Corollary 13.19.385CE662E8010

Proof. Fix $\delta >0$. On $\{ u<v-\delta \} $ the weighted inequality gives $\mu _ u\leq \mathrm{e}^{-\lambda \delta }\mu _ v=\mathrm{e}^{-\lambda \delta }\mu _{v-\delta }$. Theorem 13.19.1, applied to $v-\delta $ with $c=\mathrm{e}^{-\lambda \delta }<1$, gives $u\geq v-\delta $; now let $\delta \downarrow 0$. If the second assertion failed, the theorem would apply with the same $c<1$ to every translate $v+A$, $A\in \mathbb R$, and would give $u\geq v+A$ for all $A$. This is impossible at a point where $u$ and $v$ are both finite. $\square$