Proposition 14.24.22C3AA01F259B

Proof. Choose $\delta \in (0,1]$ with $\omega _ q\geq \delta \omega $. For $0<t<1$, put $b_ t=(1-t)v+tq$, $m_ t=\min _ X(u-b_ t)$, and $E_ s=\{ u<b_ t+m_ t+s\} $. Then $\omega _{b_ t}\geq t\delta \omega $. Write $\mu =F\omega ^ n$, $d=\kappa ^{-1/n}-1$, and $\gamma =nd$. By (5.28.9),

On $L=\{ F\leq \kappa G\} $ this is at least $(1+\gamma t)\mu $, since $G^{1/n}\geq (1+d)F^{1/n}$ there, and on $R=\{ F>\kappa G\} $ it is at least $(1-nt)\mu $. Choose $s>0$ so small that the reverse comparison factor in (14.19.7), with $\varepsilon =t\delta $, is at most $1+\gamma t/2$. Modified comparison on $E_ s$, with $x=\mu (E_ s\cap L)$ and $y=\mu (E_ s\cap R)$, gives

The open set $E_ s$ is nonempty by the definition of $m_ t$. As $\omega _{b_ t}^ n\geq (t\delta )^ n\omega ^ n$, (14.19.7) forces $\mu (E_ s)>0$, hence $y>0$.

Let $a=\min _ X(u-v)$ and $\Lambda =\| v-q\| _\infty $. Since $u-b_ t=u-v+t(v-q)$, we have $|m_ t-a|\leq t\Lambda $ and $E_ s\subset \{ u-v<a+2t\Lambda +s\} $. Given $\eta $, choose $t$ and then $s$ with $2t\Lambda +s<\eta $. Since $y=\mu (E_ s\cap R)>0$, this proves (14.24.3). Interchanging $u$ and $v$ gives the assertion for the near-maximum set. $\square$