Lemma 9.5.4FDF6E2ABDF34

Proof. Fix $a,b\in B\setminus E$. If $c\in B$ and the segment $[a,c]$ meets $E$ at a point $e=a+s(c-a)$, then $0<s\leq 1$, since $a\notin E$, so $c=a+t(e-a)$ with $t=1/s\geq 1$. We show that the set of such $c$ has $N$-dimensional Lebesgue measure zero. Exhaust $E$ by the compact sets $E_ k=E\cap \overline{B_ k}$, where $B_ k$ are concentric balls increasing to $B$. Since $a\notin E_ k$, the parameter $t$ is bounded on $\{ (e,t):e\in E_ k,\ a+t(e-a)\in B\} $, say by $T_ k$, and the map $(e,t)\mapsto a+t(e-a)$ is Lipschitz on $E_ k\times [1,T_ k]$. Cover $E_ k$ by balls of radii $r_ l$ with $\sum _ lr_ l^{N-1}$ arbitrarily small. The image of each product of such a ball with $[1,T_ k]$ is covered by at most $C_ k/r_ l$ balls of radius $C_ kr_ l$, whose total volume is at most $C'_ kr_ l^{N-1}$. Hence the image of $E_ k\times [1,T_ k]$ is Lebesgue null, and so is the union over $k$. The same holds with $b$ in place of $a$. Choose $c\in B$ outside both null sets. Then the path $[a,c]\cup [c,b]$ lies in the convex set $B$ and does not meet $E$. $\square$