Let $B$ a closed subset of $[0,1]$ of positive measure. I'm interested in the asymptotic density of the rationals in $B$.
More specifically, my question is
Let $a_n$ be $$ a_n = \frac{|\{k\in \mathbb N : \frac{k}{n}\in B\}|}{n}. $$ Is it true that $\lim_{n\to \infty} a_n = \mu(B)$ where $\mu$ is the Lebesgue measure?
I get a feeling that it's true, but I'm afraid that there's some weird set not satisfying it.
There is a closed subset $B \subset [0,1]$ of positive Lebesgue measure $\mu(B)$ consisting entirely of irrational numbers. For example, enumerate $\mathbb{Q} \cap [0,1] = \{x_1,x_2,\ldots\}$ and define $$B = [0,1] - \bigcup_{i=1}^\infty \,\,(x_i - \frac{1}{10^i},x_i + \frac{1}{10^i}) $$ and so $\mu(B) \ge 1 - \sum_{i=1}^\infty \frac{2}{10^i} = \frac{7}{9}$.
But $\lim_{n \to \infty} a_n = 0 \ne \mu(B)$.