This is related to a question asked on mathstackexchange https://math.stackexchange.com/questions/831184/for-every-null-set-e-there-is-a-measurable-set-f-with-different-upper-and-lo. This question is inspired by Remark 7.4 from the paper $\textit{Structure of Null Sets in the Plane and Applications}$.

The remark in the paper states that given a set $E \subset \mathbb{R}$ of measure zero one can find a set $F\subset \mathbb{R}$ of positive measure so that all the points of $E$ have lower Lebesgue density 0 and upper Lebesgue density 1.

In the mathstackexchange post, the answer uses the uncentred balls definition for upper and lower Lebesgue density (i.e. the lower Lebesgue density is defined as $$\liminf_{x\in B,m(B)\rightarrow 0} \frac{m(F\cap B)}{m(B)}$$ and the upper Lebesgue density is defined in the same way with $\limsup$ instead of $\liminf$). My question is, is it possible to find a set $F$ using the centred balls definition of Lebesgue density? That is, given a set of $E$ of measure 0 can we find a set $F$ of positive measure so that $$\liminf_{r\rightarrow 0} \frac{m(F\cap B_r(x))}{m(B_r(x))} = 0$$ and $$\limsup_{r\rightarrow 0} \frac{m(F\cap B_r(x))}{m(B_r(x))} = 1$$ for all $x\in E$?