TITLE: Sigma algebra as number of rational numbers in set
QUESTION [6 upvotes]: This has probably been asked before, but I cannot find it.
Define $\mu$ on $(\mathbb{R}, B(\mathbb{R})$ by letting $\mu (A)$ be the number of rational numbers in $A$. Show that $\mu$ is a $\sigma$-finite under which each open subintervall of $\mathbb{R}$ has infinite meassure.
How can I construct this sequence of $(A_i)$ covering $X$? 
Can I construct sets with only one rational number in it, still covering X?
Is there such a thing as "closest" rational number, for another rational number... Part2 seems to be true since open intervalls are homeomorphic with $\mathbb{R}$?

REPLY [6 votes]: Let $\{r_n\}$ an enumeration of $\Bbb Q$, and let $A_n:=\{r_n\}\cup\Bbb Q^c$.
These ones are sets of measure $1$ and $\bigcup\limits_nA_n=\Bbb R$. 
An open interval contains infinitely many rationals.