13.14.7 Weighted lower semicontinuity74BA8C79EB05

Proof. Subtracting a common constant, we may assume that all the potentials are nonpositive. Choose a Hermitian form $\Omega \geq \theta $. For $T>0$, let $g_ T:[-\infty ,0]\to [0,1]$ be a continuous nondecreasing cutoff, equal to zero on $[-\infty ,-2T]$ and to one on $[-T,0]$, chosen so that $g_ T$ increases pointwise to one on $(-\infty ,0]$ as $T\to \infty $. Put $v_ j=\max (u_ j,-2T)$ and $v=\max (u,-2T)$. These are uniformly bounded $\Omega $-plurisubharmonic functions, and they converge in capacity.

Fix $T$. Expanding $(\theta +\mathrm{dd}^{\mathrm{c}}v_ j)^ n$ by means of $\theta +\mathrm{dd}^{\mathrm{c}}v_ j=(\Omega +\mathrm{dd}^{\mathrm{c}}v_ j)+(\theta -\Omega )$, and bounding each smooth factor by a constant multiple of $\Omega $, we see that the total variations of the signed terms are controlled by positive bounded mixed measures. By lemma 5.29.16, these have a common capacity bound. Bounded Bedford–Taylor continuity gives weak convergence term by term. It also allows the weights $b_ jg_ T(v_ j)$: their differences from the limiting weight tend to zero in capacity, so the common capacity bound makes the corresponding integrals tend to zero. For the fixed limiting weight, remove an open set of arbitrarily small capacity, extend the weight continuously from its complement, and control the discarded part by the same bound. Since the cutoff vanishes at the truncation level $-2T$, locality gives

The first integral is nonnegative and at most the corresponding integral without the cutoff, although the bounded product $(\theta +\mathrm{dd}^{\mathrm{c}}v_ j)^ n$ may be signed. Hence the limit is bounded by the lower limit in the assertion. Letting $T\to \infty $ and using monotone convergence completes the proof. $\square$