15.37.8 Domination for extended measures6EC4D5B55F22

Proof. Translate both potentials by a common constant so that $u\leq v\leq \varphi \leq 0$. Fix $\delta $ with $\varepsilon <\delta <1$, put $\beta =1-\delta $, and set

This envelope is not identically $-\infty $ by theorem 16.28.70, applied to $v$. By lemma 16.28.68,

The stretching lemma lemma 16.28.63, applied with reference zero, shows that $z=P_\theta (w/\beta )$ belongs to $\mathcal E_0$. Scaling the background and each defining competitor gives

Indeed, if $a\in \mathrm{PSH}(X,\theta )$ lies below $t z$, then $\beta a\in \mathrm{PSH}(X,\beta \theta )$ lies below $t\beta z$, for every $t\geq 1$.

Since $q\leq v\leq \varphi $, we have

off the pole sets. Thus the envelope

is not identically $-\infty $. Here the difference obstacle is taken off the pluripolar pole sets; it is quasi-continuous and finite almost everywhere. The general envelope theorem gives a finite upper bound for $k$ and the inequality $k+q\leq u$. Since $\beta z\leq k$, upward stability gives $k\in \mathcal E_{\mathrm{env}}(X,\beta \theta ,0)$.

Write $\nu _ k=\mathrm{MA}^{\mathrm{ext}}_{\beta \theta }(k)$. By contact support, $\nu _ k$ is carried by

outside the pole sets. Positivity of mixed products and ordered contact give

These inequalities are first proved on common finite truncation levels and then pass to the limit by exhaustion. On $D\cap \{ k<0\} $ we have $u=k+q<q\leq v$, so the hypothesis gives $\nu _ k(\{ k<0\} )=0$.

This vanishing implies that $\nu _ k$ is finite. Indeed, $\widehat k=\max (k,-1)$ is a bounded $\beta \theta $-plurisubharmonic function, and locality gives

We now apply theorem 15.37.6 in the background $\beta \theta $, with reference zero, to the potentials $k$ and $0$ and the coefficient zero. Its hypotheses hold: $\beta \theta $ is semipositive and big, the reference zero satisfies the model condition, $\nu _ k$ is finite and vanishes on $\{ k<0\} $, and

The scaling identity follows from the bijection $a\mapsto \beta a$ between bounded potentials in the two backgrounds. The theorem yields $k\geq 0$, and hence $u\geq q$.

It remains to let $\delta \uparrow 1$. Fix $\delta _0\in (\varepsilon ,1)$ and put $r=P_{\delta _0\theta }(v)$. For $\delta _0<\delta <1$, set $t=(\delta -\delta _0)/(1-\delta _0)$. The potential $t v+(1-t)r$ is $\delta \theta $-plurisubharmonic and lies below $v$. Therefore

Letting $t\uparrow 1$ at points where $v$ and $r$ are finite gives $u\geq v$ almost everywhere, and then everywhere by lemma 5.24.3. Together with $u\leq v$, this proves equality. $\square$

Proof. Put $w=\max (u,v)\in \mathcal E_\varphi $ and choose $b>1$ with $(1-b^{-1})^ n>c$. The stretching lemma gives $p=Q_ b(u,w)\in \mathcal E_\varphi $ and $p\leq u$. By contact support, its measure is carried by

On $D$, positivity and ordered contact imply

On $\{ u<w\} $, locality gives $\mu _ u\leq c\mu _ w$. We restrict both inequalities to $D\cap \{ u<w\} $ and to common finite truncation levels, where all measures are finite and may be subtracted. Since $(1-b^{-1})^ n>c$, it follows that $\mu _ p$ vanishes there. Exhausting the finite levels, we conclude that $\mu _ p$ is carried by $D\cap \{ u=w\} \subseteq \{ p=u\} $; that is, $\mu _ p(\{ p<u\} )=0$.

Now lemma 15.37.9, applied to $p\leq u$, both in $\mathcal E_\varphi $, gives $p=u$. Substituting this into the defining inequality $b^{-1}p+(1-b^{-1})w\leq u$ yields $w\leq u$ off the poles, hence everywhere by lemma 5.24.3. Thus $v\leq w\leq u$. $\square$

Proof. Fix $a>0$. On $\{ u<v-a\} $ the exponential inequality gives $\mu _ u\leq \mathrm{e}^{-\lambda a}\mu _ v$; here all weights are evaluated off the pluripolar pole sets. The theorem, applied to $v-a$, gives $u\geq v-a$, and we let $a\downarrow 0$. For the second assertion, suppose $c<1$. Then the theorem applies to $v+A$ for every real $A$, since $\mu _{v+A}=\mu _ v$, and gives $u\geq v+A$. This is impossible at a point where $u$ and $v$ are both finite. $\square$