Theorem 15.34.2.9F0AB366D34B Let $\theta $ be any smooth real $(1,1)$-form and $\mu $ a finite non-pluripolar measure. Suppose $f_ j\geq 0$, $f_ j\to f$ in $L^1(\mu )$, and $\lambda _ j>0$ with $\lambda _ j\to \lambda >0$. Let $u_ j,u\in \mathrm{PSH}(X,\theta )$ be nonpositive, suppose $u_ j\to u$ in capacity, and assume
\[ \mathrm{MA}^{\mathrm{ext}}_\theta (u_ j) =\mathrm{e}^{\lambda _ j u_ j}f_ j\, \mu . \]
Then $\mathrm{MA}^{\mathrm{ext}}_\theta (u)$ is finite and
\[ \mathrm{MA}^{\mathrm{ext}}_\theta (u)=\mathrm{e}^{\lambda u}f\, \mu . \]
Moreover, $\mathrm{MA}^{\mathrm{ext}}_\theta (u_ j)\rightharpoonup \mathrm{MA}^{\mathrm{ext}}_\theta (u)$ weakly, and the right-hand sides converge in total variation.
Proof.
Discarding finitely many indices, we may assume that $\lambda _ j\geq \lambda /2$. The measures $\mathrm{MA}^{\mathrm{ext}}_\theta (u_ j)$ have masses at most $\| f_ j\| _{L^1(\mu )}$, which are uniformly bounded, and since $u_ j\leq 0$, their deep sublevels satisfy
\[ \mathrm{MA}^{\mathrm{ext}}_\theta (u_ j)(\{ u_ j<-T\} ) \leq \mathrm{e}^{-\lambda T/2}\| f_ j\| _{L^1(\mu )}. \]
The tail criterion theorem 13.15.13, which holds for an arbitrary smooth background by theorem 13.15.23, therefore shows that $\mathrm{MA}^{\mathrm{ext}}_\theta (u)$ is finite and that $\mathrm{MA}^{\mathrm{ext}}_\theta (u_ j)\rightharpoonup \mathrm{MA}^{\mathrm{ext}}_\theta (u)$.
It remains to identify this limit with $\mathrm{e}^{\lambda u}f\, \mu $. By lemma 5.29.18, $u_ j\to u$ in $\mu $-measure. Since $t\mapsto \mathrm{e}^{\lambda t}$ is $\lambda $-Lipschitz for $t\leq 0$, the functions $\mathrm{e}^{\lambda u_ j}$, which are bounded by one, converge in $\mu $-measure to $\mathrm{e}^{\lambda u}$. The change of exponent is uniform for nonpositive arguments, because the mean value theorem gives
\[ \sup _{t\leq 0}|\mathrm{e}^{\lambda _ jt}-\mathrm{e}^{\lambda t}| \leq |\lambda _ j-\lambda | \sup _{s\geq 0}s \mathrm{e}^{-(\lambda /2)s}\longrightarrow 0. \]
Thus $\mathrm{e}^{\lambda _ j u_ j}\to \mathrm{e}^{\lambda u}$ in $\mu $-measure, and hence in $f\mu $-measure, since the finite measure $f\mu $ is absolutely continuous with respect to $\mu $. As these functions are bounded by one, the convergence also holds in $L^1(f\mu )$. Therefore
\[ \begin{split} \int _ X|\mathrm{e}^{\lambda _ j u_ j}f_ j-\mathrm{e}^{\lambda u}f|\, \mathrm{d}\mu & \leq \| f_ j-f\| _{L^1(\mu )}\\ & \quad +\int _ X|\mathrm{e}^{\lambda _ j u_ j}-\mathrm{e}^{\lambda u}|f\, \mathrm{d}\mu \longrightarrow 0. \end{split} \]
This proves the total-variation convergence of the right-hand sides, and comparing the two limits gives the equation for $u$.
$\square$
Pome
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