Theorem 8.13.10A532C77D7D0C

Proof. Add the same constant to $u,v$ so that $v\geq 1$. This preserves the measure hypothesis and the desired conclusion. Choose $0<\alpha <1$ with $\alpha ^ m>c$ and put $V=\alpha v$, $\eta =1-\alpha $. Suppose $S=\inf _ X(u-V)<0$. For $0<s<-S$, the set $E_ s=\{ u<V+S+s\} $ is contained in $\{ u<V\} $ and hence in $\{ u<v\} $. It has positive volume. Indeed, if its volume were zero, then $u\geq V+S+s$ almost everywhere. Both functions are bounded admissible with their canonical representatives, so 8.2.10 would give this order everywhere, contradicting the definition of $S$. The argument does not require $E_ s$ to be open or the infimum to be attained.

Choose also $s<\eta ^3/C$, where $C$ is from 8.13.8. By 8.12.3 and the measure hypothesis,

If $c>0$, take $s$ still smaller so that $c(1+Cs\eta ^{-m})<\alpha ^ m$. If $c=0$, this strict inequality already holds. All measures are finite and positive, so the display forces the positive-volume term to be zero, a contradiction. Hence $S\geq 0$ and $u\geq \alpha v$. Let $\alpha $ increase to one, with $\alpha ^ m>c$. Since $v$ is bounded, $\alpha v\to v$ uniformly, and therefore $u\geq v$. The final assertion is $c=0$. $\square$