Lemma 15.28.38298FD96E154

Proof. Subtracting a common constant, which does not change the extended measures, we may assume that all potentials are nonpositive. Pass to a subsequence along which $s_ j\to s$ and $g_ j\to g$ almost everywhere, and set

Then $0\leq a_ j\uparrow g$ almost everywhere, and $w_ j\in \mathrm{PSH}(X,h)$ decreases to $s$.

Step 1: $\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)\geq a_ j\, \mathrm{d}V$. Fix $j$. For $N\geq j$, the finite maximum $m_ N=\max _{j\leq \ell \leq N}s_\ell $ satisfies $\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)\geq a_ j\, \mathrm{d}V$: on the set $\{ m_ N=s_\ell \} $, proposition 13.14.6 gives $\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)\geq \mathrm{MA}^{\mathrm{ext}}_ h(s_\ell )\geq g_\ell \, \mathrm{d}V\geq a_ j\, \mathrm{d}V$, and these sets cover $X$. As $N\to \infty $, the functions $m_ N$ increase, are bounded above by zero, and have regularized limit $w_ j$. By lemma 5.29.19, $m_ N\to w_ j$ in capacity, so the damped measures $\mathrm{e}^{m_ N}\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)$ converge weakly to $\mathrm{e}^{w_ j}\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)$. Since $m_ N\to w_ j$ almost everywhere and $a_ j\leq g$, dominated convergence gives $\mathrm{e}^{m_ N}a_ j\to \mathrm{e}^{w_ j}a_ j$ in $L^1(\mathrm{d}V)$. Passing to the limit in $\mathrm{e}^{m_ N}\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)\geq \mathrm{e}^{m_ N}a_ j\, \mathrm{d}V$ gives

The pole set of $w_ j$ has zero volume and is not charged by $\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)$. Dividing by the positive weight off this set gives $\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)\geq a_ j\, \mathrm{d}V$.

Step 2: passage to $s$. Since $w_ j$ decreases to $s\in \mathrm{PSH}(X,h)$, lemma 5.29.19 gives $w_ j\to s$ in capacity, and the damped measures of $w_ j$ converge weakly to $\mathrm{e}^ s\mathrm{MA}^{\mathrm{ext}}_ h(s)$. Moreover $0\leq \mathrm{e}^{w_ j}a_ j\leq g$ and $\mathrm{e}^{w_ j}a_ j\to \mathrm{e}^ sg$ almost everywhere, so dominated convergence and Step 1 give

Dividing by $\mathrm{e}^ s$ on $\{ s>-N\} $ and letting $N\to \infty $ gives $\mathrm{MA}^{\mathrm{ext}}_ h(s)\geq g\, \mathrm{d}V$ off the pole set of $s$. Both measures give zero mass to the pole set, which proves the lemma. $\square$