Proposition 16.25.7.CDA70099F425 Let $u,\phi \in \mathrm{PSH}(X,\theta )$ with $u\preceq \phi $, and let $s\in \mathbb R$ and $0<t\leq 1$. Then
\[ t^ n\operatorname {Cap}_{\theta ,\phi }(\{ u<\phi -s-t\} ) \leq \mu _ u(\{ u<\phi -s\} )+\int _ X\langle \theta _\phi ^ n\rangle -\int _ X\langle \theta _ u^ n\rangle . \]
In particular, if $u\in \mathcal E(X,\theta ,\phi )$, then $\operatorname {Cap}_{\theta ,\phi }(\{ u<\phi -s\} )\to 0$ as $s\to \infty $.
Proof.
Let $v$ be a competitor for $\operatorname {Cap}_{\theta ,\phi }$, and put $q=(1-t)\phi +tv-s$. Then $\phi -s-t\leq q\leq \phi -s$, so
\[ \{ u<\phi -s-t\} \subset \{ u<q\} \subset \{ u<\phi -s\} . \]
Moreover $q\sim \phi $ and $\mu _ q\geq t^ n\mu _ v$. Proposition 16.25.4, with $p=n$, applied to $u$ and $q$, gives
\[ t^ n\mu _ v(\{ u<\phi -s-t\} )\leq \mu _ q(\{ u<q\} ) \leq \mu _ u(\{ u<\phi -s\} )+\int _ X\langle \theta _\phi ^ n\rangle -\int _ X\langle \theta _ u^ n\rangle , \]
and taking the supremum over $v$ proves the inequality. If $u\in \mathcal E(X,\theta ,\phi )$, then $\int _ X\langle \theta _ u^ n\rangle =\int _ X\langle \theta _\phi ^ n\rangle $. With $t=1$, the right side tends to zero as $s\to \infty $ by dominated convergence, because the sets $\{ u<\phi -s\} $ decrease to the pole set of $u$, which is $\mu _ u$-null.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.