Proof.
First note that $\int _ X \theta _{\varphi }^ n\geq \delta $ as well, due to the $d_ S$-continuity of non-pluripolar masses as proved in corollary 16.9.13. Replacing $\varphi $ by $P_{\theta }[\varphi ]$, we may then assume that $\varphi $ is a model potential in $\mathrm{PSH}(X,\theta )_{>0}$.
We are free to replace $(\varphi _ j)_ j$ by a subsequence. So we may assume that
2902F1B6C4A6
16.9.18
\begin{equation} d_ S(\varphi _ j,\varphi _{j+1})\leq \frac{\delta 2^{-j-4}}{D_ n C_ n^{j+1}},\quad d_ S(\varphi ,\varphi _ j)\leq \frac{\delta 2^{-j-4}}{D_ n}, \end{equation}
where $C_ n$ is the constant in corollary 16.9.9, $D_ n$ is the constant in corollary 16.9.13.
In particular, by corollary 16.9.13,
67C255B50420
16.9.19
\begin{equation} \left|\int _ X \theta _{\varphi _ j}^ n-\int _ X \theta _{\varphi }^ n\right|\leq \delta 2^{-j-4}. \end{equation}
Step 1. We handle the $\psi _ j$'s. For each $j\geq 1$ and $k\geq 1$, by 16.9.4 we have
\[ \begin{aligned} d_ S(\varphi _ j,\varphi _ j\lor \varphi _{j+1}\lor \cdots \lor \varphi _{j+k}) & \leq C_ n\, d_ S(\varphi _ j,\varphi _{j+1}\lor \cdots \lor \varphi _{j+k})\\ & \leq C_ n\, d_ S(\varphi _ j,\varphi _{j+1})\\ & \quad + C_ n\, d_ S(\varphi _{j+1},\varphi _{j+1}\lor \cdots \lor \varphi _{j+k}). \end{aligned} \]
By iteration, we find
\[ \begin{split} d_ S(\varphi _ j,\varphi _ j\lor \varphi _{j+1}\lor \cdots \lor \varphi _{j+k})\leq \sum _{a=j}^{j+k-1}C_ n^{a+1-j}d_ S(\varphi _ a,\varphi _{a+1})\\ \leq \frac{C_ n^{-j}}{D_ n}\sum _{a=j}^{j+k-1}\delta 2^{-a-4}\leq \frac{\delta 2^{-j-3}}{D_ n}. \end{split} \]
Using corollary 16.9.12, we have
\[ \varphi _ j\lor \varphi _{j+1}\lor \cdots \lor \varphi _{j+k}\xrightarrow {d_ S}\psi _ j \]
as $k\to \infty $. Hence
3B96A5372D55
16.9.20
\begin{equation} d_ S(\varphi _ j,\psi _ j)\leq \frac{\delta 2^{-j-3}}{D_ n}. \end{equation}
We conclude that $\psi _ j\xrightarrow {d_ S}\varphi $.
Moreover, we observe that
79C50CD3AC3E
16.9.21
\begin{equation} \varphi =\inf _{j\geq 1} P_{\theta }[\psi _ j] \end{equation}
Indeed, set $\tilde{\psi }_ j=P_{\theta }[\psi _ j]$. Then $(\tilde{\psi }_ j)_ j$ is a decreasing sequence of model potentials, and $d_ S(\tilde{\psi }_ j,\psi _ j)=0$ by proposition 16.9.6. Hence $\tilde{\psi }_ j\xrightarrow {d_ S}\varphi $. If $\chi =\inf _ j\tilde{\psi }_ j$, then proposition 16.4.4 and corollary 16.9.14 give $\tilde{\psi }_ j\xrightarrow {d_ S}\chi $. Thus $d_ S(\chi ,\varphi )=0$. Since both $\chi $ and $\varphi $ are model potentials, proposition 16.9.6 gives $P_{\theta }[\chi ]=P_{\theta }[\varphi ]$, hence $\chi =\varphi $, proving (16.9.21).
Step 2. We consider the $\eta _ j$'s.
For each $j\geq 1$ and $k\geq 0$, we let
\[ \eta _ j^ k\coloneqq \varphi _ j\land \cdots \land \varphi _{j+k}. \]
Using (16.9.18), (16.9.20) and corollary 16.9.13, we have
\[ \left|\int _ X\theta _{\psi _ j}^ n-\int _ X \theta _{\varphi }^ n \right|\leq \delta 2^{-j-2} \]
for all $j\geq 1$. Taking (16.9.19) into account, we have
9F803C97D29B
16.9.22
\begin{equation} \left|\int _ X\theta _{\varphi _ j}^ n-\int _ X \theta _{\psi _{j-1}}^ n \right|\leq \delta 2^{-j} \end{equation}
for $j>1$.
