# Existence of set $\sigma$ with $|\sigma| > 0$ such that $\forall\,x\in\sigma\,\forall\varepsilon > 0$ we have $|B_\varepsilon(x)\setminus\sigma| > 0$

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.

• What about $\sigma$ equals the irrational numbers in $(0,1)$? – Gregory Grant Jul 15 '17 at 3:16
• Have you tried using a Cantor set of positive measure? – Mustafa Said Jul 15 '17 at 3:16
• @Gregory Grant: That doesn't work because the rational numbers have measure zero. – carmichael561 Jul 15 '17 at 3:17
• @carmichael561 Ok I see – Gregory Grant Jul 15 '17 at 3:18
• 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 6:18