Theorem 15.2.2D70D1E469F9D

Proof. Since $\max (u,v)=v$ on the plurifine open set $\{ u<v\} $, plurifine locality shows that replacing $v$ by $\max (u,v)$ preserves the hypothesis; it does not change the conclusion. We may therefore assume $u\leq v$, and since $v\preceq u$ there is $C\geq 0$ with $u\leq v\leq u+C$. For $b>1$ consider the stretched envelopes

The function $u-(b-1)C$ is a competitor, so $p_ b\not\equiv -\infty $; and since $bu-(b-1)v\leq u$, we have $u-(b-1)C\leq p_ b\leq u$. By contact support, the measure of $p_ b$ is concentrated on $D_ b$. The potential $b^{-1}p_ b+(1-b^{-1})v$ lies below $u$ and agrees with it on $D_ b$, so ordered contact and positivity of mixed products give

Choose $b$ so large that $(1-b^{-1})^ n>c$. On $D_ b\cap \{ u<v\} $ the hypothesis then forces $\mathrm{MA}^{\mathrm{ext}}_\theta (p_ b)$ to vanish. Since the masses may be infinite, we justify this subtraction as follows. On each set where $u,v,p_ b>-k$, every measure involved is finite, being bounded by the mass of the corresponding bounded truncation. We subtract the finite terms on each such set and then take the union over $k$, so no infinite masses are ever subtracted.

It follows that the measure of $p_ b$ is carried by $D_ b\cap \{ u=v\} \subset \{ p_ b=u\} $. We apply lemma 15.2.1 to the pair $(p_ b,u)$: their singularities are equivalent, $\mathrm{MA}^{\mathrm{ext}}_\theta (p_ b)(\{ p_ b<u\} )=0$, and $P_{\delta \theta }(u)\not\equiv -\infty $. We obtain

for all sufficiently large $b$. Suppose now that $u\geq v$ fails. By lemma 5.24.3, the set $\{ u<v\} $ has positive volume. Since both potentials are finite almost everywhere, some set $E=\{ u<v-a\} $ with $a>0$ has positive volume and both functions are finite on $E$. On $E$ the obstacle inequality gives

quasi-everywhere. Letting $b$ tend to infinity through the integers shows that the quasi-plurisubharmonic function $P_{\delta \theta }(u)\not\equiv -\infty $ equals $-\infty $ on a set of positive volume, a contradiction. Thus $u\geq v$ off the pole sets, and lemma 5.24.3 gives the inequality everywhere. $\square$