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

Question

So the question was:

For all $E\subset \mathbb R$ compact, $\lim_{x\to y} \frac{d(x, E)}{|x − y|} = 0$ for $a.e.y\in E.$

This is pretty travial if $E$ is open since for every point in $E$ you can create a ball around it and have the ball be in $E$. I was first thinking of proofing the fact on $\text{int}(E)$ (which is true since it is open), however it is not guaranteed that $m(\partial E)=0$. How would I fix this?

If it helps, you can reduce to proving that for any fixed $n\in\Bbb N$ we have $\lim_{x\to y} d(x,E)/|x-y| \le \frac1n$ for $y\in E$ almost everywhere, since taking the countable intersection over $n\in\Bbb N$ preserves sets of measure $0$. — Greg Martin, Apr 01 '20 at 17:18
Click on edit to see how I reformatted parts of your question. — zhw., Apr 01 '20 at 19:20
@Greg Martin I am trying to show that the set $E_n={y \in E|\exists r>0:\frac{d(E,x)}{|x-y|}<\frac{1}{n}, \forall x \in B(y,r)}$ has $m(E_n)=m(E)$. I am still a bit unsure how I would proof $m(E/E_n)=0$? Is there a problem with my setup? — , Apr 01 '20 at 20:13
Related: To show that the set point distant by 1 of a compact set has Lebesgue measure $0$ — Dave L. Renfro, Apr 26 '20 at 13:11

score 2 · Answer 1 · edited Apr 26 '20 at 09:12

This is true for any Lebesgue point $y$ of $E.$ First, recall the definition: $y$ is a Lebesgue point of $E$ provided

$$\lim_{x\to y^-}\frac{m((x,y)\cap E)}{y-x}=1,$$

with a similar statement from the right. Let's also recall Lebesgue's result: a.e. $y\in E$ is a Lebesgue point of $E.$

Fix such a $y.$ Think of $x$ as approaching $y$ from the left. Suppose the closest point of $E\cap [x,y]$ to $x,$ is $x+\delta,$ where $\delta\ge 0.$ As $x$ moves toward $y,$ $\delta$ will likely vary.

Now $(x,x+\delta)\cap E=\emptyset.$ It follows that

$$1=\lim_{x\to y^-}\frac{m((x,y)\cap E)}{y-x} \leq \lim_{x\to y^+}\frac{y-(x+\delta)}{y-x} = 1-\lim_{x\to y^+}\frac{\delta}{y-x}.$$

This implies $\dfrac{\delta}{y-x} \to 0$ which implies $\dfrac{d(x,E)}{y-x} \to 0.$

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

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