Prove $\frac{d(y,A)}{\lvert x-y \rvert} \to 0$ as $y \to x$ where $x$ is a density point of $A$.

Question

For $A \subseteq \mathbb R^n$ and $y \in \mathbb R^n$, we define $$d(y,A) = \inf \{\lvert x-y \rvert : x \in A \}.$$ Also, for a (Lebesgue) measurable set $A \subseteq \mathbb R^n$, we say $x \in \mathbb A$ is a density point of $A$ if $$\lim_{r \to 0^+} \frac{\lambda(B_r(x) \cap A)}{\lambda(B_r(x))} = 1,$$ where $\lambda$ is the Lebesgue measure on $\mathbb R^n$. I am trying to prove that if $A \subseteq \mathbb R^n$ is measurable, then for any $x \in A$ which is a density point of $A$, we have $$\lim_{y \to x} \frac{d(y,A)}{\lvert x -y \rvert} = 0.$$

I have been stuck on this for some time and I don't feel as though I have made any progress. It seems like it should be proven using something like the Lebesgue Differentiation Theorem, but I don't know what function to apply the theorem to. Any help would be greatly appreciated. Thank you!

Answer 1

Outline of a proof

Think of "$x$ is a density point of $A$" in this way. As you consider balls centered at $x$ with radii approaching $0,$ the measures of "$A$ intersect these balls" will approach (from a percentage/ratio point of view) the measure of the balls. That is, as you use balls to contract down onto $x,$ the balls will fill up with points of $A$ so as to approach 100% of the measure of these balls.

Now if

$$\limsup_{y \to x} \frac{d(y,A)}{\lvert x -y \rvert} > 0,$$

then there would be a sequence $\{B_{{\epsilon}_n}(x)\}$ of such balls contracting down onto $x\;$ (i.e. ${\epsilon}_n \rightarrow 0)\;$ such that each $B_{{\epsilon}_n}(x)$ contains a subball of radius $\delta_n \geq K\epsilon_n$ where the subball contains no points of $A$ and where $K > 0$ is a constant possibly depending on $x$ and $A,$ but not depending on $n.$ (Why?) The presence of these subballs means that

$$\limsup_{n \rightarrow \infty} \frac{\lambda(B_{{\epsilon}_n}(x) \cap A)}{\lambda(B_{{\epsilon}_n}(x))} < 1,$$

which contradicts $x$ being a point of density of $A.$