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 $x\in E$?



For Every Measure Zero Set $E$ There Exists a Positive Measure with Lower Lebesgue Density 0 and Upper Lebesgue Density 1
Yes, we can do this by a small modification of the original argument, see the [linked question.](https://math.stackexchange.com/questions/831184/for-every-null-set-e-there-is-a-measurable-set-f-with-different-upper-and-lo) I'll describe the whole argument again here, but if you already read those answers, then the short version is that we remove successively less of each $I\_n(p)$ as we approach the boundary of that interval. This allows us (for each $x\in I\_n(p)$) to replace $I\_n(p)$ by its maximal subinterval centered at $x$ while only slightly disturbing the approximate densities.


Construct open sets $G\_1\supset G\_2 \supset \ldots \supset E$, as follows. Start with any open $G\_1\supset E$, and write $G\_1 = \bigcup I\_1(p)$ as the union of its components. For each $I\_1(p)$, we construct an open set $G\_2(p)$ with $E\subset G\_2(p)\subset I\_1(p)$. This $G\_2(p)$ will cover only a small portion of $I\_1(p)$ (possible since $|E|=0$), and, moreover, this portion will not be unduly concentrated near the boundary of $I\_1(p)$ (this is the modification). More specifically, if $I\_1(p)=(a,b)$, then we demand that
$$
|(a,a+h)\cap G\_2(p)| \le 2^{-1}h \quad\quad\quad (1)
$$
for all $h>0$, and similarly near $b$. We can do this by cutting $I\_1(p)$ into pieces $(a+2^{-k}, a+2^{-k+1})$ and treating these separately (if an endpoint is in $E$ here, I very slightly move such a point).


Then we put $G\_2=\bigcup G\_2(p)=\bigcup I\_2(q)$. We repeat this whole procedure, with $2^{-1}$ replaced by $2^{-2}$ in (1), to obtain $G\_3$ etc., and we finally set $F=\bigcup (G\_{2n-1}\setminus G\_{2n})$ (start with $G\_1$, remove $G\_2$, add $G\_3$ etc. etc.).


If $x\in E$, then $x\in I\_n(p\_n)$ for some (unique) $p\_n$ for all $n$. If $n$ is odd and we write $I\_n(p\_n)=(a,b)$ and $x$ is closer to $a$ than to $b$, then we consider $J=(a,a+2(x-a))$. Then $J\cap F\supset J\setminus G\_{n+1}$, so (1) says that $|J\cap F|/|J|$ is almost $1$ if $n$ is large.


Similarly, we find intervals for which this ratio is almost zero from the even $n$'s.


