1

How do Platonist-leaning mathematicians think about the measurability/non-measurability of subsets of $X=\mathbb{R}\cap [0,1]$? For clarity, let's use "size" for the informal concept of length/area/volume, and "measure" for usual formalized version of this concept. In the context of subsets of $X$, would most mathematicians agree, disagree or have no opinion about the following statements (feel free to just answer for yourself):

  1. If a set is Lebesgue-measurable, then its size is its Lebesgue measure. If a set is not Lebesgue measurable, then it is completely meaningless to ask about its size.
  2. Lebesgue measure has little or nothing to do with the intuitive notion of size. It is a purely formal concept, and there is no deep philosophical significance to a set being non-measurable.
  3. Lebesgue measure is a correct but incomplete formalization of the notion of size. There are canonical extensions of Lebesgue measure that allow one to meaningfully talk about the size of certain non-Lebesgue-measurable sets.
  4. For some sets, the question of their size is fundamentally meaningless. That is, there is absolutely no reasonable way to assign them a size, even if one extends beyond Lebesgue measure.
  5. There are multiple, conflicting notions of "size" which are all compatible with Lebesgue measure. Some non-Lebesgue-measurable sets may be assigned different sizes, depending on which notion one has in mind.

Please feel free to add to this list, if you feel that something is missing!

Ari Herman
  • 751
  • no, measurability depends of the $\sigma$-algebra defined in the space –  Jun 14 '20 at 21:01
  • which part are you answering "no" to? – Ari Herman Jun 14 '20 at 21:01
  • 3
    I believe @Masacroso is answering "no" to your title question (which seems not-so-related to the questions you're actually asking). – peek-a-boo Jun 14 '20 at 21:07
  • 1
    Not all notions of size are compatible with Lebesgue measure. For that matter, there’s more than one intuitive notion of size. – Brian M. Scott Jun 14 '20 at 21:09
  • also point 2 and 3 are not true: the Lebesgue measure is an extension of the intuition of size, in the sense of length, area or volume. Point 4 depends of the axiom of choice, and point 5 could be arguable –  Jun 14 '20 at 21:10
  • @ Brian M. Scott I'm aware that there are notions of size that are entirely different from Lebesgue measure (the obvious example being cardinaility). But are you saying that there are notions of size intended to formalize length/area/volume, but which are incompatible with Lebesgue measure? If you, would you care to elaborate on these? – Ari Herman Jun 14 '20 at 21:20
  • @ peek-a-boo Thanks for the comment, I hopefully have improved the title. – Ari Herman Jun 14 '20 at 21:22
  • 1
    @AriHerman: No. It was not at all obvious to me that you were restricting yourself to notions of size intended to formalize length, area, and volume (and as it happens, I’m not much interested in those notions). – Brian M. Scott Jun 14 '20 at 21:29
  • @Brian M. Scott Thanks, I'll try to make that clearer. – Ari Herman Jun 14 '20 at 21:30
  • @Masacroso Could you elaborate on why (2) and (3) are "not true"? To me, they seem at least partially a matter of opinion, so I was surprised to get such a definitive response. – Ari Herman Jun 14 '20 at 21:36
  • 1
    @AriHerman because for the intuitive notion of volume we are agree on any semi-ring of sets, there is no other natural choice of length in the semi-ring of intervals. Then extending this notion of length from a semi-ring to a $\sigma$-algebra the extension is unique, at least in euclidean spaces that are naturally $\sigma$-finite. Then in reality there is no choice or freedom to choose a different model of length, at least for the more simple and intuitive assumptions about how a measure must be (i.e. countable additive and no negative) –  Jun 14 '20 at 21:42
  • Okay, I understand why you are saying that (2) is not true. However, what you said doesn't seem to apply to (3). Why can't there be reasonable models of length that extend Lebesgue measure? That is, that agree with Lebesgue measure on the Lebesgue measurable sets, but also assign measures to some non-Lebesgue-measurable sets? – Ari Herman Jun 14 '20 at 21:52

1 Answers1

2

I cannot talk for "Platonist-leaning mathematicians", but here is my take.

  1. The Lebesgue measure $\lambda$ is determined on all Lebesgue-measurable sets if you prescribe that

    • $\lambda(X)=1$
    • $\lambda(\varnothing)=0$
    • $\lambda$ is translation-invariant
    • for a disjoint sequence $\{E_n\}\subset X$, $$\tag1\lambda(\bigcup_nE_n)=\sum_n\lambda(E_n).$$

So yes, if a set if Lebesgue measurable, its size is its Lebesgue measure. For non-measurable Lebesgue sets, one can easily define Lebesgue's outer measure (and its definition is fairly intuitive). But things like $(1)$ fail, so it is hard to defend that the outer measure of a (non-measurable) set is its size when things like joining two disjoint sets will give you a "size" that is not the sum of the sizes.

