The Lebesgue density theorem in $\mathbb{R}^n$ may be stated as follows. For a Lebesgue-measurable $A\subseteq\mathbb{R}$ and $r>0, x\in\mathbb{R}^n$, define
$$ \chi\_{A,r}(x)=\frac{\mu(A\cap B\_r(x))}{\mu(B\_r(x))},$$
where $\mu(\cdot)$ is the Lebesgue measure and $B\_r(x)$ is the closed ball about $x$ with radius $r$.
Then the limit $\chi\_A(x)=\lim\_{r\to0}\chi\_{A,r}(x)$ exists for $\mu$-almost all $x\in\mathbb{R}^n$ and is either 0 or 1, depending on $x$'s membership in $A$.


I would like to prove a uniform version of this claim, something along the following lines: for all $R>0$,
$$ \lim\_{r\to0} \mathrm{ess}\sup\_{x\in B\_R(0)} |\chi\_{A,r}(x)-\chi\_A(x)|=0.$$


If the above is known to be false (example?), I'd be happy with a "high probability version", something like: for all $\epsilon>0$ there is a $B'\subset B\_R(0)$ with $\mu(B\_R(0)\setminus B')<\epsilon$ such that
$$ \lim\_{r\to0} \mathrm{ess}\sup\_{x\in B'} |\chi\_{A,r}(x)-\chi\_A(x)|=0.$$
The latter looks like it's provable using the techniques suggested here:
<https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/>
but if it's already known, I'd be grateful for a reference.



A uniform Lebesgue density theorem
For the first version of your claim, consider $n=1$, $A = [0,1]$. For any $r \in (0,1)$, $\chi\_{A,r}(x)$ is near $1/2$ in a neighbourhood of $0$, so 
$$\text{ess-sup}\_{x \in B\_R(0)} |\chi\_{A,r}(x) - \chi\_A(x)| \ge 1/2$$


