Prove that $\lim_{x\to y} \frac{d(x, F )}{|x−y|} = 0$ for a.e. $y \in F$.

Question

Let $F \subset \mathbb{R}$ be a closed set and define the distance from $x \in \mathbb{R}$ to $F$ by $d(x,F)= \inf_{y \in F} |x−y|.$

Prove that

$$\lim_{x\to y} \frac{d(x, F )}{|x−y|} = 0$$ for a.e. $y \in F$.

Hint: Consider Lebesgue points of F.

I am not sure where to begin on this question. Any suggestions? Thanks.

Answer 1

The quickest way to obtain the result you seek is to appeal to Rademacher's Theorem since the distance function is $1$-Lipschitz and since this function attains its minimum at $y\in F$ the derivative can only be $0$. I suspect this is not available to you (just thought it might be interesting) so we proceed as follows:

If $x\in F$ then we have $\frac{d(x,F)}{\lvert x-y\rvert}=0$ so we only worry about $x\in F^{c}$. In this case we denote by $z_{x}\in F\cap[x,y]$ the closest point to $x$ in $[x,y]$ that belongs to $F$. Notice that it must be that $(x,z_{x})\subset F^{c}$ by the definition of $z_{x}$. So we have:

$0\le\frac{d(x,F)}{\lvert x-y\rvert}\le\frac{\lvert x-z_{x}\rvert}{\lvert x-y\rvert}=\frac{m((z_{x},x))}{\lvert x-y\rvert}\le\frac{m(F^{c}\cap(x,y))}{m((x,y))}\to0$.

if $y$ is a Lebesgue point of $F$. Thus, at Lebesgue points of $F$ the derivative exists and is $0$. Since almost every point is a Lebesgue point we conclude that the derivative exists almost everywhere.