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.