31
$\begingroup$

Is axiom of choice required to show the existence of non-measurable sets? Is there a Lebesgue non-measurable set that can be constructed without axiom of choice?

Related question on MO says it is consistent: https://mathoverflow.net/questions/73902/axiom-of-choice-and-non-measurable-set

$\endgroup$
9
  • 5
    $\begingroup$ Solovay proved that there is a model of Set Theory without Choice where every subset of the real numbers is measurable. In that model, the Axiom of Dependent Choice is true. So you need at least something stronger than DC in order to establish the existence of non-measurable sets. $\endgroup$
    – Arturo Magidin
    Apr 19, 2012 at 17:00
  • 3
    $\begingroup$ Yes, you can construct non-measurable sets using non-principal ultrafilters and the existence of such a thing is strictly weaker than AC: terrytao.wordpress.com/2008/10/14/… $\endgroup$
    – Qiaochu Yuan
    Apr 19, 2012 at 17:01
  • $\begingroup$ That's very interesting. I had thought that the existence of ultrafilters was equivalent to AC. Thanks! $\endgroup$
    – Neal
    Apr 19, 2012 at 17:03
  • 1
    $\begingroup$ @Neal: AC is equivalent to the statement "Every lattice has an ultrafilter." It is strictly stronger than "There is a non-principal ultrafilter over $\mathbb{N}$." This is perhaps the source of your confusion. $\endgroup$
    – Cameron Buie
    Apr 19, 2012 at 17:12
  • 3
    $\begingroup$ @Cameron: I actually believe that the main source of confusion is the fact that people are usually ignorant to how much choice is really needed, and they simply use Zorn's lemma to do things. This gives the impression that many things require full choice while some require none. $\endgroup$
    – Asaf Karagila ♦
    Apr 19, 2012 at 17:14

1 Answer 1

Reset to default
33
$\begingroup$

The answer is, you cannot.

It is consistent with ZF that the real numbers are a countable union of countable sets, this implies that every set of reals is Borel and therefore measurable. Of course, in such model it is nearly impossible to develop the analysis we know.

However it is consistent relative to an inaccessible cardinal that there is a model of ZF+DC where all the sets of real numbers are Lebesgue measurable, and DC allows us to do most of classical analysis too.

Non-measurable sets can be generated by free ultrafilters over $\mathbb N$ too, which as remarked is a strictly weaker assumption that the axiom of choice. If there are $\aleph_1$ many real numbers and DC holds then there is an non-measurable set as well, which implies that ZF+DC($\aleph_1$) also implies the existence of non-measurable sets of real numbers - however this is not enough to imply the existence of free ultrafilters over the natural numbers!

Several other ways to generate non-measurable sets of real numbers:

  1. The axiom of choice for families of pairs;
  2. Hahn-Banach theorem;
  3. The existence of a Hamel basis for $\mathbb R$ over $\mathbb Q$.

There are several other ways as well, but none are quite close to the full power of the axiom of choice.

One important remark is that we can ensure that the axiom of choice holds for the real numbers as usual, but breaks in many many severe ways much much further in the universe (that is counterexamples will be sets generated much later than the real numbers in the von Neumann hierarchy). This means that the axiom of choice is severely negated - but the real numbers still behave as we know them.

The above constructions and to further read about ways to construct non-measurable sets cf. Horst Herrlich, Axiom of Choice, Lecture Notes in Mathematics v. 1876, Springer-Verlag (2006).

$\endgroup$
4
  • $\begingroup$ Can you provide a source of the Hahn-Banach way to generate a non-measurable? $\endgroup$
    – leo
    Apr 19, 2012 at 18:41
  • 1
    $\begingroup$ @leo: The HB theorem is enough to prove the Banach-Tarski paradox. $\endgroup$
    – Asaf Karagila ♦
    Apr 19, 2012 at 18:48
  • $\begingroup$ @AsafKaragila "the real numbers are a countable union of countable sets" is that a typo? Did you mean countable union of open sets? Or are the real numbers countable in ZF theory? If they are can you link to some text with more info on the topic. I'm interested in reading about this. $\endgroup$
    – gjl
    Jan 30, 2020 at 17:51
  • 4
    $\begingroup$ @gjl: Neither of your suggestions is correct. It is not a typo, and the real numbers are uncountable (Cantor's argument requires no choice whatsoever). And as we know, once you eliminate the impossible, the remaining—however improbable—must be the truth. Indeed, the key point is that a countable union of countable sets need not be countable without assuming some choice. In fact, a countable union of finite sets can be uncountable. As far as reading material goes, there's no a lot I can suggest that isn't extremely technical. $\endgroup$
    – Asaf Karagila ♦
    Jan 30, 2020 at 17:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .