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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:

  1. The outer measure of an interval is its length.
  2. The outer measure is countably sub-additive.
  3. If $A$ is countable, $m^\ast A=0$.
  4. 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?

Ramen Nii-chan
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  • The proof that a countable union of countable sets is countable requires the axiom of choice. See e.g. this answer. –  Jul 05 '21 at 23:04
  • Related: https://math.stackexchange.com/questions/304378/do-you-need-the-axiom-of-choice-to-accept-cantors-diagonal-proof and https://math.stackexchange.com/questions/2046808/what-zf-axioms-does-cantors-diagonal-argument-require – Joe Jul 05 '21 at 23:07
  • This may also be of interest. –  Jul 05 '21 at 23:15
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    AC not needed to show $[0,1]$ is uncountable. However AC is needed to show the countable union of countable sets is countable. – GEdgar Jul 05 '21 at 23:17

1 Answers1

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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.)

Eric Wofsey
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    +1: you added your second paragraph just as I was typing up a comment pointing out that $m^*(A) = 0$ when $A$ is countable follows directly from the definition of the outer measure and doesn't need AC. – Rob Arthan Jul 05 '21 at 23:27
  • So, countable choice is not needed to prove that $[0,1]$ is uncountable, but Royden has chosen a proof that uses countable choice. Am I right? – Ramen Nii-chan Jul 06 '21 at 07:24
  • @WilliamMcGonagall: If the proof invokes the sub-additivity of outer measure, then yes, that proof uses countable choice. – Eric Wofsey Jul 06 '21 at 15:47
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    @William McGonagall: Royden has chosen a proof that uses countable choice --- Excerpt from a 21 July 2006 sci.math post: "$\ldots$ I think one of the biggest sources for this confusion is caused by non-foundational textbooks that glibly throw around comments about AC, such as measure theory texts when nonmeasurable sets are constructed." Here I'm talking about being unclear about proving existence vs. some type of constructive existence, uncountable choice vs. countable choice, AC not needed for finitely many choices, etc. – Dave L. Renfro Jul 06 '21 at 17:19