For a general decreasing weight $w$, writing $w(s)=\int _ s^\infty (-w'(t))\, \mathrm{d}t$ and integrating the finite-depth estimate (13.15.4) in $t$ bounds both the total mass and the mass on deep sublevel sets. Above any fixed depth, the damped measure is a weighted product of bounded potentials, which is continuous under convergence in capacity; the tail bound then removes the restriction to a fixed depth.
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.
$\square$
For $w(t)=\mathrm{e}^{-pt}$, $p>0$, the integral $B_ w(0)$ is finite because the exponential dominates the polynomial factor. Thus the exponentially damped measures of theorem 13.15.7 are continuous under convergence in capacity. By that theorem, their tails also satisfy
\[ \nu _{p,u}(\{ u<-T\} ) \leq \mathrm{e}^{-pT/2}\nu _{p/2,u}(X) \leq C_ p\mathrm{e}^{-pT/2}. \]
For polynomial weights, the general estimate takes the following explicit form.
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.
$\square$
For $n\geq 2$, the capacity exponent is $1-n/(q+2)$. The threshold $q>n-2$ comes from the mixed-mass estimate in proposition 13.15.2.
Logarithmic factors give weights at the threshold. For $n\geq 3$ and $a>1$, put $r=n-2>0$ and
\[ w(t)=(1+t)^{-r}[\log (\mathrm{e}+t)]^{-a}. \]
Differentiation gives
\[ (1+t)^ r(-w'(t)) =\frac{r}{(1+t)[\log (\mathrm{e}+t)]^ a} +\frac{a}{(\mathrm{e}+t)[\log (\mathrm{e}+t)]^{a+1}}. \]
The ratios $(\mathrm{e}+t)/(1+t)$ are bounded. The substitution $y=\log (\mathrm{e}+t)$ therefore bounds the integrals of the two terms over $[T,+\infty )$ by fixed multiples of $[\log (\mathrm{e}+T)]^{1-a}$ and $[\log (\mathrm{e}+T)]^{-a}$, respectively. Since $a>1$ and $\log (\mathrm{e}+T)\geq 1$, we obtain $B_ w(T)\leq C_{n,a}[\log (\mathrm{e}+T)]^{1-a}$. Hence, by proposition 13.15.9, the damped measures $(1-u)^{-(n-2)}[\log (\mathrm{e}-u)]^{-a}\mu _ u$ have uniformly bounded mass, their mass on $\{ u<-T\} $ is at most a constant times $[\log (\mathrm{e}+T)]^{1-a}$, and they are continuous under convergence in capacity. In dimensions one and two, $r_ n=0$ and $B_ w(T)=w(T)$, so every weight satisfying the monotonicity, differentiability and vanishing conditions of proposition 13.15.9 can be used.
Damping can restore convergence when every undamped measure has infinite mass. When the bounded mass property fails, take the potential $U$ constructed in the proof of theorem 13.14.18, whose current has integrable coefficients by proposition 13.14.21, and normalize it by $\sup _ XU=0$. For $0<\varepsilon <1$, the potentials $\varepsilon U$ converge to zero in capacity as $\varepsilon \downarrow 0$: by (5.29.4), the capacity of $\{ \varepsilon U<-\delta \} $ is at most $C\varepsilon /\delta $ for each fixed $\delta >0$ and small $\varepsilon $. Consequently, for $p>0$ and $q>r_ n$,
\[ \mathrm{e}^{p\varepsilon U}\mu _{\varepsilon U}\rightharpoonup \omega ^ n, \qquad (1-\varepsilon U)^{-q}\mu _{\varepsilon U}\rightharpoonup \omega ^ n, \]
although $\mu _{\varepsilon U}(X)=+\infty $ for every $0<\varepsilon <1$. The full currents also converge to $\omega $ in the coefficient $L^1$ norm, and their Gauduchon traces converge in total variation.
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