Step 2.1. We claim that for a fixed $j\geq 1$, for any $k\in \mathbb {N}$, we have $\eta _ j^ k\in \mathrm{PSH}(X,\theta )$ and
A82EF3572727
16.9.23
\begin{equation} \int _ X \theta _{\eta _ j^ k}^ n\geq \int _ X \theta _{\varphi _ j}^ n-\sum _{a=1}^{k} \delta 2^{-j-a}. \end{equation}
We argue by induction on $k\geq 0$. The case $k=0$ is trivial. When $k>0$, assume that the case $k-1$ is known. Then
\[ \begin{aligned} \int _ X \theta _{\eta _ j^{k-1}}^ n+\int _ X \theta _{\varphi _{j+k}}^ n \geq \int _ X \theta _{\varphi _ j}^ n-\sum _{a=1}^{k-1}\delta 2^{-j-a}+\int _ X \theta _{\psi _{j+k-1}}^ n-\delta 2^{-j-k}\\ \geq \int _ X \theta _{\varphi _ j}^ n-\delta 2^{-j}+\int _ X \theta _{\psi _{j+k-1}}^ n>\int _ X \theta _{\psi _{j+k-1}}^ n, \end{aligned} \]
where the first inequality follows from the inductive hypothesis and (16.9.22).
Observe that
\[ \eta _ j^{k-1}\lor \varphi _{j+k}\leq \psi _{j+k-1}, \]
it follows from proposition 16.3.25 that $\eta _ j^ k\in \mathrm{PSH}(X,\theta )$. By theorem 16.3.35, we deduce that
\[ \begin{aligned} \int _ X \theta _{\eta _ j^ k}^ n\geq & \int _ X \theta _{\varphi _{j+k}}^ n+ \int _ X \theta _{\eta _ j^{k-1}}^ n-\int _ X \theta _{\psi _{j+k-1}}^ n\\ \geq & \int _ X \theta _{\varphi _ j}^ n-\sum _{a=1}^{k} \delta 2^{-j-a}, \end{aligned} \]
where the second inequality follows from the inductive hypothesis and (16.9.22). Therefore, (16.9.23) follows.
Step 2.2. It follows by induction from proposition 16.3.29 that for any $j\geq 1$, $k\geq 0$,
\[ P_{\theta }\left[\eta _ j^ k\right]=\eta _ j^ k. \]
By proposition 16.4.4, we have
\[ \lim _{k\to \infty } \int _ X \theta _{\eta _ j^ k}^ n=\int _ X \theta _{\eta _ j}^ n \]
for any $j\geq 1$. Letting $k\to \infty $ in (16.9.23), we find that
\[ \int _ X \theta _{\eta _ j}^ n\geq \int _ X \theta _{\varphi _ j}^ n-\delta 2^{-j}>0 \]
for $j\geq 1$. Observe that we also have
\[ \int _ X \theta _{\eta _ j}^ n\leq \int _ X \theta _{\varphi _ j}^ n \leq \int _ X \theta _{\psi _ j}^ n \]
for $j\geq 1$ by theorem 13.13.11. It follows from these inequalities and (16.9.19) that
\[ \lim _{j\to \infty }\int _ X\theta _{\eta _ j}^ n=\int _ X\theta _{\varphi }^ n. \]
For $k\geq 1$, we have
\[ \eta _ j^ k=\varphi _ j\land \eta _{j+1}^{k-1}\leq \eta _{j+1}^{k-1}. \]
Letting $k\to \infty $ gives $\eta _ j\leq \eta _{j+1}$. Hence $(\eta _ j)_ j$ is increasing. Hence corollary 13.13.12 gives
\[ \int _ X \theta _{\eta }^ n=\lim _{j\to \infty }\int _ X \theta _{\eta _ j}^ n=\int _ X \theta _{\varphi }^ n, \]
where $\eta =\operatorname *{sup}_{j\geq 1}{}\! {}^{*} \eta _ j$. For fixed $m$, if $j\geq m$, then $\eta _ j\leq \varphi _ j\leq \psi _ m$, whereas if $j<m$, then $\eta _ j\leq \eta _ m\leq \psi _ m$. Thus $\eta _ j$ is a candidate in the definition of $P_{\theta }[\psi _ m]$ for all $j,m$, and (16.9.21) gives $\eta \leq \varphi $. It follows from proposition 16.8.7 that $\eta \sim _ P \varphi $. By corollary 16.9.12 and proposition 16.9.6, we have $\eta _ j\xrightarrow {d_ S}\varphi $.
$\square$
Pome
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