15.2 Domination with a smaller background2ACF52860663

Proof. Choose $C$ with $v-C\leq u$. Since $v_\delta \leq v$, the obstacle $u-v_\delta $ is bounded below by $-C$ off the pole sets, and it is finite almost everywhere. Hence the envelope

is bounded by lemma 5.40.2, and by contact support (theorem 16.28.5) its measure is carried by $D=\{ p+v_\delta =u\} $. Moreover $p+v_\delta \leq u$ everywhere, and

is a sum of two positive currents. Positivity of mixed products and ordered contact therefore give

this inequality is first proved on common finite truncation levels and then extended by exhaustion. Off the pole sets, every point of $D\cap \{ p<0\} $ satisfies $u=p+v_\delta <v_\delta \leq v$. The hypothesis therefore gives $\mathrm{MA}_{(1-\delta )\theta }(p)(\{ p<0\} )=0$. Since $(1-\delta )\theta $ is semipositive and big, bounded domination (theorem 16.28.56, with proposition 16.28.55), applied to $p$ and the constant $0$ with coefficient zero, yields $p\geq 0$. Hence $u\geq p+v_\delta \geq v_\delta $. $\square$

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. $\square$

Proof. Fix $t>0$. On $\{ u<v-t\} $ the assumed inequality gives $\mathrm{MA}_\theta ^{\mathrm{ext}}(u)\leq \mathrm{e}^{-\lambda t}\mathrm{MA}_\theta ^{\mathrm{ext}}(v-t)$. The theorem, applied to $u$ and $v-t$ with $c=\mathrm{e}^{-\lambda t}$, gives $u\geq v-t$; now let $t\downarrow 0$. For two solutions of the same equation in the stated type, apply this result in both directions. The weighted measures are compared off the common pole sets, which these measures do not charge, so no value of an infinite weight is needed. $\square$