Measurable Set $E$ so that $m(E\cap(a, b)) \gt 0$ for all $a, b \in \mathbb{R} $ and $m(E) \lt \infty$

Question

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?

Answer 1

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.

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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.