Proof.
1 $\implies $ 2. By theorem 16.9.30, we have
\[ \lim _{j\to \infty } \int _ X \theta _{\varphi _ j}^ n=\int _ X \theta _{\varphi }^ n>0. \]
We may therefore assume that $\int _ X \theta _{\varphi _ j}^ n>0$ for all $j \geq 1$. It follows from proposition 16.6.22 that
\[ \int _ X \theta _{\varphi _ j}^ n=\operatorname{vol}\theta _{\varphi _ j} \]
for any $j\geq 1$. Using theorem 16.9.57, we conclude (18.1.3).
2 $\iff $ 3. By definition,
\[ \operatorname{vol}\theta _{\varphi }= \int _ X\theta _{P_{\theta }[\varphi ]_{\mathcal I}}^ n. \]
Moreover, $\varphi -\sup _ X\varphi \leq P_{\theta }[\varphi ]_{\mathcal I}$. Thus, if 2 holds, theorem 16.3.15 gives $P_{\theta }[\varphi ]_{\mathcal I}\leq P_{\theta }[\varphi ]$; the reverse inequality follows from proposition 16.6.18. Conversely, if 3 holds, then 2 follows from proposition 16.3.10.
3 $\implies $ 1. Note that the condition in 1 characterizes the closure of analytic singularities in $\mathrm{PSH}(X,\theta )$.
Step 1. We first assume that $\theta _{\varphi }$ is a Kähler current. We will prove the following more general result in this case: Without assuming 3, $P_{\theta }[\varphi ]_{\mathcal{I}}$ always lies in the closure of analytic singularities.
Let $(\varphi _ j)_ j$ be a quasi-equisingular approximation of $\varphi $ in $\mathrm{PSH}(X,\theta )$. We will show that $\varphi _ j\xrightarrow {d_ S}P_{\theta }[\varphi ]_{\mathcal{I}}$. Let
\[ \psi =\inf _{j\in \mathbb {Z}_{>0}} P_{\theta }[\varphi _ j]. \]
Since $\varphi \leq \varphi _ j$ for all $j>0$, we have
\[ \varphi -\sup _ X\varphi \leq P_{\theta }[\varphi _ j] \]
for all $j$, and hence
\[ \varphi -\sup _ X\varphi \leq \psi . \]
In particular, $\int _ X\theta _\psi ^ n>0$.
Now proposition 16.9.6, proposition 16.4.4 and corollary 16.9.14 give
\[ \varphi _ j\xrightarrow {d_ S}\psi . \]
Moreover, observe that $\psi $ is $\mathcal{I}$-model by proposition 16.6.29 and proposition 16.6.22. Hence, if $\varphi \sim _{\mathcal I}\psi $, then
\[ P_\theta [\varphi ]_{\mathcal I}=P_\theta [\psi ]_{\mathcal I}=\psi . \]
So it suffices to show that $\varphi \sim _{\mathcal I}\psi $. It only remains to argue that $\psi \preceq _{\mathcal{I}} \varphi $. For this purpose, take $\lambda >0$. We need to show that
\[ \mathcal{I}(\lambda \psi )\subseteq \mathcal{I}(\lambda \varphi ). \]
By the strong openness theorem 5.16.21, we may take $\lambda '>\lambda $ such that $\mathcal{I}(\lambda \psi )=\mathcal{I}(\lambda ' \psi )$. It follows from the definition of the quasi-equisingular approximation that
\[ \mathcal{I}(\lambda '\psi )\subseteq \mathcal{I}(\lambda '\varphi _ j)\subseteq \mathcal{I}(\lambda \varphi ) \]
for large enough $j$, where the first inclusion uses $\psi \leq P_\theta [\varphi _ j]$ and $P_\theta [\varphi _ j]\sim _{\mathcal{I}}\varphi _ j$. Our assertion follows.
It follows from the proof that we may take $\varphi _ j$ so that $\theta _{\varphi _ j}$ is a Kähler current for all $j\geq 1$.
Step 2. We handle the general case.
Assume 3 holds. By 13.13.47, we can find $\psi \in \mathrm{PSH}(X,\theta )$ so that $\theta _{\psi }$ is a Kähler current and $\psi \leq \varphi $. We let
\[ \psi _ j=(1-j^{-1})\varphi +j^{-1}\psi \]
for each $j\in \mathbb {Z}_{>1}$. Then $(\psi _ j)_ j$ is an increasing sequence converging almost everywhere to $\varphi $. Then
\[ P_{\theta }[\psi _ j]_{\mathcal{I}}\xrightarrow {d_ S}P_{\theta }[\varphi ]_{\mathcal{I}}=P_{\theta }[\varphi ] \]
by proposition 16.6.30, corollary 16.9.12. From Step 1, we know that each $P_{\theta }[\psi _ j]_{\mathcal{I}}$ lies in the closure of analytic singularities, hence so is $P_{\theta }[\varphi ]\sim _ P \varphi $. Therefore, 1 follows.
$\square$
Pome
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