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In theoretical classes (for engineering), such as on random processes or statistical learning theory, sometimes great emphasis is put on measurability. For instance, I have heard somewhat extensive explanations of Borel $\sigma$-algebras or the Vitali set.

At one point I asked the point of all this discussion. Does it have practical implications? I was told, "not really, but the formalism is crucial for the integrity and advancement of the field."

As I've learnt a bit more on the topic, I've understood that such sets (e.g., non-measurable with respect to the Lebesgue measure) are contingent on the acceptance of the axiom of choice. I further understand this axiom as being able to prove existence without construction—e.g., for the Vitali set, partitioning $[0, 1]$ into infinitely many "bins" and choosing a representative from each bin. An honest-to-goodness proof about this set would need to describe how such a representative is chosen; since there are infinitely many bins, in order to have a finite proof, instead of manually choosing representatives, you must construct a rule to accept one of the bins and output a representative. The axiom of choice just says that such a rule exists; accepting it essentially allows you to complete the proof of existence while skipping the step of actually constructing a rule for a given set of sets. Lastly, I understand that no one has yet constructed such a rule (though I don't know if this would be impossible—the answer in this MO post says it "seems unprovable").

The short of this is that, not only will a non-measurable set ever come up in practice, no one can even point you to one in concept. No hard examples can be given—proofs about them just say they exist provided your acceptance of choice. They're just out there in the aether, existing only if you believe them to exist but completely invisible both in practice and in concept. It's like the invisible teapot in space—you can believe it's there but it has basically no implications whatsoever.

Provided this understanding is correct, my question is: why do we care to handle the "edge cases" of non-measurable sets, especially in engineering classes like the ones I was in, but even in pure math? In what sense is it "crucial to the integrity and advancement of the field"?

Further, and in a similar vein, why do we bother with the axiom of choice? I understand it is independent of ZF but has great utility, i.e., extending finite proofs to infinite cases; but it seems like a shortcut to a real proof that comes at the cost of creating mathematical spectres that you now have to worry about.

Note: in contrast to this question, my question includes two questions. 1) What real benefit does the axiom of choice bring? (From my view, just assuming something exists but not having any examples is not beneficial. The difference from @Lee Mosher's comment, "From this point of view the axiom of choice is just another existence axiom", is that, for instance, assuming infinite sets exists is immediately appealing because we previously had ideas of such sets. It's not intuitive in the same way that we can always necessarily make infinity choices at once.) 2) Supposing choice, non-measurable sets seem like absolute trivia. Perhaps the most arcane textbooks should mention them, but what benefit does it bring to discuss sets you cannot give an example of and which you only assume exist? The mentioned question only asks, "Why do we accept the existence of non-measurable sets?" which is close, but not identical, to 1) and does not touch on 2).

Further, the answer in said question says that non-measurable sets are not exclusive to systems with choice, while the answer in this question says otherwise.

I hope it is therefore obvious that this question is not remotely a duplicate of the first linked question. Asking "Why would you accept non-measurable sets?" and being told, "because choice implies them" (the contents of the supposed duplicate), is a completely separate question from "Supposing you accept choice, what is the merit or exigency for discussing non-measurable sets? In what way is it necessary to handle them for the advancement of the field (say, statistical ML)? Where is the real benefit coming in?"

