Theorem 13.15.9.EE93EE2035F6 Let $w:[0,+\infty )\to (0,+\infty )$ be continuously differentiable and nonincreasing, with $w(t)\to 0$ as $t\to +\infty $. For $T\geq 0$ define
\[ B_ w(T)=\int _ T^\infty (1+t)^{r_ n}(-w'(t))\, \mathrm{d}t. \]
Assume $B_ w(0)<+\infty $. For nonpositive $u\in \mathrm{PSH}(X,\omega )$, the damped measures $\nu _{w,u}=w(-u)\mu _ u$ satisfy
\[ \nu _{w,u}(X)\leq C_\omega B_ w(0),\qquad \nu _{w,u}(\{ u<-T\} )\leq C_\omega B_ w(T). \]
For every Borel set $E$ and $T\geq 0$,
\[ \nu _{w,u}(E)\leq w(0)(1+T)^ n\operatorname {Cap}_\omega (E)+C_\omega B_ w(T). \]
In particular, these measures are uniformly absolutely continuous with respect to capacity. If nonpositive $u_ j,u\in \mathrm{PSH}(X,\omega )$ satisfy $u_ j\to u$ in capacity, then $\nu _{w,u_ j}\rightharpoonup \nu _{w,u}$.
Proof.
Write $s=-u$. Since $w(s)=\int _ s^\infty (-w'(t))\, \mathrm{d}t$ for finite $s$, and since $\mu _ u$ does not charge the pole set $\{ s=+\infty \} $, Tonelli's theorem and lemma 13.15.3 give
\[ \int _ X w(s)\, \mathrm{d}\mu _ u =\int _0^\infty (-w'(t))\mu _ u(\{ s\leq t\} )\, \mathrm{d}t \leq C_\omega B_ w(0). \]
Restricting the integral to $\{ s>T\} $ gives the same bound with $B_ w(T)$ in place of $B_ w(0)$, since the inner set is then $\{ T<s\leq t\} $, which is empty unless $t>T$.
For the capacity bound, the second estimate in (13.15.4) gives
\[ \int _{E\cap \{ s\leq T\} }w(s)\, \mathrm{d}\mu _ u \leq w(0)(1+T)^ n\operatorname {Cap}_\omega (E). \]
Adding the tail estimate for $\{ s>T\} $ proves the asserted bound. Since $B_ w(T)\to 0$, choosing first $T$ large and then the capacity small gives uniform absolute continuity.
To prove continuity, fix a continuous nonincreasing $\eta :[0,+\infty )\to [0,1]$, equal to one on $[0,1]$ and zero on $[2,+\infty )$, and put $\eta _ T(s)=\eta (s/T)$. For fixed $T>0$, set $v_{j,T}=\max (u_ j,-2T)$ and $g_ T(t)=\eta _ T(-t)w(-t)$ on $[-2T,0]$. Since $g_ T(-2T)=0$, locality gives
D2588BCCC6F7
13.15.10
\begin{equation} \eta _ T(-u_ j)w(-u_ j)\mu _{u_ j} =g_ T(v_{j,T})\, \mathrm{MA}_\omega (v_{j,T}). \end{equation}
Truncation preserves convergence in capacity. The functions $g_ T(v_{j,T})$ are uniformly bounded and quasi-continuous, and they converge in capacity to the corresponding function for $u$. The weighted continuity of bounded products in section 5.27, section 5.29 therefore gives weak convergence of the measures in (13.15.10) for each fixed $T$. The measure $(1-\eta _ T(-u_ j))\nu _{w,u_ j}$ is carried by $\{ u_ j<-T\} $, so by the tail bound its integral against a continuous test function $f$ is at most $C_\omega B_ w(T)\| f\| _\infty $, uniformly in $j$; the same holds for $u$. Letting $T\to \infty $ completes the proof.
$\square$
Pome
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