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Let $B \subset \mathbb R^n$ be the unit ball. Consider a Borel measurable set $E \subset B$ with positive Lebesgue measure $|E|>0$ (say $|E| = |B|/2$).

Then, Lebesgue's density theorem, says that for a.e. $x\in E$ $$ \lim_{r \downarrow 0} \frac{|B(x,r)\backslash E|}{|B(x,r)|} = 0. $$

We can restate it as follows: for a.e. $x\in E$, for all $\epsilon>0$ there exists $r_0 = r_0(x, \epsilon)>0$ such that $$ |B(x,r)\backslash E| \leq \epsilon |B(x,r)|, \quad 0<r<r_0(x,\epsilon) . $$ I am particularly interested in the dependence $\epsilon(r, x)$.

I have a question about this. Probably it has been studied but I have not been able to find any reference.

Given $E$, can we prove some uniformity for $\epsilon$ in a positive measure set (maybe of measure smaller than $|E|$)? That is, can we find some $r_*>0$ and $\phi$ continuous with $\phi(0)=0$ such that $$ \epsilon(r,x) \leq \phi(r), \quad 0<r<r_* $$ for all $x \in \tilde E$ for some Borel set $\tilde E\subset E$ with $0<|\tilde E|\leq |E|$.

Edit: Initially I had two questions but I have decided to delete one.

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  • $\begingroup$ Did you see mathoverflow.net/questions/405361/…? $\endgroup$
    – Boris Bukh
    Oct 3 at 12:54
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    $\begingroup$ Ad 1: The good news is that you have $|B_r \ E | \le |B_s \ E| \le |B_r \ E| + |B_s \ B_r |$ for $r<s$, which is optimal. If you play around this, you should get some estimates on the local Lipschitz constant of $\epsilon(r)$. The bad news is that the estimate will be something like $Lip(\epsilon) \approx 1/r$. So as long as you start early enough you can grow as fast as you want. Ad 2: I think for any fixed $r$, I can find $E$ such that $\epsilon(r) > 1/10$ or so, by a variation of the usual dense union of balls counterexample. There still might be some room if you are $E$-dependent though. $\endgroup$
    – mlk
    Oct 3 at 12:57
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    $\begingroup$ @BorisBukh That link is recursive... $\endgroup$
    – mlk
    Oct 3 at 12:58
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    $\begingroup$ Sorry. mathoverflow.net/questions/242919/… $\endgroup$
    – Boris Bukh
    Oct 3 at 13:25
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    $\begingroup$ Tao writes about this (in dimension 1) here: google.com.my/amp/s/terrytao.wordpress.com/2007/06/18/…. Maybe you might find something of interest. $\endgroup$
    – Nate River
    Oct 4 at 3:53
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Let $$f_n(x) = \sup_{r \in {\mathbb Q} \cap [\frac{1}{n+1},\frac{1}{n})} \frac{|B(x,r)\setminus E|}{|B(x,r)|}\,,$$ so that $f_n(x) \to 0$ for a.e. $x \in E$. By Egorov's theorem [1], for every $\epsilon>0$ there is a subset $\tilde{E} \subset E$ with $|E \setminus \tilde{E}| <\epsilon$, such that $f_n(x) \to 0$ uniformly on $\tilde{E}$. It follows that $$ \lim_{r \downarrow 0} \frac{|B(x,r)\backslash E|}{|B(x,r)|} = 0 $$ uniformly in $\tilde{E}$, even when the limit is considered for real $r>0$.

[1] https://en.wikipedia.org/wiki/Egorov%27s_theorem

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