Pome0D2E3E7B4DFDOn deep sublevel sets the damping weight tends to zero, so convergence of the weighted measures can leave their undamped mass uncontrolled. Uniformly vanishing deep tails supply the additional control: the following condition gives a finite limit measure and weak convergence of the undamped measures $\mu _{u_ j}$. It can equivalently be expressed by an increasing integrability weight.
Theorem 13.15.13.8B2A9680933C Let $u_ j,u\in \mathrm{PSH}(X,\omega )$ be nonpositive and suppose $u_ j\to u$ in capacity. Assume
Then $\mu _ u$ is finite and $\mu _{u_ j}\rightharpoonup \mu _ u$.
Proof. Choose $T_0$ such that every tail $\mu _{u_ j}(\{ u_ j<-T\} )$ with $T\geq T_0$ is at most one. By lemma 13.15.3, $\mu _{u_ j}(X)\leq C(1+T_0)^{r_ n}+1$, so the tail condition already gives a uniform bound for the total masses.
With the cutoffs $\eta _ T$ of (13.15.10), the fixed-level argument in the proof of theorem 13.15.9, applied with $w\equiv 1$, gives
The masses of the measures on the left are bounded by the uniform bound for the total masses. Since $\eta _ T(-u)$ increases to one off the pole set as $T\to \infty $, monotone convergence proves $\mu _ u(X)<+\infty $.
For a continuous test function $f:X\to \mathbb R$, the error made by inserting $\eta _ T(-u_ j)$ is at most $\| f\| _\infty \mu _{u_ j}(\{ u_ j<-T\} )$, which is small for large $T$, uniformly in $j$. For the limit potential, the corresponding error tends to zero because $\mu _ u$ is finite and does not charge the pole set. Let first $j\to \infty $ with $T$ fixed, and then $T\to \infty $. $\square$
Let $\chi :[0,+\infty )\to [1,+\infty )$ be continuous, nondecreasing and unbounded. For a nonpositive potential $u\in \mathrm{PSH}(X,\omega )$ define
with the value $+\infty $ allowed. Since $\chi (T)\to +\infty $, a uniform bound for this functional forces the sublevel tails to vanish uniformly.
Corollary 13.15.14.7312F05A4396 Let $u_ j,u\in \mathrm{PSH}(X,\omega )$ be nonpositive. If $u_ j\to u$ in capacity and $\sup _ jJ_{\chi ,\omega }(u_ j)\leq C$, then
Proof. Since $\chi \geq 1$, the total masses are at most $C$, and
Thus theorem 13.15.13 applies. For the bound on $J_{\chi ,\omega }(u)$, use (13.15.10) with the compactly supported continuous function $s\mapsto \chi (s)\eta _ T(s)$ in place of $w\eta _ T$. The corresponding weighted integrals converge for each fixed $T$, and they are at most $C$. Let $T\to \infty $ and use monotone convergence. $\square$
Proposition 13.15.15.ACA3FA133A3B For a family of nonpositive potentials in $\mathrm{PSH}(X,\omega )$, the uniform tail condition of theorem 13.15.13 is equivalent to a uniform bound on $J_{\chi ,\omega }$ for some continuous nondecreasing unbounded function $\chi \geq 1$.
Proof. One implication was proved in corollary 13.15.14. Conversely, write the family as $(u_ i)_{i\in I}$. As in the proof of theorem 13.15.13, the tail condition and lemma 13.15.3 give a uniform bound for the total masses. Choose numbers $T_ k\to +\infty $, with $T_{k+1}>T_ k+1$, such that $\mu _{u_ i}(\{ -u_ i>T_ k\} )\leq 2^{-k}$ for every $i\in I$. Let $\rho _ k$ be a continuous nondecreasing function, equal to zero on $[0,T_ k]$ and to one on $[T_ k+1,+\infty )$, and define
The sum is locally finite, so $\chi $ is continuous, nondecreasing and unbounded. Tonelli's theorem gives
so this weight gives a uniform integrability bound for the whole family. $\square$
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