Lemma 13.18.192ABF533A916

Proof. Adding constants to $u$ and $v$ only translates the competitors, so we may assume $v\leq 0$. Write $V=\overline{\operatorname {vol}}_\theta (v)$. Since $D<b^{-n}\delta $, the definition of the upper volume gives $w\in \mathrm{PSH}(X,\theta )$ with $w\sim u$ and

After a translation of $w$ we may also assume $w\leq v$. For positive integers $j$, put

The function $v-bj$ is a competitor for $h_ j$, and the defining convex inequality together with $w_ j\leq v$ gives $h_ j\leq v$. Thus $v-bj\leq h_ j\leq v$, so $w_ j\sim v$ and $h_ j\sim v$, and $h_ j$ decreases as $j$ increases. We claim that the decreasing limit $h$ of the $h_ j$ is not identically $-\infty $. If it were, theorem 5.25.2 would give $\sup _ Xh_ j\to -\infty $.

By theorem 15.8.6, the measure of $h_ j$ is carried by the contact set $C_ j=\{ b^{-1}h_ j+(1-b^{-1})v=w_ j\} $. Positivity of the mixed products and ordered contact for $b^{-1}h_ j+(1-b^{-1})v\leq w_ j$ give

Fix a positive integer $k$. If $h\equiv -\infty $, then $h_ j\leq -bk$ everywhere for all large $j>k$. Since $v\leq 0$, every point of $C_ j$ then satisfies $w_ j\leq -k$, hence $w\leq -k$. Moreover $w_ j=w$ on $\{ w>-k\} \subset \{ w>v-j\} $, so locality gives

Here the first inequality uses $h_ j\sim v$, and the last uses $w_ j\sim v$. Letting $k\to \infty $ contradicts (13.18.2), since the finite non-pluripolar measure $\mathrm{MA}_\theta (w)$ gives no mass to the pole set of $w$. This proves the claim. Consequently $h\in \mathrm{PSH}(X,\theta )$, and $b^{-1}h+(1-b^{-1})v\leq w$. If $w\leq u+C$, then $h-bC\in \mathcal A_ b(u,v)$, which proves the lemma. $\square$