Corollary 15.37.11267BAD315E3C

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. $\square$