When the exponents converge to a positive limit, the factor $\mathrm{e}^{\lambda u}$ makes the Monge–Ampère mass of deep sublevels uniformly small. If the potentials converge in capacity, the tail criterion of theorem 13.15.13 then identifies the limit of their Monge–Ampère measures, and convergence of the right-hand sides identifies the equation.
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 $u$.
$\square$
Additive normalizations of the potentials can be absorbed into the densities: translating each $u_ j$ by a constant multiplies $f_ j$ by an exponential factor, and the theorem applies as long as the modified densities still converge in $L^1(\mu )$. Convergence in capacity is a hypothesis; the theorem says nothing about sequences that converge only in $L^1$, and it does not keep the limit in a prescribed envelope class.
The proof also shows where the difficulty lies as $\lambda _ j\downarrow 0$: the tail bound $\mathrm{e}^{-\lambda _ jT}\| f_ j\| _{L^1(\mu )}$ no longer tends to zero uniformly in $j$ as $T\to \infty $. A zero-exponent limit therefore needs another source of uniform tail control, such as the integrability condition of corollary 13.15.14; a uniform bound on the total masses does not by itself control these tails.
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