I've been studying Royden and I have found something that looks like a contradiction to me, and can't find where it is that my reasoning is wrong. On Chapter 3 he defines the outer measure $m^*$ of a set $A$ as:
$m^*A = \inf_{A \subset I_n} \sum l(I_n)$
where $\{I_n\}$ is a countable collection of open intervals, and $l$ is the length.
What troubles me is that corollary 3 on chapter 3 says: "If $A$ is countable, $m^*A = 0$". And right bellow problem 5 states: "Let $A$ be the set of rational numbers between 0 and 1, and let $\{I_n\}$ be a finite collection of open intervals covering $A$. Then $\sum l(I_n) \geq 1$".
My question is, how can every open covering of the rationals between 0 and 1 have length grater or equal to 1 if at the same time, by virtue of being countable, the outer measure of the rationals is 0 (corollary 3) and hence:
$m^*(\mathbb{Q}\cap[0,1]) = \inf_{\mathbb{Q} \cap [0,1] \subset I_n}\sum l(I_n) = 0$.
Isn't there a contradiction in saying $\inf_{\mathbb{Q} \cap [0,1] \subset I_n} \sum l(I_n) = 0$, and for all $\{I_n\}$, $\sum l(I_n) \ge 1$?
The important difference between the two is that one covering is allowed to be countable, while the other is finite. The fact that allowing countable coverings gives $\mathbb Q \cap [0,1]$ measure 0 is the same as for all other countable sets; that a finite covering has measure at least 1 is a good exercise.