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I have this claim concerning a Lebesgue density point that I need to understand rigorously: Let $0$ be a density point of a closed set $A\subset \mathbb{R}^n$. Then for any $x\in\mathbb{R}^n$ there exists a $y\in A$ such that $\lim_{|x|\to 0}\frac{|x-y|}{|x|} =0$.

Conceptually I believe this makes sense, as $0$ is a density point, if you are really close to $0$, then there is a lot of set $A$ there as well. This means, given $x$ close to $0$, it should be really easy to find a $y\in A$ that is much closer to $x$, then $x$ is to $0$. The issue, of course, is making this rigorous. Any help would be much appreciated.

Recall: A (Lebesgue) density point of a measurable set $A$ is any $x$ such that, $$ \lim_{r\to 0} \frac{|A\cap B(x,r)|}{|B(x,r)|} = 1 $$

Highlander13
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  • Likely useful: Lebesgue points of density and similar notions – Dave L. Renfro Mar 31 '23 at 20:12
  • Thank you for this suggestion! I will look at it carefully later but I was able to solve this using a similar problem statement that was easier to work with. – Highlander13 Mar 31 '23 at 20:35
  • @DaveL.Renfro in fact the solutions to your linked post appear to be identical to my own, thanks for finding that. – Highlander13 Mar 31 '23 at 21:36
  • The question you linked in your answer also has an answer by me, which I didn't remember. I knew I'd written about this before (mostly just indicating the idea without formally writing a lot of symbolic stuff), but I didn't realize I'd done this twice before. When I found the question I linked to in my earlier comment, I didn't look further, thinking that was it was all I had previously written from this "distance to set" point of view. If interested, non-centered vs centered density -- see 1 and 2. – Dave L. Renfro Mar 31 '23 at 22:37
  • I didn't even notice you provided the answer to the other question! Thanks for the extra links. – Highlander13 Apr 01 '23 at 13:05

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Just realized this is basically the same as the question here: Prove $\frac{d(y,A)}{\lvert x-y \rvert} \to 0$ as $y \to x$ where $x$ is a density point of $A$.

For completeness sake, here is my proof for the linked question: We want to show $$ \lim_{|x|\to 0}\frac{d(x,A)}{\Vert x\Vert}=0. $$ Suppose to the contrary that the limit does not equal 0, that is, there exists $\varepsilon_0>0$ and $x_k\to0$ such that for all $k\in\mathbb{N}$, $$ \frac{d(x_k,A)}{\Vert x_k\Vert}\geq \varepsilon_0. $$ Let $\delta_k:=d(x_k,A)$, so that we have for all $k$, $\frac{\delta_k}{\Vert x_k\Vert}\geq \varepsilon_0$. Also note that $\delta_k>0$ implies $B(x_k, \delta_k/2)\cap A = \emptyset$. Now consider the balls $B(0,\Vert x_k\Vert)$, and note by the density of $0$,

\begin{equation} \lim_{k\to\infty}\frac{|A\cap B(0,\Vert x_k\Vert)|}{|B(0,\Vert x_k\Vert|} = 1. \end{equation}

We can also note that there exists $y_k\in B(0,\Vert x_k \Vert) \cap B(x_k,\delta_k/2)$** such that $$ B(y_k, \delta_k/4)\subset B(0,\Vert x_k \Vert) \cap B(x_k,\delta_k/2) $$

As $B(x_k, \delta_k/4)\cap A = \emptyset$ implies $A\cap B(0,\Vert x_k\Vert) \subset B(0,\Vert x_k \Vert)\setminus B(x_k, \delta_k/4)$ we have, $$ |A\cap B(0,\Vert x_k\Vert)| \leq |B(0,\Vert x_k \Vert)\setminus B(y_k, \delta_k/4)| = |B(0,\Vert x_k \Vert)| - |B(y_k, \delta_k/4)| $$ and hence, $$ \frac{|A\cap B(0,\Vert x_k\Vert)|}{|B(0,\Vert x_k\Vert|} \leq 1 - \frac{|B(y_k,\delta_k/4)|}{|B(0,\Vert x_k \Vert)|} = 1- \frac{\delta_k^n}{4^n\Vert x_k\Vert^n}\leq 1- \frac{\varepsilon_0^n}{4^n} $$ contradicting the definition of density point. The result then follows by noting, as $A$ is closed, for each $x\in\mathbb{R}^n$ there exists a $y\in A$ such that $$ d(x, A) = \Vert x-y\Vert $$ **To find this $y_k$ explicitly, draw a line segment from $0$ to $x_k$ and call $z_k$ the point where this line segment intersects the boundary of $B(x_k,\delta_k/2)$. Then $y_k$ is the midpoint on the line segment between $x_k$ and $z_k$. I am sure I could explicitly find a formula but I like this geometric description.

Highlander13
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