Proposition 16.4.4.048490BEAC8C Let $(\epsilon _ i)_{i\in I}$ be a decreasing net in $\mathbb {R}_{\geq 0}$ with limit $0$. Take a Kähler form $\omega $ on $X$. Consider a decreasing net $(\varphi _ i)_{i\in I}$ with $\varphi _ i\in \mathrm{PSH}(X,\theta +\epsilon _ i \omega )$ satisfying
B1E48C713966
16.4.5
\begin{equation} P_{\theta +\epsilon _ i\omega }\left[\varphi _ i\right]=\varphi _ i \end{equation}
for every $i\in I$. Set $\varphi \coloneqq \inf _{i\in I}\varphi _ i$. Then $P_{\theta }[\varphi ]=\varphi $, and
243BB43027AE
16.4.6
\begin{equation} \lim _{i\in I}\int _ X \left(\theta + \epsilon _ i \omega \right)^ n_{\varphi _ i}=\int _ X \theta _{\varphi }^ n. \end{equation}
Moreover, if $\int _ X \theta _{\varphi }^ n>0$, then for any prime divisor $E$ over $X$, we have
C4257CD1FB0B
16.4.7
\begin{equation} \lim _{i\in I} \nu \left(\varphi _ i,E\right)=\nu (\varphi ,E). \end{equation}
Proof.
We first observe that $\varphi \in \mathrm{PSH}(X,\theta )$. In fact, for each $i_0\in I$, the tail $(\varphi _ i)_{i\geq i_0}$ lies in $\mathrm{PSH}(X,\theta +\epsilon _{i_0}\omega )$ and decreases to $\varphi $. Since $\sup _ X\varphi _ i=0$ by (16.4.5), the limit is not identically $-\infty $. Thus $\varphi \in \mathrm{PSH}(X,\theta +\epsilon _{i_0}\omega )$ for every $i_0\in I$. Since $\epsilon _{i_0}\to 0$, it follows that $\varphi \in \mathrm{PSH}(X,\theta )$. For every $i\in I$ we have
\[ P_{\theta }[\varphi ]\leq P_{\theta +\epsilon _ i\omega }[\varphi ]\leq P_{\theta +\epsilon _ i\omega }[\varphi _ i]=\varphi _ i. \]
Therefore, $P_{\theta }[\varphi ]\leq \varphi $. The reverse inequality is trivial. Hence $P_{\theta }[\varphi ]=\varphi $.
By the monotonicity theorem theorem 13.13.11 and the multi-linearity of the non-pluripolar product proposition 13.11.4 6, we have
\[ \varliminf _{i\in I} \int _ X \left(\theta + \epsilon _ i \omega \right)^ n_{\varphi _ i} \geq \varliminf _{i\in I} \int _ X \left(\theta + \epsilon _ i \omega \right)^ n_{\varphi } = \int _ X \theta _{\varphi }^ n. \]
We now argue the reverse inequality.
Fix $i_0\in I$. Since passing to the tail $i\geq i_0$ does not change the limsup, we have
\[ \begin{aligned} \varlimsup _{i\in I} \int _ X \left(\theta + \epsilon _ i \omega \right)^ n_{\varphi _ i}=& \varlimsup _{i\geq i_0} \int _{\left\{ \varphi _ i =0\right\} } \left(\theta + \epsilon _ i \omega \right)^ n_{\varphi _ i} \\ \leq & \varlimsup _{i\geq i_0} \int _{\left\{ \varphi _ i =0\right\} } \left(\theta + \epsilon _{i_0} \omega \right)^ n_{\varphi _ i}\\ \leq & \int _{\left\{ \varphi =0\right\} } \left(\theta + \epsilon _{i_0} \omega \right)^ n_{\varphi }, \end{aligned} \]
where in the first line we used (16.4.5) and theorem 16.3.12, and in the last line we used the fact that $\varphi _ i \searrow \varphi $ on the tail $i\geq i_0$, and on this tail, $\varphi _ i \to \varphi $ in $L^1$ as well by corollary 5.17.5, and therefore, corollary 13.13.16 is applicable. Taking the limit with respect to $i_0$, we arrive at the desired conclusion:
\[ \varlimsup _{i\in I} \int _{X} \left(\theta + \epsilon _ i \omega \right)^ n_{\varphi _ i} \leq \varliminf _{i_0\in I}\int _{\{ \varphi =0\} } \left(\theta + \epsilon _{i_0} \omega \right)^ n_{\varphi } = \int _{\{ \varphi =0\} } \theta ^ n_{\varphi } \leq \int _{X} \theta ^ n_{\varphi }, \]
where the equality follows from the multi-linearity of the non-pluripolar product proposition 13.11.4 6 and the last inequality follows from theorem 16.3.12. This finishes the proof of (16.4.6).
It remains to argue (16.4.7). Fix $i_0\in I$. By proposition 5.17.7, there is a constant $B>0$ such that
\[ \nu (\psi ,E)\leq B,\quad \psi \in \mathrm{PSH}(X,\theta +\epsilon _{i_0}\omega ). \]
By 13.13.30 and (16.4.6), for any $\delta \in (0,1)$ there exists $i_{\delta }\geq i_0$ such that for all $i\geq i_{\delta }$ there exists $\psi _ i \in \mathrm{PSH}(X,\theta +\epsilon _ i\omega )$ satisfying $(1-\delta ) \varphi _ i + \delta \psi _ i \leq \varphi $. This implies that for $i\geq i_{\delta }$ we have
\[ (1-\delta )\nu (\varphi _ i,E) + \delta \nu (\psi _ i,E) \geq \nu (\varphi ,E) \geq \nu (\varphi _ i,E). \]
Since $\psi _ i\in \mathrm{PSH}(X,\theta +\epsilon _{i_0}\omega )$ for $i\geq i_{\delta }$, we get
\[ \varliminf _{i\in I}\nu (\varphi _ i,E)\geq \frac{\nu (\varphi ,E)-\delta B}{1-\delta }. \]
Letting $\delta \searrow 0$ gives $\varliminf _{i\in I}\nu (\varphi _ i,E)\geq \nu (\varphi ,E)$. The reverse inequality follows from $\varphi \leq \varphi _ i$, and hence we conclude (16.4.7).
$\square$
Pome
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