  1. As (clearly, I hope!) said in 1, Lebesgue measure has everything to do with "size". It's defined in terms of "size" and extended logically from there.

  2. One of those "canonical extensions" (I don't know what they are) of Lebesgue measure would have to assign measure to non-measurable sets that does not agree with Lebesgue's outer measure. So I cannot see where "meaningful" would come from: you would have a "size" of a set that does not agree with the size obtained by covering it with smaller and smaller segments. The latter is the notion of "size" on which all of Calculus is built, so you seem to be willing to stir quite a few things here (all of calculus, basically). How would this be "meaningful"?

  3. Lebesgue measure is super-common-sense way to assign "size" to subsets of $X$, so I cannot imagine where you are going here.

  4. "There are multiple, conflicting notions of "size" which are all compatible with Lebesgue measure". Don't agree. See 1.

The only "natural" way of assigning measure to non-Lebesgue-measurable sets is by denying the Axiom of Choice, using something like that Solovay Model. So now you have extended Lebesgue measure to all non-measurable sets. And you cannot exhibit any of them, because you don't have the Axiom of Choice. So now you have a "natural" measure on all subsets of $X$; all the sets of $X$ where this would make a difference are not accessible to you, and meanwhile you have broken huge parts of analysis by moving to an ad-hoc model of set theory that gives you something useless, at the cost of losing lots of useful things.

