Theorem 15.37.65023B8CBC940

Proof. It suffices to prove $u\geq w$ for $w=\max (u,v)$. Upward stability (proposition 16.28.62) gives $w\in \mathcal E_\varphi $, and locality gives $\mu _ u\leq c\mu _ w$ on $\{ u<w\} $. For $b>1$ put $q_ b=Q_ b(u,w)$, which belongs to $\mathcal E_\varphi $ by lemma 16.28.63. It satisfies $q_ b\leq u$, and by contact support its measure is carried by

On $D_ b$ the potential $b^{-1}q_ b+(1-b^{-1})w$ touches $u$ from below, so positivity of mixed products and ordered contact give

Choose $b$ so large that $(1-b^{-1})^ n>c$. Comparing this inequality with $\mu _ u\leq c\mu _ w$ on $D_ b\cap \{ u<w\} $ forces both $\mu _ w$ and $\mu _{q_ b}$ to vanish there. Since $\mu _ w(X)$ may be infinite, this subtraction is first carried out on common finite truncation levels of the potentials, where all measures are finite; exhausting the complements of the pole sets then gives the assertion for the extended measures.

It follows that $\mu _{q_ b}$ is carried by $D_ b\cap \{ u=w\} \subseteq \{ q_ b=u\} $. Since $q_ b\leq u$, ordered contact applied once more gives

with coefficient one, independent of $b$.

As $b\to \infty $ through the integers, the potentials $q_ b$ decrease, because the obstacles $bu-(b-1)w=u-(b-1)(w-u)$ do. Suppose that they decreased to $-\infty $ identically. Then theorem 5.25.2 would give $M_ b:=\sup _ Xq_ b\to -\infty $, and (15.37.7) would imply

because $\mu _ u$ is finite and does not charge the pole set of $u$. This is impossible, so the decreasing limit $q$ is not identically $-\infty $. Now fix $a>0$. On $\{ u<w-a\} $, the defining inequality of $q_ b$ gives $q_ b\leq u-(b-1)a$, so $q=-\infty $ at every point of this set where $u$ is finite. Since neither $q$ nor $u$ is identically $-\infty $, their pole sets have zero smooth volume, and hence so does $\{ u<w-a\} $. Thus $u\geq w$ almost everywhere, and then everywhere by lemma 5.24.3. In particular $u\geq v$. $\square$