From this question and this question (and their answers) I gather that it is consistent with ZF without The Axiom of Choice to assume that there exist countable sets $A_n$, $n\in \mathbb N$, such that $\mathbb R=\bigcup_{n\in \mathbb N} A_n$. Given any positive measure $\mu$ on the Borel $\sigma$-algebra of $\mathbb R$ which satisfies $\mu(\{x\})=0$ for all singletons $\{x\}$, this implies $$ \mu(\mathbb R)=\sum_{n\in \mathbb N} \mu(A_n)=0. $$ In particular Lebesgue measure cannot exist. However, when I looked through a construction of Lebesgue measure, I was not able to pinpoint any particular argument that (could not be refined to an argument that) required any choice. This has left me a bit puzzled

My question is if I have misunderstood the situation, or if indeed some amount of choice is necessary to construct Lebesgue measure on $\mathbb R$. In the second case, I would be very interested in knowing 'how much' choice is needed.


2 Answers 2


That is correct, if the real numbers are a countable union of countable sets, then measure theory as we know it goes out the window. The same can be said if the real numbers are a countable union of countable union of countable sets, and so on. Essentially for the same reason.

To your question, the Lebesgue measure is the completion of the Borel measure. So we first need to ensure that the Borel measure is non-trivial. But if each singleton gets measure $0$, and the countable union of measure zero sets is a measure zero set, then it is impossible to have any set of positive measure which is:

  1. A singleton,
  2. A countable union of singletons,
  3. A countable union of sets which are countable union of singletons,

and of course the list can proceed, but in the case $\Bbb R$ already satisfies the third criteria, there's no point really.

  • $\begingroup$ I've often wondered why you somehow get a bunch if information for free as soon as you are able to identify some measure related to a problem. I guess the answer is that you make essential use of some weak form if choice, and this is where the extra information derives from. $\endgroup$ Apr 20, 2015 at 10:04
  • $\begingroup$ It's the $\sigma$-additivity that kills it without choice. You can salvage a considerable portion of the basic parts of measure theory in a situation like that, but it becomes very complicated and requires us to talk about Borel codes for sets, rather than just Borel sets. You can find information in the 5th volume of Fremlin's monstrous Measure Theory book (which is available freely from his website). $\endgroup$
    – Asaf Karagila
    Apr 20, 2015 at 13:37

As far as I know, all constructions of Lebesgue measure use at least a countable choice principle. For example, one approach to the Hahn-Hopf extension theorem (which allows us to go from finite unions of intervals to arbitrary Borel sets) is to construct an outer measure, but the proof of $\sigma$-subadditivity uses choice principles. (Given disjoint sets $A_n$, we can cover each of them by countably many intervals with suitable total length, but to cover the union we need to pick a covering for each $n$.)

Can you sketch the approach that you think does not use AC? At least one textbook made a horrible blunder in this respect, so I'm not surprised that an otherwise sound mathematician could miss an application of the axiom. (The usual proof of Artin's theorem requires some setup before the general principle of choice can be applied, but the author in question did not even realise that the principle is needed. To make matters worse, the introduction holds forth on the superiority of Zorn's lemma.)

  • $\begingroup$ I think I had in mind one of the more "direct" constructions of one-dimensional Lebesgue measure, starting from the definition $\lambda([a,b])=b-a$ and extending to larger and larger classes of sets. At some point, an abstract extension theorem is needed, which blatantly uses the countable axiom of choice, as you point out. Being inexperienced in the ways of the choice, I simply did not realise! $\endgroup$ Mar 9, 2017 at 12:52

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