Construct an open set containing all rational points in interval with given measure.

Question

In a recent exam I was asked to construct an open set $M \subseteq [0,1]$ mit $\mathbb{Q} \cap [0,1] \subseteq M$ with $\lambda[M] \in (0, \frac{1}{2})$ where $\lambda$ is the uniform distribution on $[0,1]$. Since every rational number $q$ is assumed to be in $M$ and $M$ is open there exists $\epsilon > 0$ with $B_\epsilon(q) \subseteq M$. Doesn't this already imply $M = [0,1]$? What am I missing?

Answer 1

Enumerate the rationals $q_0, \ldots, q_n, \ldots$. Then consider the balls $B_{2^{-n-3}}(q_n)$. Then $\lambda(\bigcup_{n \in \mathbb{N}} B_{2^{-n-3}}(q_n)) \leq \sum_{n \in \mathbb{N}} \lambda(B_{2^{-n-3}}(q_n))) = \sum_{n \in \mathbb{N}} 2^{-n-3} = \frac{1}{4}$.

$M = \bigcup_{n \in \mathbb{N}} B_{2^{-n-3}}(q_n)$ is one of the sets you are searching.