Lemma 5.44.12.88E063207C1B Let $\nu $ be a non-pluripolar probability measure with $A_\theta (\nu )<\infty $, and let $\nu _ j$ be its smoothings defined by (5.44.11). Then
\[ \sup _ j A_\theta (\nu _ j)<\infty . \]
Moreover, let $z_ j,z$ be quasi-plurisubharmonic functions with a common upper bound and a common smooth lower curvature bound, and suppose that $z_ j\to z$ both in $L^1(X)$ and in $L^1(\nu )$. Then
2D0225A9A28D
5.44.13
\begin{equation} \int _ X|z_ j-z|\, \mathrm{d}\nu _ j\longrightarrow 0. \end{equation}
Consequently, for every $\lambda >0$,
\[ \mathrm{e}^{\lambda z_ j}\nu _ j\rightharpoonup \mathrm{e}^{\lambda z}\nu . \]
Proof.
On each coordinate ball $B_ a$ choose a fixed smooth function $q_ a$ with $\mathrm{dd}^{\mathrm{c}}q_ a\geq \theta $. If $w\in \mathrm{PSH}(X,\theta )$ and $\sup _ Xw=0$, then $w+q_ a$ is plurisubharmonic, and the submean inequality gives
\[ -S_{a,j}w\leq -w+S_{a,j}q_ a-q_ a\leq -w+C, \]
with a constant independent of $w$ and $j$. Integrating and using (5.44.11) proves $A_\theta (\nu _ j)\leq A_\theta (\nu )+C$.
For the second assertion, the tail envelopes
\[ H_ j=\left(\sup _{k\geq j}z_ k\right)^* \]
decrease to $z$ by lemma 5.25.1. Enlarge the functions $q_ a$ so that $H_ j+q_ a$, $z_ j+q_ a$ and $z+q_ a$ are all plurisubharmonic. Since $H_ j\geq z_ j$ and $H_ j\geq z$, we have $|z_ j-z|\leq 2H_ j-z_ j-z$, and the identity (5.44.11) and the submean inequality imply
\[ \begin{split} \int _ X|z_ j-z|\, \mathrm{d}\nu _ j & \leq \sum _ a\int _ X\eta _ a S_{a,j}(2H_ j-z_ j-z)\, \mathrm{d}\nu \\ & \leq 2\sum _ a\int _ X\eta _ a \bigl (S_{a,j}(H_ j+q_ a)-(z+q_ a)\bigr )\, \mathrm{d}\nu +\int _ X(z-z_ j)\, \mathrm{d}\nu . \end{split} \]
All these integrals are well defined: each function on the larger side of a submean inequality is bounded above, and its negative part is controlled by that of the integrable function on the smaller side plus a constant.
Radial averages of a plurisubharmonic function decrease when the radius decreases. Therefore $S_{a,j}(H_ j+q_ a)$ decreases to $z+q_ a$ on the support of $\eta _ a$: to identify the limit, fix an index $k$, use $H_ j\leq H_ k$ for $j\geq k$, let $j\to \infty $ in the radius, and then let $k\to \infty $; the reverse inequality is the submean inequality. The nonnegative integrand in the second line of the last display is dominated on this compact support by a constant minus $z$, which is $\nu $-integrable, so its integral tends to zero by dominated convergence. The final integral tends to zero by convergence in $L^1(\nu )$. This proves (5.44.13).
Since the exponential is Lipschitz on the common range of the functions, (5.44.13) also gives $\int _ X|\mathrm{e}^{\lambda z_ j}-\mathrm{e}^{\lambda z}|\, \mathrm{d}\nu _ j\to 0$. If $\zeta >0$ is smooth, then $\zeta \mathrm{e}^{\lambda z}=\mathrm{e}^{\lambda z+\log \zeta }$ is bounded and quasi-plurisubharmonic. For any fixed bounded quasi-plurisubharmonic function $F$, the same local submean argument and dominated convergence in (5.44.11) give
\[ \int _ XF\, \mathrm{d}\nu _ j\longrightarrow \int _ XF\, \mathrm{d}\nu . \]
Applying this to $F=\zeta \mathrm{e}^{\lambda z}$ gives $\int _ X\zeta \mathrm{e}^{\lambda z_ j}\, \mathrm{d}\nu _ j\to \int _ X\zeta \mathrm{e}^{\lambda z}\, \mathrm{d}\nu $ for every smooth $\zeta >0$. Since every smooth function is a difference of two positive smooth functions, the convergence holds for all smooth test functions, and uniformly bounded masses together with uniform approximation of continuous functions by smooth ones prove weak convergence.
$\square$
Pome
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