Pome905590071E2FThe extraction lemma provides the lower envelope required by the integral criterion. The same fixed measure then controls the measures of deep sublevel sets, so convergence in capacity passes to the Monge–Ampère measures themselves.
Theorem 15.37.19.6FBF04A6D4DE Suppose $u_ j\in \mathcal E_\varphi $, $u_ j\to u\in \mathrm{PSH}(X,\theta )$ in $L^1(X)$, and $\mu _{u_ j}\leq \nu $ for a fixed finite non-pluripolar measure $\nu $. Then
Proof. Start with an arbitrary subsequence. By lemma 15.37.13, a further subsequence has a common lower envelope $h\in \mathcal E_\varphi $, and $u\in \mathcal E_\varphi $. Inequality (15.37.2) and Hartogs' lemma give a common upper bound $u_ j\leq \varphi +C$.
After a common translation, assume $u_ j,u\leq 0$. Since $\mu _{u_ j}\leq \nu $, lemma 5.44.2 gives
Thus proposition 15.37.17 gives convergence in capacity along this subsequence. For the measures themselves, the common lower bound gives
since $\nu $ is finite and does not charge the pole set of $h\not\equiv -\infty $. The tail criterion for smooth backgrounds, theorem 13.15.23, therefore gives $\mu _{u_ j}\rightharpoonup \mu _ u$, and weak limits preserve domination by the fixed finite measure $\nu $. Every subsequence of the original sequence has a further subsequence with these conclusions and the same limit $u$, so both convergences hold for the original sequence. $\square$
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