Lebesgue points of density and similar notions

Let $F\subset \mathbb{R}^d$ and $\delta(x)=d(x,F)=\inf\{|x-y|:y\in F\}$ be the distance from $x$ to $F$. It is easy to show that $\delta(x+y)\leq |y|$ for all $x\in F$. Prove the more refined estimate: $$\lim_{|y|\rightarrow0}\frac{\delta(x+y)}{|y|}=0,\text{ for a.e. }x\in F .$$

Notes A hint is given that says "Assume $x$ is a point of density of $F$ and use the conclusion: If $E$ is a measurable subset of $\mathbb{R}^d$ then almost every $x\in E$ is a point of density of $E$, and almost every $x\in E^c$ is not a point of density of $E$.

I have proved the above result (under notes) using the Lebesgue differential theorem in a previous question. But I am having difficulty relating the limit in the question to that in the definition of a Lebesgue point of density, although they seem very similar. Any help is appreciated. Thanks!

Solution 1:

I think we only need to assume that almost every $x \in E$ is a point of Lebesgue density of $E,$ which does not require that $E$ is a Lebesgue measurable set. [Indeed, if we can assume that $E$ is a Lebesgue measurable set, then we can conclude much more than what you said, namely that at almost every $x \in E^c$ the Lebesgue density of $E$ at $x$ is equal to $0$ (instead of "is less than $1$"). But I don't see that we need to know anything about the Lebesgue density of $E$ at points not in $E.$]

The result you want follows from the following more precise result.

Theorem: Let $E \subseteq {\mathbb R}^N$, $x \in {\mathbb R}^N$, and assume that $x$ is a point of Lebesgue density of $E.$ Then $$\lim_{|y| \rightarrow 0}\frac{\delta(x+y)}{|y|} = 0$$ Proof: If not, then $\;\limsup_{|y| \rightarrow 0}\frac{\delta(x+y)}{|y|} > 0,\;$ from which it follows that there exists $\epsilon > 0$ and a sequence $\{y_n\}$ with $y_n \rightarrow x$ such that for each $n$ we have $\delta(x+y_n) \geq \epsilon |y_n|.$ The key idea is to notice that this last inequality tells us there is a sequence of arbitrarily small balls centered at $x,$ each of which contains a "sufficiently big" subball having empty intersection with $E,$ where "sufficiently big" is big enough to contradict the assumption that the Lebesgue density of $E$ at $x$ is equal to $1.$ Specifically, the balls centered at $x$ have radii $|y_n| + \epsilon |y_n|$ and the corresponding subballs are centered at $x+y_n$ and have radii $\epsilon |y_n|.$ Note that the ratio of each subball radius to the corresponding ball radius is ${\epsilon}/({1 + \epsilon}),$ and so the ratio of the volume of each subball to the corresponding volume of the ball is $\left(\frac{\epsilon}{1 + \epsilon}\right)^N.$ Because this ratio of volumes is a positive constant (that depends only on $F,$ $x,$ and $N),$ and no points of $E$ belong to any of the subballs, we cannot have the Lebesgue density of $E$ at $x$ equal to $1,$ which gives a contradiction.

To Summarize: From the assumption "the Lebesgue density of $E$ at $x$ equals $1,$" it follows that, as we shrink down to $x$ using balls centered at $x,$ the ratios of the measures of $E^c$ intersect the balls to the measures of the balls must approach $0$ (because the ratios of the measures of $E$ intersect the balls to the measures of the balls must approach $1$). However, $\delta(x+y_n) \geq \epsilon |y_n|$ provides us with a sequence of balls shrinking down to $x$ in which the ratios of the measures of $E^c$ intersect the balls to the measures of the balls is bounded above $0.$ [Note that the measure of $E^c$ intersect any of the balls is greater than or equal to the measure of the corresponding subball.]

(ADDED NEXT DAY) Something I overlooked: If we don't assume that $E$ is Lebesgue measurable then, in various places above, "measure" and "density" need to be replaced with "outer measure" and "outer density".

Solution 2:

Let $x\in F$ be s.t. $x$ is a point of (Lebesgue) density. In other words, we have: $$\lim_{m(B)\to 0,x\in B}\frac{m(B\cap F)}{m(B)}=1.$$ In particular, we can consider the collection of balls $\left\{B_{2|y|}(x)\right\}_{|y|>0}$, and then we have $$\lim_{|y|\to 0}\frac{m(B_{2|y|}(x)\cap F)}{m(B_{2|y|}(x))}=1. $$ By definition of the distance function $\delta(z)$, for any $z\in \mathbb{R}^d$ we have that $B_{\delta(z)}(z)\cap F=\emptyset$. This gives:

$$\frac{m(B_{2|y|}(x)\cap F)}{m(B_{2|y|}(x))}=\frac{m(B_{2|y|}(x)\cap F)-m(B_{\delta(x+y)}(x+y)\cap F)}{m(B_{2|y|}(x))} \enspace (*)$$ Note that for all $z\in B_{\delta(x+y)}(x+y)$, $|x-z|=|x+y-y-z|\leq |(x+y)-z|+|y|<\delta(x+y)+|y|\leq 2|y|$. Thus, $B_{\delta(x+y)}(x+y)\subset B_{2|y|}(x)$. With this in mind we can rewrite $(*)$ and bound it: \begin{align*} (*)=\frac{m\left((B_{2|y|}(x)\setminus B_{\delta(x+y)}(x+y))\cap F\right)}{ m\left( B_{2|y|}(x)\right)} &\leq \frac{m\left((B_{2|y|}(x)\setminus B_{\delta(x+y)}(x+y))\right)}{ m\left( B_{2|y|}(x)\right)}\\ &=\frac{|B_{2|y|}(x)|}{|B_{2|y|}(x)|}-\frac{|B_{\delta(x+y)}(x+y)|}{|B_{2|y|}(x)|}\\ &=1-\frac{\delta(x+y)^d v_d}{2^d|y|^d v_d}. \end{align*} where $v_d=|B_1(0)|$. Taking the limit of both sides as $|y|\to 0$ gives the desired result for $x$, and since almost every point of $F$ is a point of Lebesgue density, the statement is proven.