I used to think that in the development of Lebesgue measure, the axiom of choice is only needed to prove the existence of non-measurable sets, but I am surprised by the proof of the uncountability of $[0,1]$ in Royden's Real Analysis. Briefly speaking, the proof has four steps:
- The outer measure of an interval is its length.
- The outer measure is countably sub-additive.
- If $A$ is countable, $m^\ast A=0$.
- The set $[0,1]$ is not countable.
However, in an answer on this site, I read that "it is consistent with that $\mathbb R$ is a countable union of countable sets". Since the outer measure is countably sub-additive, the outer measure of a countable union of countable sets must be zero. So, in Royden's proof, some form of axiom of choice must have been used. I guess it is used in the proof of the countable sub-additivity of the outer measure, but I am not sure. Any idea?
Sub-additivity of outer measure uses countable choice. To briefly recall the argument, suppose $(A_n)$ is a sequence of sets and you want to show that $\mu^*(\bigcup A_n)\leq \sum \mu^*(A_n)$ where $\mu^*$ denotes the outer measure. You prove this by, for each $n$, choosing a family of intervals which cover $A_n$ and have total length not too much bigger than $\mu^*(A_n)$. For any individual $n$, the existence of such a family follows from the definition of $\mu^*$, but countable choice is needed to simultaneously choose such a family for every value of $n$.
(Incidentally, this use of countable choice can be eliminated if you merely want to prove that every countable set has outer measure $0$ and hence that $[0,1]$ is uncountable. Indeed, in that case you only need to use sub-additivity of outer measure in the special case where each $A_n$ is a singleton. In that case, you can just explicitly define the family of intervals you need to choose; namely you just take one interval around your single point which is sufficiently small.)