Let $(\mathbb{R},\mathcal{M},m)$ be the set of real numbers with Lebesgue measure $m$.
Fix $\epsilon\in[0,1]$.
Does there exist an everywhere dense set $X\subset \mathbb{R}$ such that for any interval $I=[a,b]$, we have that $m(X\cap [a,b])=\epsilon\cdot m([a,b])$?
Motivation: Obviously, if $\epsilon =0$, then $X=\mathbb{Q}$ suffices, and similarly, if $\epsilon =1$, then $X=\mathbb{R}-\mathbb{Q}$ works. Originally, I was wondering if there was a way of somehow "refining" the set of irrationals so that when we intersect this refined set with any interval, we get only a "fraction" of the interval's measure rather than all of it, or none of it. We don't have to limit ourselves to refining the irrationals; any everywhere dense set would be sufficient.
Any help would be greatly appreciated. Thank you.
For any measurable set $X$, $\frac {m(X\cap (x-\epsilon, x+\epsilon))} {2\epsilon}$ tends to $0$ or $1$ for almost all $x$. Hence, the conlusion holds only for $\epsilon =0$ or $1$.
Ref: https://en.wikipedia.org/wiki/Lebesgue%27s_density_theorem