Lemma 17.5.144D025B7103EA

Proof. Put $g_ j=\inf _{k\geq j}f_ k$, so that $g_ j\uparrow f$ $\nu $-almost everywhere. We first prove that $\mu _{a_ j}\geq g_ j\nu $ for the tail envelopes $a_ j=(\sup _{k\geq j}q_ k)^*$, and then let $j\to \infty $. For $l\geq j$, the finite maximum $\max \{ q_ j,\ldots ,q_ l\} $ has measure at least $g_ j\nu $. Indeed, let $r_1,r_2\in \mathrm{PSH}(X,\theta )$ satisfy $\mu _{r_1},\mu _{r_2}\geq g_ j\nu $, and put $r=\max \{ r_1,r_2\} $. The set $\{ r_1\geq r_2\} $ is the contact set $\{ r_1=r\} $, so lemma 16.13.1, applied to $r_1\leq r$, gives $\mu _ r\geq \mu _{r_1}\geq g_ j\nu $ on $\{ r_1\geq r_2\} $; in the same way $\mu _ r\geq g_ j\nu $ on $\{ r_2>r_1\} $. Hence $\mu _ r\geq g_ j\nu $. Since $\mu _{q_ k}\geq f_ k\nu \geq g_ j\nu $ for $k\geq j$, induction on $l$ proves the claim.

The suprema $\sup _ Xq_ k$ are bounded, by proposition 10.3.1, so the regularized upper envelope $a_ j$ is $\theta $-psh, and it equals $\sup _{k\geq j}q_ k$ almost everywhere (theorem 5.10.2, applied in charts after adding smooth local potentials of $\theta $). The finite maxima $\max \{ q_ j,\ldots ,q_ l\} $ therefore increase almost everywhere to $a_ j$ as $l\to \infty $, and their measures converge weakly to $\mu _{a_ j}$ by corollary 13.8.9. The fixed measure $g_ j\nu \leq \mu _{q_ j}$ is finite, so the inequality passes to the weak limit, by lemma 17.5.3(2):

By lemma 17.4.3, $a_ j$ decreases pointwise to $q$.

To let $j\to \infty $ we truncate. Fix $C>0$ and an index $j_0$. The truncations $a_ j^ C=\max \{ a_ j,V_\theta -C\} $ have uniformly bounded difference from $V_\theta $: they are at least $V_\theta -C$, and since the suprema of the $a_ j$ are bounded, minimality of $V_\theta $ bounds them above by $V_\theta +C'$. They decrease to $q^ C=\max \{ q,V_\theta -C\} $, so they converge to $q^ C$ in extended capacity, by lemma 13.7.3 applied in charts after adding smooth local potentials of $\theta $. Since $q^ C$ has bounded difference from $V_\theta $, it has the mass of $V_\theta $ (corollary 13.8.8), that is, $q^ C\in \mathcal E(X,\theta ,V_\theta )$. Corollary 13.9.4, with $V_\theta $ in place of $\phi $, therefore gives $\mu _{a_ j^ C}\to \mu _{q^ C}$ weakly.

On the set $A_ C=\{ q>V_\theta -C\} $ we have $a_ j\geq q>V_\theta -C$, so $a_ j^ C=a_ j$ on the plurifine open set $\{ a_ j>V_\theta -C\} \supset A_ C$, and locality gives $\mu _{a_ j^ C}\geq \mathbf1_{A_ C}g_{j_0}\nu $ for $j\geq j_0$. Since $\mathbf1_{A_ C}g_{j_0}\nu $ is a fixed finite measure, the inequality passes to the weak limit, by lemma 17.5.3(2), and gives $\mu _{q^ C}\geq \mathbf1_{A_ C}g_{j_0}\nu $. On the plurifine open set $A_ C$ we have $q^ C=q$, so locality gives $\mu _ q\geq \mathbf1_{A_ C}g_{j_0}\nu $. Thus $\mu _ q(E)\geq \int _{E\cap A_ C}g_{j_0}\, \mathrm d\nu $ for every Borel set $E$. Let $j_0\to \infty $ and then $C\to \infty $. The sets $A_ C$ increase to the set where $q$ and $V_\theta $ are finite, whose complement is pluripolar and hence $\nu $-null, so monotone convergence gives $\mu _ q(E)\geq \int _ Ef\, \mathrm d\nu $, that is, $\mu _ q\geq f\nu $. $\square$