Is it possible to construct a set $E$ so that for all $a, b \in \mathbb{R}, a < b $ the Lebesgue measure, donated here with $m$, of $E\cap(a, b)$ is always positive $$m(E\cap(a, b)) \gt 0$$ and also satisfies $m(E) \lt \infty$ ?

I would say no, because $a, b$ are arbitrary and if $E \ne \mathbb{R}$ one can always find an $a$ and $b$ so that $E\cap(a, b) = \emptyset$. But I can't come up with a proof to verify this. Has anyone an idea?

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    $\begingroup$ The statement ''if $E \ne \mathbb{R}$ one can always find an $a$ and $b$ so that $E\cap(a, b) = \emptyset$'' is simply false: any dense set $E$ intersects every interval. I'm sure there is an answer to your question on this site if you search for it. $\endgroup$ – Umberto P. Jan 26 '18 at 15:23
  • $\begingroup$ @NateEldredge thanks nate $\endgroup$ – Chiray Jan 26 '18 at 15:25
  • $\begingroup$ @NateEldredge that question is related, but this question is much simpler. $\endgroup$ – Umberto P. Jan 26 '18 at 15:25

Let $\{r_n\}$ be an enumeration of rationals, let $$E = \bigcup (r_n-\frac{1}{2^n}, r_n+\frac{1}{2^n})$$ then $E$ satisfies your requirement.

A more interesting question would be finding $E$ such that both $$m(E\cap (a,b))> 0 \qquad m(E^c\cap(a,b))> 0$$ for any open interval $(a,b)$. Such set also exist.


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