Martin Argerami
  • 217,281
  • Thank you for such a thorough answer! I think you're saying that any reasonable notion of size should, when defined, coincide with outer measure. That's not obvious to me; could you expand a bit? Take the set $A'={\frac{x}{2}:x\in A}\cup{\frac{x+1}{2}:x\notin A}$, where $A\subseteq [0,1]$ is non-measurable. Note that $A'$ is non-measurable, yet it can be obtained by partitioning $[0,1/2]$ into two pieces and translating one of them by 1/2. To me, it seems plausible to say that $A'$ has size 1/2, yet I think we can make the outer measure of $A'$ equal 1, by choosing the right $A$. – Ari Herman Jun 14 '20 at 23:58
  • 1
    But that's precisely the problem. Do you know what Banach-Tarski's Paradox is? You can take a ball in $\mathbb R^3$, divide it in 5 parts, translate the five parts, and get two balls. That shows you that such sets cannot possibly have "size" in a reasonable sense (say, invariant by translation). – Martin Argerami Jun 15 '20 at 01:30
  • Banach-Tarski implies that no translation invariant extension of Lebesgue measure is defined on ALL subsets. But are you saying, that there is no way to extend the Lebesgue at all without losing translation invariance? According to this post, that is not true. So, I don't think Banach-Tarski directly implies that there COULDN'T be non-Lebesgue-measurable sets that look like $A'$ and can be adjoined to Lebesgue measure without violating translation invariance. – Ari Herman Jun 15 '20 at 08:02
  • 1
    But what's the point? Yes, you can manufacture a contrived set that you cannot describe in words and requires explicit use of AC to be constructed, that is not Lebesgue-measurable and you can assign it "measure zero" in a translation-invariant way. What have you gained exactly? – Martin Argerami Jun 15 '20 at 14:13
  • I think the use of AC is not so relevant, since AC is accepted by most mathematicians. As for what one "gains" by doing this, I'm not sure how to answer. If there are ways to extend Lebesgue measure without getting paradoxes, isn't that interesting? It keeps open the possibility of finding natural, alternative ways to measure "size", which could be used to extend Lebesgue measure (without causing major problems). For example, I think it's at least arguable that all subsets of cardinality $<2^{\aleph_0}$ ought to have size $0$, even though some of these are not Lebesgue measurable. – Ari Herman Jun 15 '20 at 15:28
  • Actually, it looks like ZFC+MA implies that sets of size $<2^{\aleph_0}$ are measurable and have measure $0$. So, it is definitely consistent with ZFC that there exists a translation invariant extension of Lebesgue measure that assigns measure $0$ to every set with cardinality $<2^{\aleph_0}$. I think this shows that there are plausible assumptions one could adopt regarding "size", which extend but do not conflict with Lebesgue measure. – Ari Herman Jun 15 '20 at 20:03
  • 1
    Sorry, I don't really see where you are going. You started by talking about "size" and possible alternatives to Lebesgue measure. And now you are talking about using a new specific axiom so that some exotic sets become nullsets... What's the point? – Martin Argerami Jun 15 '20 at 22:21
  • Yeah, we may have gone on a tangent. In your original answer, you seemed to be saying that Lebesgue measure is sort of the optimal/only way to formalize the notion of size, right? I've been trying to push back on that by arguing that Lebesgue measure can be extended in ways that make intuitive sense and don't violate translation invariance. Your response seems to be that those extensions are not interesting, because the sets involve are too exotic or non-constructive. Do you agree with that summary? – Ari Herman Jun 15 '20 at 22:44
  • Kind of. I agree with the second part. The first part, I would say that Lebesgue measure is what you get if you want the length of intervals to be the difference between the endpoints, and the measure to behave reasonably. – Martin Argerami Jun 15 '20 at 22:59
  • Yes, that makes sense. So, from fairly obvious principles, one can argue that informal "size" and Lebesgue measure coincide on measurable sets. In other words, these obvious principles imply that any "good" measure will be a (not necessarily proper) extension of Lebesgue measure. But then you are going further and saying that we cannot properly extend Lebesgue measure in any meaningful way, right? In other words, non-measurable sets really do not have a size (i.e. there can be no reasonable way to answer how big a non-measurable set is, even for special cases). Do I understand your correctly? – Ari Herman Jun 15 '20 at 23:43
  • 1
    Yes. Lebesgue's outer measure (cover the set as tight as you can with countable unions of intervals) always exists. For example, in the question you linked there is an example where they take sets with outer measure 1 and make them nullsets in the extension of the measure. The extended measure then fails regularity, which is another desirable property. So, "any meaningful way" is probably too strong because I cannot claim to know everything that's possible and not. But Lebesgue measure is limited to the Lebesgue $\sigma$-algebra for a reason. – Martin Argerami Jun 16 '20 at 00:19
  • "Meaningful" was probably a poor word choice on my part, but I think I understand you. You're saying that it is implausible that some future mathematician will invent an extension of Lebesgue measure that is so obviously "right" (i.e. matches our intuition about how "size" should behave), that it will be readily adopted by mathematicians. In other words, any proper extension of Lebesgue measure is going to have some significant problem (e.g. not being translation invariant), that would prevent it's being universally accepted as the correct formalization of "size". Do you agree? – Ari Herman Jun 16 '20 at 09:18
  • 1
    Somewhat. The more important point is that any set that will reasonably appear in any mathematician's work is already covered by Lebesgue measure. Non-Borel sets do not appear naturally unless you are working on axioms or something like that, and in that case you would usually not care about measure. – Martin Argerami Jun 16 '20 at 13:57
  • Okay...so, while there may be reasonable ways of extending Lebesgue measure, doing so would be pointless, because non-Lebesgue-measurable sets don't play a significant role in mathematics (except perhaps in foundations)? I disagree slightly about non-Borel sets not appearing naturally (what about a basis for $\mathbb{R}$ as $\mathbb{Q}$-vector space); but certainly the majority of sets that one sees ARE Borel. I think your perspective is really interesting and much more pragmatic than what I was expecting to see. Thanks for taking the time to discuss it! : ) – Ari Herman Jun 16 '20 at 14:59
  • 1
    No problem :) Note that you'll be hard pressed to explain why you would ever want to use Lebesgue measure on a $\mathbb Q$-basis of $\mathbb R$. – Martin Argerami Jun 16 '20 at 15:02
  • Haha, that may truly be a matter of taste...to me it's very natural to wonder about the possible sizes of a $\mathbb{Q}$-basis of $\mathbb{R}$! – Ari Herman Jun 16 '20 at 15:11
  • What huge part of analysis is broken in Solovay's model? It satisfies dependent choice which is more than enough to develop analysis. In fact, some set theorists at the time conjectured that it is going to be widely adopted, since most of "conventional", i.e.separable, mathematics still works out perfectly. – Jonathan Schilhan Jul 01 '20 at 18:27
  • Losing Hahn-Banach, Krein-Milman, ultrafilters, the Stone-Cech compactification, and who knows what else, would kill the majority of Functional Analysis. Not knowing if your vector space has a basis, or if your ideal is contained in a maximal ideal would be deal-breakers for many algebraists and analysts. And even the theorems that can be saved, usually require more complicated proofs, from what I gather. So it looks like there is little to gain a a lot to lose. Our own Asaf Karaglia has written about this. – Martin Argerami Jul 01 '20 at 20:15
  • That's why I wrote "separable" mathematics. I really don't know any classical theorem about separable Hilbert/Banach spaces that would fail in Solovay's model and I would be surprised if there are. The way I remember the story is that mathematicians wanted to keep the strong machinary of Choice, in order not to think about the details and have more complicated proofs, as you say. – Jonathan Schilhan Jul 02 '20 at 08:44
  • So you think there are gains from not having Hahn-Banach on $\ell^\infty(\mathbb N)$ or on $B(\ell^2(\mathbb N))$ (none of which is separable, even though they are super natural and strongly related to "separable mathematics"), from not having the Stone-Cech compactification, from not having Krein-Milman, from not having Banach-Alaoglu? – Martin Argerami Jul 02 '20 at 22:56