PomeTheorem 15.37.10.FB5C83859579 Let $u\in \mathcal E_\varphi $ and $v\in \mathrm{PSH}(X,\theta )$ with $v\preceq \varphi $. If $0\leq c<1$ and
\[ \mu _ u\leq c\mu _ v\quad \hbox{on }\{ u<v\} , \]
then $u\geq v$.
Theorem 15.37.10FB5C83859579
Proof. Put $w=\max (u,v)\in \mathcal E_\varphi $ and choose $b>1$ with $(1-b^{-1})^ n>c$. The stretching lemma gives $p=Q_ b(u,w)\in \mathcal E_\varphi $ and $p\leq u$. By contact support, its measure is carried by
\[ D=\{ b^{-1}p+(1-b^{-1})w=u\} . \]
On $D$, positivity and ordered contact imply
\[ b^{-n}\mu _ p+(1-b^{-1})^ n\mu _ w\leq \mu _ u. \]
On $\{ u<w\} $, locality gives $\mu _ u\leq c\mu _ w$. We restrict both inequalities to $D\cap \{ u<w\} $ and to common finite truncation levels, where all measures are finite and may be subtracted. Since $(1-b^{-1})^ n>c$, it follows that $\mu _ p$ vanishes there. Exhausting the finite levels, we conclude that $\mu _ p$ is carried by $D\cap \{ u=w\} \subseteq \{ p=u\} $; that is, $\mu _ p(\{ p<u\} )=0$.
Now lemma 15.37.9, applied to $p\leq u$, both in $\mathcal E_\varphi $, gives $p=u$. Substituting this into the defining inequality $b^{-1}p+(1-b^{-1})w\leq u$ yields $w\leq u$ off the poles, hence everywhere by lemma 5.24.3. Thus $v\leq w\leq u$. $\square$
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