Theorem 16.28.56 (Guedj–Lu)C67CEBABC968

Proof. If the conclusion failed, we could choose $a>0$ such that $E=\{ u<v-a\} $ is nonempty. This set has positive volume by lemma 5.24.4. Fix $C$ with $|u|,|v|\leq C$.

For $b>1$ define the bounded envelope

Since $\theta \geq 0$, constants below the obstacle are competitors. By the bounded contact theorem, theorem 16.28.4, the measure $\mathrm{MA}_\theta (w_ b)$ is concentrated on $D_ b$. The potential $b^{-1}w_ b+(1-b^{-1})v$ lies below $u$ quasi-everywhere, hence everywhere by lemma 5.24.3, and it equals $u$ on $D_ b$ outside a pluripolar set. Ordered contact and positivity of the mixed terms give

Choose $b$ so large that $(1-b^{-1})^ n>c$. Then the hypothesis forces $\mathrm{MA}_\theta (w_ b)=0$ on $D_ b\cap \{ u<v\} $. Since $\theta $ is non-collapsing, $w_ b$ has positive total mass, so some of this mass is carried by $D_ b\cap \{ u\geq v\} $. At a point of this set outside the exceptional pluripolar set,

Consequently $\sup _ Xw_ b\geq -C$ for all large $b$.

On the other hand, on the fixed set $E$ the obstacle bound gives $w_ b\leq C-(b-1)a$ quasi-everywhere. Hence

for all sufficiently large $b$. This contradicts the uniform $L^1$ bound for normalized potentials in theorem 5.25.2. $\square$