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).


marked as duplicate by bof, José Carlos Santos, Asaf Karagila general-topology Nov 2 '18 at 17:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ 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 $(0,1)\setminus E$. $\endgroup$ – bof Nov 2 '18 at 12:25
  • $\begingroup$ 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? $\endgroup$ – Asaf Karagila Nov 2 '18 at 17:33

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.


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


Not the answer you're looking for? Browse other questions tagged or ask your own question.