Outer measure proof for rational numbers

Question

I saw this problem solved for particular cases like $(0,1)$ but never for general.

If $A \subset \mathbb{Q} \cap (a,b)$ and $a<b$ (set of all rational numbers in $(a,b)$)

Claim: For every $\epsilon >0$ $\exists$ an open subset $U \subseteq (a,b)$ such that $$A \subseteq U \text{ and } \mu^{*}(A) < \epsilon$$

Please help! Thank you.

Answer 1

HINT: Enumerate elements of $A$ such that $A=\{a_1,a_2,\dots\}$. Let $U_k$ be an open interval containing $a_k$ with measure not exceeding $\epsilon/2^k$. $$ U=\bigcup_k U_k. $$