Benjamin Havens
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  • That's a helluva engineering maths class you have there. Maybe the maths lecturer just pitched the level off a bit :) – Paul Apr 03 '25 at 17:20
  • You cannot just ignore the fact that non-measurable sets exist (at least under AoC). Mathematics is a precise art of formalism. Intuitive approach already lead to many nonsensical conclusions, and there was a deliberate effort to make maths formal. So, even if those sets are not practical, they do exist and cannot just be blindly ignored. – freakish Apr 03 '25 at 17:20
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    An appeal to the axiom of choice is an honest-to-goodness proof in a system in which you include the axiom of choice. Constructing non-measurable sets without the AoC is harder (and may not be possible in this case---I don't know that part of the theory all that well). – Xander Henderson Apr 03 '25 at 17:20
  • If you are only concerned about real world uses of maths then you probably don't need to care about non-measurable sets. There is a lot that you don't need to care e.g. that pi is irrational; you'll never prove that with a real circle. Why do mathematicians care? Hard to say but we care about lots of weird stuff with no real world use. Some of us particularly like the areas with no real world use. – badjohn Apr 03 '25 at 17:20
  • The axiomatic approach to mathematics is peppered with existence axioms. I wonder if your objection applies to them all. For example, (almost) every single axiom of set theory asserts the existence of something: an empty set exists; an infinite set exists; a set exists with this property or that property. From this point of view the axiom of choice is just another existence axiom. – Lee Mosher Apr 03 '25 at 17:21
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    I think we just need a logically correct way to develop our theory, so that the theorems we want to use are strictly true, without exception. We would hope that we can assign a nonnegative measure to each subset of $\mathbb R^n$ in such a way that measure is translation invariant and if we have a finite collection of disjoint subsets of $\mathbb R^n$, then the measure of their union would be the sum of their measures. But surprisingly the Banach-Tarski construction shows that this is not possible. Our best workaround is to assign a measure only to sufficiently “nice” subsets of $\mathbb R^n$. – littleO Apr 03 '25 at 17:23
  • @XanderHenderson I appreciate that if choice is accepted, formally appealing to it constitutes a "real" proof. The difficulty I have with this is that claiming something exists while simultaneously insisting that, not only do you not have an example of it, but the only reason you think such an example exists is an assumption you made, doesn't feel very "proof-y". – Benjamin Havens Apr 03 '25 at 18:01
  • @BenjaminHavens I am not sure why that doesn't feel "proofy" to you. All of mathematics is about starting from a set of axioms, which are assumed truths, and deducing new truths from those axioms. The axiom of choice is no different from other axioms in this sense. But perhaps you are more interested in constructivist mathematics? The AoC is a little more complicated there... – Xander Henderson Apr 03 '25 at 18:06
  • @XanderHenderson I suppose this is getting epistemological, but there is a difference in my estimation between a mathematical proof, establishing truth contingent on axioms, and proving something to someone. There is a corresponding difference between, "you can have a collection of things" (oh yeah, I've seen collections of things before) and "you can always construct a rule to choose things from infinitely many sets" (I know I can choose things from finite collections, but lots of familiar intuitions break down at infinity). – Benjamin Havens Apr 03 '25 at 18:10
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    To a mathematician, I don't think that there is a difference between "mathematical proof" and "establishing truth contingent on axioms". Everything in mathematics is contingent on the axioms. – Xander Henderson Apr 03 '25 at 18:14
  • @XanderHenderson Given that the question is, in part, a critique of the axiom of choice, perhaps it should be clear that even if something is a mathematical proof, depending on the axioms it uses, I might not consider it an epistemological proof. One is a question of form, the other of utility. – Benjamin Havens Apr 03 '25 at 18:29
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    @BenjaminHavens Every mathematical proof is contingent on which axioms are used. – Xander Henderson Apr 03 '25 at 18:31
  • @badjohn for clarity, I am not only casting doubt on practical utility, but on theoretical utility as well, as the question emphasizes a few times. Just because I may never have a physical object exactly 5m long doesn't mean the concept of being 5m long is useless. In contrast, not only will I never encounter a non-measurable set, you can't even give me an example of one in concept. – Benjamin Havens Apr 03 '25 at 18:32
  • @XanderHenderson With respect, duh. That has nothing to do with what I said. – Benjamin Havens Apr 03 '25 at 18:32
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    @BenjaminHavens Then I don't know what you are saying. Mathematics makes no epistemological claims beyond "it is possible to deduce contingently true statements from a set of axioms". – Xander Henderson Apr 03 '25 at 18:40
  • @XanderHenderson Mathematics functions within epistemology. A mathematical proof says, "If you believe these things (premise), you must believe this other thing (conclusion) to be logically consistent." Given that the question casts doubt on AoC, it should be clear why I don't consider a proof involving AoC as dispositive, even if it is technically a proof. The question asks both, why should I accept the premise of AoC, and, if I do, what benefit does it bring, even theoretically, to navel-gaze by considering sets I cannot even interact with in principle? – Benjamin Havens Apr 03 '25 at 18:48
  • (1) That isn't a mathematical question, you might consider asking something similar on [philosophy.se]. (2) Which is why I suggested that you look at constructivism, above. – Xander Henderson Apr 03 '25 at 18:50
  • @XanderHenderson A mathematician is more qualified to say how advancing some field requires handling pathological sets than a philosopher. Further, a mathematician can give practical reasons for accepting AoC, which is more what I would like to understand. Thank you for this discussion, cheers. – Benjamin Havens Apr 03 '25 at 18:58
  • @BenjaminHavens Many of the people who are active on [philosophy.se] are philosophers of mathematics. They are, quite frankly, much more qualified to answer your questions than most people here, since most working mathematicians don't care about the axiom of choice. Asking most working mathematicians about the advantages of AoC is like asking a fish about the advantages of water. One of the most qualified people here to answer that question is Asaf, and he has indicated that he believes that your question has already been answered. – Xander Henderson Apr 03 '25 at 19:01

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We care about measurable vs. non-measurable sets because the basic procedures of measure theory need you to be working at all times with a collection of sets fitting the condition called measurability. By ignoring this issue, you are not making sure that the arguments being used are at all times using measurable sets and thus are making sense.

Here is an analogy: why do we bother doing calculus on real numbers rather than just rational numbers? It's not as if a physical measurement, which always comes with an error range, is going to depend on the distinction between rationality and irrationality. So do you think it would be a good idea to use calculus only with fractions?

The axiom of choice is used all over the place in mainstream pure math in ways that don't look to the non-expert to be about the axiom of choice: it is used to show every nonzero commutative ring has a maximal ideal and to show every field has an algebraic closure, for instance, and these are both very important results in algebra that mathematicians rely on to do other things.

KCd
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  • Funny you should mention calculus on the rational numbers. I have a set of course notes I've been working on which develop everything algebraically (without limits), but which only applies to rational functions (ratios of polynomials) with rational coefficients. You can make a lot of progress in this direction, but youi miss out on transcendental functions, which makes it very hard to solve things like $y' = y$ or $y'' = - y$. Real numbers only really pop up when you want to extend the theory to "infinite polynomials" (e.g. Taylor series). – Xander Henderson Apr 03 '25 at 18:09
  • @XanderHenderson It also pops up before power series when you want to integrate: antiderivatives of rational functions need not be rational functions. – KCd Apr 03 '25 at 20:42
  • Indeed, though my approach is to kind of cheat a little in two ways: (1) I only really develop a theory of differentiation for rational functions, with a solid emphasis on polynomials---the theory of integration still works in a "big picture" kind of way, but the theorems which allow for computation are only proved for polynomial functions, and then (2) the first transcendental function that shows up is the logarithm, which is defined via an integral---that is the motivating example which leads to Taylor series. – Xander Henderson Apr 03 '25 at 21:27
  • The point being that the whole project is a kind of thought experiment about how far you can push calculus without actually having to deal with the continuum. – Xander Henderson Apr 03 '25 at 21:28