# Set where the speed of convergence is uniform in Lebesgue's density theorem

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 $$|B(x,r)\backslash E| \leq \epsilon |B(x,r)|, \quad 0 I am particularly interested in the dependence $$\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 $$\epsilon(r,x) \leq \phi(r), \quad 0 for all $$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.

• Did you see mathoverflow.net/questions/405361/…? Oct 3 at 12:54
• 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.
– mlk
Oct 3 at 12:57
• @BorisBukh That link is recursive...
– mlk
Oct 3 at 12:58
• Oct 3 at 13:25
• 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. Oct 4 at 3:53

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 $$r>0$$.