Theorem 13.15.11.B72AF5B37CB7 For every $q>r_ n$, the damped measures
\[ \sigma _{q,u}:=(1-u)^{-q}\mu _ u \]
have uniformly bounded mass for $u\in \mathrm{PSH}(X,\omega )$, $u\leq 0$. More precisely,
\[ \sigma _{q,u}(\{ u<-T\} ) \leq C\frac{q}{q-r_ n}(1+T)^{r_ n-q}\quad (T\geq 0), \]
where $C$ depends only on $\omega $. There is also a constant $C_ q$ such that, for every Borel set $E\subset X$,
\[ \sigma _{q,u}(E)\leq C_ q\operatorname {Cap}_\omega (E)^{(q-r_ n)/(q+n-r_ n)}. \]
Proof.
For $w(t)=(1+t)^{-q}$, direct integration gives
\[ B_ w(T)=q\int _ T^\infty (1+t)^{r_ n-q-1}\, \mathrm{d}t =\frac{q}{q-r_ n}(1+T)^{r_ n-q}. \]
Thus proposition 13.15.9 gives the mass and tail estimates. For the capacity bound put $\delta =\operatorname {Cap}_\omega (E)$. If $0<\delta \leq 1$, choose $T=\delta ^{-1/(q+n-r_ n)}\geq 1$ in that theorem. Both $T^ n\delta $ and $T^{r_ n-q}$ equal $\delta ^{(q-r_ n)/(q+n-r_ n)}$, so its two terms have the required bound. If $\delta =0$, let $T\to \infty $; if $\delta >1$, use the total mass bound.
$\square$
Pome
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