What would be the advantage of accepting non-measurable sets?

I personally feel that non-measurable sets only exist because of infamous Banach-Tarski paradox...

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    $\begingroup$ If you believe in the axiom of choice then you have to accept the existence of non measurable sets. $\endgroup$
    – Seth
    Jul 17, 2012 at 13:26
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    $\begingroup$ Would a more accurate way to phrase the question be "I personally don't feel that non-measurable sets exist because of the Banach-Tarski paradox?" As the answers say, they exist if (or because) you accept some form of choice. You can use this to select which axioms you use. There is often a bias toward "nice" examples. If you ask somebody to describe a function, they will often go to the board and draw a differentiable curve, though we know most functions are not even continuous. $\endgroup$ Jul 17, 2012 at 13:39
  • $\begingroup$ I wonder if user1894 is working in some branch of mathematics that does not (or claims not to) us the Axiom of Choice? $\endgroup$
    – GEdgar
    Jul 17, 2012 at 14:47
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    $\begingroup$ @Gedgar: In Garnir's paper where he proved automatic continuity for a class of topological vector spaces (including Banach spaces) he describes the three parts of functional analysis (in his view, from the early 1970's) as constructive (in ZF); classical (in ZFC); and Solovayan (in Solovay-like models). I wonder what he would say about this today... $\endgroup$
    – Asaf Karagila
    Jul 18, 2012 at 22:30

2 Answers 2


One correction to your question is that non-measurable sets actually proved to exist by Vitali in 1905, his construction of a non-measurable set is now called a Vitali set (assuming the axiom of choice). The Banach-Tarski paradox appeared about two decades later in 1923.

There is no immediate advantage in accepting the existence of non-measurable sets. In fact it "harms" us in some way, it means that we have to be more careful in how we define measure and so on.

However there is a great advantage in accepting the axiom of choice, or at least the ultrafilter lemma (which is a weakened version of choice), both implying the existence of non-measurable sets. In fact much weaker claims than the axiom of choice imply the existence of non-measurable sets. To name a few:

  1. The [weak] ultrafilter lemma,
  2. The Hahn-Banach theorem (which also implies the Banach-Tarski paradox),
  3. The real numbers can be well-ordered,
  4. Every family of pairs has a choice function.

To read more, you can try Herrlich's wonderful chapter about measurability in his book The Axiom of Choice.

Whether or not to accept such existence boils down, in essence, to what you are trying to do. If you want to do finitistic mathematics, deal with finitely generated objects and a limited collection of their subsets then there is no harm in not assuming the axiom of choice.

However if you wish to deal with infinitely generated objects, such as $\ell_2(\mathbb N)$ or other measure theoretic necessities, then the axiom of choice is usually needed to allow a "smooth" transition from finitely generated objects to infinitely generated objects.

The key problem is provability, a lot of properties depend on the axiom of choice and we simply cannot prove their truth value without it. So you end up having to assume a lot more than simply saying "assume choice". In this aspect, assuming the axiom of choice helps both to decide a lot of properties (but not all, of course) as well allows immediate generalizations of the proofs to higher cardinalities.

To read more:

  1. Advantage of accepting the axiom of choice
  2. Is Banach-Alaoglu equivalent to AC?
  3. Foundation for analysis without axiom of choice?
  4. Axiom of choice and calculus
  5. Number Theory in a Choice-less World
  6. Can one construct a non-measurable set without Axiom of choice?

I'm sure you will get very competent answers discussing the relation between non-measurable sets and the axiom of choice and the like and the reltion between finitely additive and $\sigma$-additive measures.

Here I want to give a fairly practical reason why it might be preferable to have a measure not defined on the whole power set but a smaller $\sigma$-algebra.

There is no intuitive notion of measure or volume for arbitrary sets of points. We do have, however, a fairly good notion of volume for certain geometrical objects, such as rectangular blocks. It is natural to extend our notion of volume from elementary objects to more complicated objects by approximating the more complicated objects by simpler objects. For example, we can approximate the volume of a ball by approximating it by the disjoint union of many very small cubes. Now there is no reason why we should be able to approximate every set of points meaningfully by simple objects we know the volume of. So if we want to assign a notion of volume or measure to every set, we are going to have to make some ad-hoc choices.

The argument has even more bite in probability theory. Say we want to describe a probability measure on the real line. The usual way to do this in practice is by specifying a cumulative distibution function, which essentially pins down the measure of each interval and then using a result that says that a probability measure on the Borel $\sigma$-algebra is uniquely determined by its values on each interval. If we want to assign a measure to every set, we have to specify lot of values for sets that do not even occur in practice.

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    $\begingroup$ While I think that the point itself is very good, I think that it builds too much on post-choice intuition. I wouldn't be surprised if you went back in time and asked Lebesgue in 1903 whether or not all sets are measurable and his answer would have been positive. It seems to me that "the ancient ones" conceived the universe of mathematics as one approximated by geometrically definable sets. $\endgroup$
    – Asaf Karagila
    Jul 17, 2012 at 16:26
  • $\begingroup$ @Asaf: I think one of the motivations of Bruno de Finetti for pushing finitely additive probabilities was that one can define them on the powerset. Nowadays, we know that purely finitely additive probabilities are horrendously non-constructive objects. $\endgroup$ Jul 17, 2012 at 17:35
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    $\begingroup$ @Asaf: Lebesgue might have held that belief until he died in 1941! When Tarski tried to publish a result showing how another result (for any infinite set $X$ there is a bijection between $X$ and $X\times X$) implied the axiom of choice, Lebesgue rejected the paper, stating that an implication between two false statements was not interesting. The other referee, Frechet, also rejected the paper since he didn't think an implication between two true statements was a new result! See this, p. 209. $\endgroup$ Jul 18, 2012 at 16:30
  • $\begingroup$ @Per: You can also see this cited in this of answer of mine. There is still quite some difference between well-ordering the line and non-measurable sets. I agree that he may have actually thought so until his death. $\endgroup$
    – Asaf Karagila
    Jul 18, 2012 at 17:04
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    $\begingroup$ @RonMaimon I'm looking forward to seeing your definition of "arbitrary shape". $\endgroup$ May 31, 2015 at 16:03

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