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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.

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    $\begingroup$ What about $\sigma$ equals the irrational numbers in $(0,1)$? $\endgroup$ – Gregory Grant Jul 15 '17 at 3:16
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    $\begingroup$ Have you tried using a Cantor set of positive measure? $\endgroup$ – Mustafa Said Jul 15 '17 at 3:16
  • $\begingroup$ @Gregory Grant: That doesn't work because the rational numbers have measure zero. $\endgroup$ – carmichael561 Jul 15 '17 at 3:17
  • $\begingroup$ @carmichael561 Ok I see $\endgroup$ – Gregory Grant Jul 15 '17 at 3:18
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    $\begingroup$ Also called a "fat Cantor set", a certain kind of closed nowhere dense subset of [0,1] that has positive measure. $\endgroup$ – DanielWainfleet Jul 15 '17 at 6:18

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