Lemma 14.24.114FB9051BD00E

Proof. Put $a=\min _ Xw<b=\max _ Xw$, and let $k$ be the function of proposition 14.24.5. It is Lipschitz and nonnegative, and $k(a)=0$. Choose $\epsilon _ j\downarrow 0$ with $a+\epsilon _ j<b$. For $\delta >0$, set

If $L$ is a Lipschitz constant for $k$, then $k(t)\leq L(t-a)$ on this interval, so $J_ j(\delta )\to \infty $ as $\delta \downarrow 0$. Choose $\delta _ j>0$ with $J_ j=J_ j(\delta _ j)\geq j$. Let $p_ j=0$ on $(-\infty ,a]$, $p_ j=1$ on $[a+\epsilon _ j,\infty )$, and

The pushforward identity (14.24.7) gives

Every point of $U$ has a coordinate neighborhood $V$ as in proposition 14.24.5, since $F$ is continuous. By (14.24.8), therefore, $\mathrm{d}(p_ j(w))\to 0$ in $L^1_{\mathrm{loc}}(U)$. The bounded functions $p_ j(w)$ converge pointwise and in $L^1_{\mathrm{loc}}$ to $\mathbf1_{\{ w>a\} }$, whose distributional gradient therefore vanishes on $U$. Since $U$ is connected, this indicator function is almost everywhere constant on $U$. As $w$ is nonconstant, $\{ w>a\} $ is a nonempty open set, and its intersection with the dense open set $U$ is nonempty and open; hence the constant is one. Thus $\{ w=a\} \cap U$ has zero volume, and since $F=0$ off $U$, the first assertion follows. Applying the argument to $-w$ gives the assertion for the maximum set. $\square$