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.
Best Answer
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