4

This might be silly, but I am not sure:

Does there exist a Lebesgue measurable subset $E \subseteq (0,1)$ such that

  1. $E$ and $(0,1) \setminus E$ both have positive Lebesgue measure.

  2. $E$ and $(0,1) \setminus E$ both have empty interiors.

If we relax condition $1$, then $E=Q\cap (0,1)$ works. If we relax condition $2$, then the fat Cantor set does the job. (Its complement have non-empty interior though).

Asaf Shachar
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  • 1
    In my answer to this question I constructed an $F_\sigma$ set $M\subseteq\mathbb R$ such that $0\lt m(M\cap I)\lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $\mathbb R\setminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=M\cap(0,1)$ and $(0,1)\setminus E$. – bof Nov 02 '18 at 12:25
  • What irrationals on $(0,\frac12)$ with rationals from $(\frac12,1)$ aren't good enough for you? Since when did you become so picky? Are things really that bad since I left? – Asaf Karagila Nov 02 '18 at 17:33

2 Answers2

5

Let $E$ be the union of $(0,1) \setminus \mathbb Q)\cap (0,\frac 1 2]$ and $\mathbb Q\cap (\frac 1 2,1)$. Then $E$ and $(0,1)\setminus E$ both have positive measure and they have no interior.

Kavi Rama Murthy
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2

Let $E$ be a fat Cantor set $C$ together with the $C^\complement\cap\mathbb Q$.

José Carlos Santos
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