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 $x\in E$?

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    $\begingroup$ You want $E, F\subset\mathbb R$, as in the linked question? $\endgroup$ Commented Jun 16, 2014 at 19:24
  • $\begingroup$ @ChristianRemling Ahhh! Thank you. That is very important. Yes, $E,F \subset \mathbb{R}$. I will edit the question. $\endgroup$
    – k3thomps
    Commented Jun 16, 2014 at 19:26

1 Answer 1


Yes, we can do this by a small modification of the original argument, see the linked question. 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.

  • $\begingroup$ Thanks for your insight! I guess the wiggle room that you take advantage of is that we used the measure to find the sets $G_i$, but this is done in some black box sort of way and we don't really know anything about them. Your idea is to still let the measure pick the sets $G_i$, but you only let it pick sets from the universe of sets satisfying condition (1). Is this correct? $\endgroup$
    – k3thomps
    Commented Jun 17, 2014 at 15:24
  • $\begingroup$ Yes, and I can do this by cutting $I$ into smaller pieces. Say $I$ is $[0,1]$. Then, for each large $k$, I choose an open set $H_k$ with $E\subset H_k \subset (2^{-k},2^{-k+1})$ with $|H_k|\le 2^{-n-k}$ (if we are at step $n$). This just uses (outer) regularity of Lebesgue measure. Then $G_{n+1}(p)$ is the union of these $H_k$'s. $\endgroup$ Commented Jun 17, 2014 at 17:11
  • $\begingroup$ Very nice. Thank you for your answer! $\endgroup$
    – k3thomps
    Commented Jun 17, 2014 at 17:46

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