so
\[ \mathds {1}_{\left\{ P_{\theta +\omega }(f)=f \right\} }\, \theta _{P_{\theta +\omega }(f)}^ n=\mathds {1}_{\left\{ P_{\theta +\omega }(f)=f \right\} }\, \theta _{f}^ n.{}^{\href{#sgpt-a0000018446}{8}} \]
However, Step 2 implies that
\[ \mathds {1}_{\left\{ P_{\theta +\omega }(f)=\varphi \right\} }\, \theta _{\varphi }^ n=\mathds {1}_{\left\{ P_{\theta +\omega }(f)=\varphi \right\} }\, \theta _{P_{\theta +\omega }(f)}^ n. \]
Combining these results, we conclude (13.13.59).
$\square$
Pome
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