How To Think About Measurability in $\mathbb{R}$

Question

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!

Answer 1

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.

2. 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.

3. 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"?

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

5. "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.