Proposition 10.31.190BEB0F48780F

Proof. The limit $v$ is not identically $-\infty $ and satisfies $\sup _ Xv=0$, and $v_ j,v\leq V\leq 0$. By lemma 5.44.12, $\sup _ jA_\theta (\nu _ j)<\infty $, which gives the uniform estimate

In particular the relative first moments $\int _ X(V-v_ j)\mu _{v_ j}$ are bounded, and proposition 10.31.8, with fixed reference $w=V$, puts $v$ in $\mathcal E_ V$.

By proposition 10.31.15, applied to the fixed measure $\nu $, the $v_ j$ are uniformly integrable with respect to $\nu $. For each fixed $T$, the bounded functions $\max (v_ j,-T)$ have a common lower curvature bound in a Hermitian background $\Omega \geq \theta $ and converge in $L^1$ for smooth volume to $\max (v,-T)$; hence lemma 5.44.2 gives their convergence in $L^1(\nu )$. These truncations need not be $\theta $-plurisubharmonic; they are used only in the larger Hermitian background. Uniform integrability, followed by $T\to \infty $, gives $v_ j\to v$ in $L^1(\nu )$, and the limit is integrable because $A_\theta (\nu )<\infty $.

The smoothing lemma, in the form (5.44.13), now yields

Therefore the self-measure convergence theorem theorem 10.31.11 gives convergence in capacity. Finally, (10.31.20) gives the sublevel estimate

and the masses of the $\mu _{v_ j}$ are bounded by $M_ V$. The tail criterion theorem 13.15.23 therefore gives the asserted weak convergence. $\square$