Theorem 16.22.10 (Domination principle)0F4D01059518

Proof. After reducing to the case $u\leq v$, we show that the extrapolation envelopes $E_ a(u,v)$ agree with $u$ almost everywhere for $\mu _ u$. Hence they do not collapse as $a\to \infty $, while their limit equals $-\infty $ quasi-everywhere on $\{ u<v\} $, so that this set is pluripolar.

The potential $\max \{ u,v\} $ is more singular than $\chi $ and less singular than $u$, so

thus $\max \{ u,v\} \in \mathcal E(X,\theta ,\chi )$. Since $\{ u<\max \{ u,v\} \} =\{ u<v\} $, we may replace $v$ by $\max \{ u,v\} $ and assume that $u\leq v$. Subtracting the same constant from $u$ and $v$, we may also assume $v\leq 0$. Let $m>0$ be the common mass of $u$, $v$ and $\chi $.

For $a>1$ let $\eta _ a=E_ a(u,v)$, $\gamma _ a=a^{-1}\eta _ a+(1-a^{-1})v$ and $D_ a=\{ \gamma _ a=u>-\infty \} $. By theorem 16.22.4, $\eta _ a$ is a potential of mass $m$; in particular its obstacle $au+(1-a)v$ has a competitor, and lemma 16.22.3(2), applied to the pair $u\leq v$, gives $\eta _ a\leq u$, $\gamma _ a\leq u$ and

Since $u\leq v$, the hypothesis means that $\mu _ u$ is carried by $\{ u=v\} $, so $\mu _{\eta _ a}$ is carried by $D_ a\cap \{ u=v\} $. At a point of this set, $a^{-1}\eta _ a+(1-a^{-1})u=u$, that is, $\eta _ a=u$. Hence $\mu _{\eta _ a}$ is carried by $\{ \eta _ a=u\} $, and lemma 16.13.1, applied to $\eta _ a\leq u$, gives $\mu _{\eta _ a}\leq \mathbf1_{\{ \eta _ a=u\} }\mu _ u\leq \mu _ u$. Both measures have mass $m$, so they are equal, and $\eta _ a=u$ holds $\mu _ u$-almost everywhere. In particular

where the last integral is positive because $m>0$ and $\mu _ u$ does not charge the pole set of $u$.

As $a$ increases, the obstacle $v+a(u-v)$ decreases, so $\eta _ a$ decreases. If $\eta _ a$ decreased to $-\infty $ identically as $a\to \infty $ through the integers, dominated convergence, with $0\leq \mathrm e^{\eta _ a}\leq 1$, would make the left side of the last display tend to zero, a contradiction. Hence the limit $\eta $ along the integers is a potential. Each $\eta _ a$ satisfies $\eta _ a\leq v+a(u-v)$ outside a pluripolar set; let $N$ be the union of these sets over the integers $a\geq 2$. At a point outside $N$ where $u$ is finite and $u<v$, we get $\eta \leq v+a(u-v)$ for every integer $a$, hence $\eta =-\infty $. Thus $\{ u<v\} $ is contained in the union of $N$ and the pole sets of $u$ and $\eta $, which is pluripolar. Therefore $u\geq v$ quasi-everywhere, and hence everywhere. $\square$