# Can a set of positive measure and its complement both have empty interior? [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 2 '18 at 17:32

• 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 2 '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 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 $$C^\complement\cap\mathbb Q$$.