Lemma 18.28.21D7B3984EBF7A

Proof. Step 1. We first prove (18.28.22) when $\varphi = V_{\theta }$.

Let $S\subseteq X$ be a proper analytic subset, such that $V_{\theta }$ is locally bounded on $X\setminus S$. Such $S$ exists by 13.13.8. Fix holomorphic balls $B\Subset B'\Subset X\setminus S$.

By proposition 5.14.3 and 18.28.9, we can take an increasing sequence $(\eta _ j)_ j$ in $\mathrm{PSH}(X,\theta )$ such that

Note that $P_{\theta ,K}[V_{\theta }](v)\sim V_{\theta }$ by 18.28.15. Take $C>0$ so that

Then we may replace $\eta _ j$ by $\eta _ j\lor (P_{\theta ,K}[V_{\theta }](v)-C)$ and assume that $\eta _ j\sim V_{\theta }$ for each $j$. In particular, $\eta _ j$ is locally bounded on $X\setminus S$.

For each $j\geq 1$, let $\gamma _ j\in \mathrm{PSH}(X,\theta )$ be the balayage of $\eta _ j$ on $B$, defined in proposition 5.17.9. Note that $(\gamma _ j)_ j$ is increasing due to proposition 5.14.22.

For the first equality in (18.28.22), we may assume that $B\Subset X\setminus (S \cup K)$. It suffices to show that $\theta _{P_{\theta ,K}[V_{\theta }](v)}^ n$ does not charge $B$.

Since $B$ lies outside $K$ by assumption, we also have $\gamma _ j|_ K=\eta _ j|_ K \leq v$ for all $j\geq 1$. In particular, $\gamma _ j\leq P_{\theta ,K}[V_{\theta }](v)$.

It follows that $\gamma _ j \nearrow P_{\theta ,K}[V_{\theta }](v)$ almost everywhere as well. Testing against non-negative continuous functions supported in $B$ and using theorem 13.13.9, we find that $\theta ^ n_{P_{\theta ,K}[V_{\theta }](v)}$ does not charge $B$, as desired.

We next prove the second equality in (18.28.22). The proof is essentially the same as that of 13.13.21.

It suffices to show that $\theta ^ n_{P_{\theta ,K}[V_{\theta }](v)}$ does not charge the set $K\cap \{ P_{\theta ,K}[V_{\theta }](v)|_ K<v\} $.

Let $x\in (X\setminus S) \cap K$ be a point such that $P_{\theta ,K}[V_{\theta }](v)(x)<v(x)-\epsilon $ for some $\epsilon >0$. We may assume that $x\in B$ and $B'$ is small enough, so that

1. B2B18BBFA88D

   $P_{\theta ,K}[V_{\theta }](v)|_{\overline{B'}}<v(x)-\epsilon $.

2. D0D5ED334E0E

   $v|_{\overline{B'}\cap K}>v(x)-\epsilon $.

By 1, we have $\gamma _ j|_{\partial B}\leq v(x)-\epsilon $, hence by proposition 5.14.22, $\gamma _ j|_ B\leq v(x)-\epsilon $. By 2, we have $\gamma _ j|_{B\cap K}\leq v|_{B\cap K}$. On the other hand, $\gamma _ j|_{K\setminus B}=\eta _{j}|_{K\setminus B}\leq v|_{K\setminus B}$. Therefore, $\gamma _ j|_ K\leq v$. It follows again that $\gamma _ j\leq P_{\theta ,K}[V_{\theta }](v)$. Therefore, we conclude as in the previous paragraph that $\gamma _ j\nearrow P_{\theta ,K}[V_{\theta }](v)$. Thus, testing against non-negative continuous functions supported in $B$, we conclude that $\theta ^ n_{P_{\theta ,K}[V_{\theta }](v)}$ does not charge $B$ again by theorem 13.13.9.

Step 2. We handle the general case.

By corollary 18.28.17,

Now corollary 16.3.14 gives

where in the second inequality we have used Step 1. Hence (18.28.23) follows, and a fortiori (18.28.22) follows as well. $\square$