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Is there a Borel set $\sigma\subset [0,1]$ of positive Lebesgue measure such that for all $x\in\sigma$ and all $\varepsilon > 0$ we have that $|B_\varepsilon(x)\setminus\sigma| > 0$?

Here, $|\cdot|$ denotes the Lebesgue measure and $B_r(x)$ is the interval with center $x$ and length $r > 0$.

It seems to me that the Smith-Volterra-Cantor set might be a candidate for this, but I cannot prove it.

Friedrich Philipp
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    What about $\sigma$ equals the irrational numbers in $(0,1)$? – Gregory Grant Jul 15 '17 at 03:16
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    Have you tried using a Cantor set of positive measure? – Mustafa Said Jul 15 '17 at 03:16
  • @Gregory Grant: That doesn't work because the rational numbers have measure zero. – carmichael561 Jul 15 '17 at 03:17
  • @carmichael561 Ok I see – Gregory Grant Jul 15 '17 at 03:18
  • There's a "well-known" exercise in measure theory to construct a Borel subset $A$ of $[0,1]$ such that $0<m(A\cap I)<m(I)$ for all intervals $I\subset [0,1]$. It's an exercise in Folland's book, and if I remember correctly Rudin has a characteristically slick paper on this problem. – carmichael561 Jul 15 '17 at 03:21
  • @carmichael561 Thanks. I will have a look into the book. I need in addition that $m([0,1]\setminus A) > 0$. – Friedrich Philipp Jul 15 '17 at 03:31
  • @carmichael561 I do not find it in the book. Can you give me the exact reference, please? – Friedrich Philipp Jul 15 '17 at 03:42
  • It's exercise 33 in Chapter 1 of Folland's Real Analysis. I also found this MSE post: https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval – carmichael561 Jul 15 '17 at 03:45
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    Also called a "fat Cantor set", a certain kind of closed nowhere dense subset of [0,1] that has positive measure. – DanielWainfleet Jul 15 '17 at 06:18
  • @DanielWainfleet The Smith-Volterra-Cantor set which I mentioned in my post is such a set. Can you prove that it satisfies my needs? – Friedrich Philipp Jul 15 '17 at 11:53
  • The complement $\tau=[0,1]$ \ $\sigma$ of a Smith-Volterra-Cantor set $\sigma$ is an open subset of $\mathbb R$ that is dense in $[0,1].$ For any $x\in [0,1]$ and any $\epsilon >0$ the set $U=B_{\epsilon}(x)\cap [0,1]$ is an interval of positive length so $V= int_{\mathbb R}(U)$ is an open real interval of positive length, and is a subset of [$0,1]$. So $\tau \cap V$ is a subset of $\mathbb R $ and it is not empty (by the denseness of $\tau$ in $[0,1])$. So $\tau \cap U$ has positive measure. And of course $\tau \cap V\subset B_{\epsilon}(x)$ \ $\sigma.$ – DanielWainfleet Jul 15 '17 at 14:01
  • Thanks so much, Daniel. The density does the job. I would acknowledge an answer. – Friedrich Philipp Jul 15 '17 at 22:38

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