Theorem 13.19.1B390F0847B3A

Proof. Put $\ell =\underline{\operatorname {vol}}_\theta (\varphi )>0$ and $\mathcal E_\varphi =\mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$. We may replace $v$ by $\max (u,v)$: by locality, the hypothesis is unchanged on $\{ u<v\} $, and by upward stability (proposition 16.28.62) the new $v$ lies in $\mathcal E_\varphi $. Thus we may assume $u\leq v$. Translate $u,v$ by the same constant so that $v\leq \varphi $, and then translate all three potentials so that $\varphi \leq 0$. The idea is to stretch $u$ away from $v$; the stretched potentials keep lower volume $\ell $, and this is incompatible with a strict inequality $u<v$ on a set of positive volume. Set

Lemma 16.28.63 gives $q_ b\in \mathcal E_\varphi $ and $q_ b\leq u$. Since $b^{-1}q_ b+(1-b^{-1})v\leq u$ with contact set $D_ b$, contact support and positivity of the mixed products give

Here $\mu _{q_ b}$ is carried by $D_ b$. Choose $b$ so large that $(1-b^{-1})^ n>c$. On $D_ b\cap \{ u<v\} $ the hypothesis bounds the right-hand side by $c\mu _ v$, and subtracting finite measures shows that $\mu _{q_ b}$ vanishes there. Thus $\mu _{q_ b}$ is carried by $D_ b\cap \{ u=v\} \subset \{ q_ b=u\} $. Since $q_ b\leq u$, ordered contact for this pair and theorem 13.18.5, which gives $\underline{\operatorname {vol}}_\theta (q_ b)=\ell $, now yield

Suppose now that $u<v$ somewhere. Then there is $\delta >0$ such that $\{ u<v-\delta \} $ has positive Lebesgue measure; otherwise $u\geq v$ almost everywhere, and hence everywhere by lemma 5.24.3. On this set the convex inequality gives $q_ b\leq u-(b-1)\delta $. The family $q_ b$ decreases in $b$ and is bounded above by $u$. Its limit is $-\infty $ on a set of positive measure, so it cannot be $\theta $-plurisubharmonic and is therefore identically $-\infty $. Hence theorem 5.25.2 gives $\sup _ Xq_ b\to -\infty $. The sets $\{ q_ b=u\} \subset \{ u\leq \sup _ Xq_ b\} $ then have $\mu _ u$-mass tending to zero, because $\mu _ u$ is finite and puts no mass on the pole set of $u$. This contradicts (13.19.2). $\square$