PomeProposition 8.7.7.7E8FF9A60609 If continuous admissible $u,v$ satisfy $\mathbf1_{\{ u<v\} }H_ m(u)=0$, then $u\geq v$.
Proposition 8.7.77E8FF9A60609
Proof. Replace $v$ by $\max (u,v)$, so $u\leq v$. For $b>1$ put $p_ b=P(bu-(b-1)v)$ and $z_ b=b^{-1}p_ b+(1-b^{-1})v\leq u$. On the contact set $C_ b=\{ p_ b=bu-(b-1)v\} $ one has $z_ b=u$; the contact inequality and polarized positivity give
\[ \mathbf1_{C_ b}b^{-m}H_ m(p_ b) \leq \mathbf1_{C_ b}H_ m(z_ b) \leq \mathbf1_{C_ b}H_ m(u). \]
The envelope measure is supported on $C_ b$ and has positive mass. It vanishes on $C_ b\cap \{ u<v\} $ by the hypothesis, so some point of $C_ b\cap \{ u=v\} $ exists. Hence $\sup _ Xp_ b\geq \inf _ Xv$, while $p_ b\leq u$. The estimate (8.7.5) is consequently uniform for $p_ b$. On the open set $E_ a=\{ u<v-a\} $, for $a>0$, one has $p_ b\leq v-ba$, and therefore
\[ C_\omega \geq \int _ X(\sup _ Xp_ b-p_ b)\omega ^ n \geq (ba-\operatorname {osc}_ Xv)\int _{E_ a}\omega ^ n. \]
Letting $b\to \infty $ gives zero volume for $E_ a$. Since it is open it is empty, and $a\downarrow 0$ gives $u\geq v$. $\square$
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