Proof.
The equality part in (18.28.52) follows from corollary 18.28.14. It remains to prove the weak convergence.
Let $\tau $ be the weak limit of a subsequence of $(\beta ^ k_{v,\varphi ,\omega ^ n})_ k$. In view of (18.28.48), it suffices to show that
14347C41585E
18.28.53
\begin{equation} \tau \leq \theta _{P_{\theta ,X}[\varphi ](v)}^ n. \end{equation}
We can obtain an upper bound of $\tau $ via the observation:
\[ \beta ^ k_{v,\varphi ,\omega ^ n} \leq \beta ^ k_{v,V_{\theta },\omega ^ n}. \]
After replacing $\varphi $ by $V_{\theta }$, the corresponding problem is handled by [Theorem 1.2, Ber11], giving us
\[ \beta ^ k_{v,V_{\theta },\omega ^ n} \rightharpoonup \theta _{P_{\theta ,X}[V_{\theta }](v)}^ n=\mathds {1}_{\left\{ v = P_{\theta ,X}[V_{\theta }](v)\right\} }\, \theta _ v^ n \quad \textup{as }k\to \infty , \]
where the last identity is due to proposition 16.3.41.
We obtain that
1B7624ED3973
18.28.54
\begin{equation} \tau \leq \mathds {1}_{\left\{ v = P_{\theta ,X}[V_{\theta }](v)\right\} }\, \theta _ v^ n. \end{equation}
Now observe that it suffices to show that
C700200CB6BE
18.28.55
\begin{equation} \tau \left\{ P_{\theta ,X}[\varphi ](v) < v\right\} =0. \end{equation}
Assuming this, then (18.28.54) further gives
\[ \tau \leq \mathds {1}_{\left\{ P_{\theta ,X}[\varphi ](v)= v\right\} }\, \theta _ v^ n= \theta _{P_{\theta ,X}[\varphi ](v)}^ n, \]
where the equality follows from proposition 16.3.41, and (18.28.53) follows.
It now remains to argue (18.28.55). Fix a Kähler form $\omega $ on $X$. Observe that
\[ P_{\theta +\epsilon \omega ,X}[\varphi ](v) \sim \varphi \]
for all $\epsilon \geq 0$ by 18.28.15 and proposition 16.6.22. The family $P_{\theta +\epsilon \omega ,X}[\varphi ](v)$ is decreasing as $\epsilon \to 0+$. Let
\[ \chi \coloneqq \inf _{\epsilon >0}P_{\theta +\epsilon \omega ,X}[\varphi ](v). \]
Then $\chi \in \mathrm{PSH}(X,\theta )$ since $\chi \geq P_{\theta ,X}[\varphi ](v)$. Moreover $\chi \leq v$ and $\chi \preceq \varphi $. Hence
\[ \chi \leq P_{\theta ,X}[\varphi ](v). \]
Thus, equality holds, and
\[ P_{\theta +\epsilon \omega ,X}[\varphi ](v)\searrow P_{\theta ,X}[\varphi ](v). \]
In other words,
\[ \left\{ P_{\theta ,X}[\varphi ](v) < v\right\} =\bigcup _{\epsilon \in \mathbb {Q}_{>0}}\left\{ P_{\theta +\epsilon \omega ,X}[\varphi ](v) < v\right\} . \]
So in order to show (18.28.55), we may fix $\epsilon >0$ and argue that
CE5263973CB7
18.28.56
\begin{equation} \tau \left\{ P_{\theta +\epsilon \omega ,X}[\varphi ](v) < v\right\} =0. \end{equation}
Let $k\geq 1$, and $s \in \mathrm{H}^0(X,L^ k \otimes T \otimes \mathcal I(k\varphi ))$ be a section such that $N^ k_{v,\omega ^ n}(s) \leq 1$. Then by [Lemma 4.1, Ber11], there exists $C>0$ such that
\[ h^ k\otimes h_ T(s,s)\mathrm{e}^{-kv} \leq B^ k_{v,\varphi ,\omega ^ n} \leq B^ k_{v,V_{\theta },\omega ^ n} \leq k^ n C. \]
This implies that
\[ \frac{1}{k}\log h^ k \otimes h_ T(s,s) \leq v + \frac{\log C }{k} + n \frac{\log k}{ k }. \]
We define $\varphi _ k$ as in proposition 5.20.17. Take $\alpha _ k\nearrow 1$ as in proposition 5.20.17. After adding a constant to $\varphi $, we may assume that $\varphi \leq 0$ on $X$. After replacing $\alpha _ k$ by a subsequence, we also have $\alpha _ k\varphi \in \mathrm{PSH}(X,\theta +\epsilon \omega )$ for all $k>0$, and $\alpha _ k\varphi \searrow \varphi $. Then
\[ \frac{1}{k}\log h^ k \otimes h_ T(s,s)\preceq \varphi _ k\preceq \alpha _ k\varphi . \]
We notice that $\frac{1}{k}\log h^ k \otimes h_ T(s,s) \in \mathrm{PSH}(X, \theta + \epsilon \omega )$ for all $k \geq k_0(\epsilon )$. Indeed,
\[ \frac{1}{k}\log h^ k\otimes h_ T(s,s) \in \mathrm{PSH}\left(X,\theta + k^{-1} c_1(T,h_ T)\right) \]
by proposition 5.20.4.
In particular,
\[ \frac{1}{k}\log h^ k \otimes h_ T(s,s)-\frac{\log C}{k}-n\frac{\log k}{k}\leq P_{\theta +\epsilon \omega ,X}[\alpha _ k \varphi ](v). \]
Now taking supremum over all candidates $s$, we obtain that
725395815923
18.28.57
\begin{equation} B^ k_{v,\varphi ,\omega ^ n} \leq C k^ n \mathrm{e}^{k \left(P_{\theta +\epsilon \omega ,X}[\alpha _ k \varphi ](v)-v\right)},\quad k \geq k_0. \end{equation}
Now we prove (18.28.56). First observe that $\alpha _ k\varphi \xrightarrow {d_ S}\varphi $ by 16.9.25. Thus, by proposition 18.28.35,
\[ P_{\theta +\epsilon \omega ,X}[\alpha _ k \varphi ](v) \searrow P_{\theta +\epsilon \omega ,X}[\varphi ](v) \]
and we get that
\[ \bigcup _{k=1}^{\infty }\left\{ P_{\theta +\epsilon \omega ,X}[\alpha _ k \varphi ](v) < v\right\} =\left\{ P_{\theta +\epsilon \omega ,X}[\varphi ](v) < v\right\} . \]
It suffices to show that $\tau $ does not put mass on the set $\{ P_{\theta +\epsilon \omega ,X}[\alpha _ m \varphi ](v) < v\} $ for each fixed $m$. Fix a positive rational $\delta >0$, it suffices to show that $\tau $ puts no mass on $F\coloneqq \{ P_{\theta +\epsilon \omega ,X}[\alpha _ m \varphi ](v) < v-\delta \} $. Note that $F$ is open, and (18.28.57) gives $\beta ^\ell _{v,\varphi ,\omega ^ n}(F)\to 0$. Therefore, $\tau $ puts no mass on $F$ as well, proving (18.28.56).
$\square$
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