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I've been reading a bit about Lebesgue measurability in real analysis. The common example of Vitali Sets came up to demonstrate why we can't have any old subset of $\mathbb{R}$ be Lebesgue measurable.

I've seen another example of non-measurable sets involving rational rotations of all the points about a circle.

To me, the set of points about the unit circle and the set of points on the $[0,1]$ seem equivalent, in which case, rational rotations are the same as the rational shifts of Vitali's theorem. In both cases, we have an uncountable set that that is partitioned into a countably infinite set via some quotient group (e.g., $\frac{\mathbb{R}}{\mathbb{Q}}$) (plus Axiom of Choice!), which leads us to having to conclude the measure is $0$ or $\infty$, neither of which equal the (known) Lebesgue measure of the set being partitioned.

Is there a theorem or sufficient/necessary set of conditions that can back this up?

EDIT: Per the comments below by @user85667 I think I need to be more precise when I say "isomorphic" or “similar to” Vitali Sets. Here is the particular sense I am thinking:

Conjecture

All non-Lebesgue-measurable sets on $\mathbb{R}$ are dense elements of an uncountable quotient group.

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    I don't know which notion of isomorphism you intend, but take into account that if you take an arbitrary non-empty set of measure zero that is disjoint with your non-measurable set, you get a new non-measurable set. – plop Aug 16 '22 at 00:17
  • @user85667 so maybe isomorphic up to measure zero sets? –  Aug 16 '22 at 00:21
  • @user85667 maybe I should say "equivalent mod sets of measure zero" or "almost isomorphic" (akin to "almost surely")? –  Aug 16 '22 at 00:28
  • You can also take unions with disjoint measurable sets. The result is also non-measurable. – plop Aug 16 '22 at 00:31
  • @user85667 so sounds like the answer is strictly "no" but what other types of sets can we use apart from those generated by this "countable shifting + AC" construction? –  Aug 16 '22 at 00:38
  • Not necessarily. It is just that isomorphism can mean different things. Maybe you mean as sets (in other words having the same cardinality), or as topological spaces, or some other sense. Even if we choose, for example the first, the answer depends on which axioms you assume. In ZFC+CH, yes, all will have the same cardinality. Dropping CH, we cannot prove if they do. – plop Aug 16 '22 at 00:53
  • @user85667 hmm -- this is interesting. So I'm imagining something more specific than cardinality -- its the idea that all non-measurable sets on a space $\Omega$ can be transformed to Vitali Sets on $[0.1]$ –  Aug 16 '22 at 03:54
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    The question of what you call "isomorphism" is the question about which transformations are you allowing to be used. The axioms assumed are also important. The example above of transformation=bijection and assuming ZFC+CH, gives a positive answer, all non-measurable sets can be proven isomorphic, in this sense. A change on the transformations allowed, or the axioms, can turn the answer to negative, unprovable, or unknown. – plop Aug 16 '22 at 13:34
  • @user85667 thank you — I tried to make my conjecture more precise (although it maynow be trivial) –  Aug 16 '22 at 13:41
  • Well, note that how the group acts on $\mathbb{R}$ needs to be specified for the question to be interesting. It is the same problem. Assuming ZFC+CH, you can pick a bijection sending any non-measurable set to Vitali's set. Then you can pull back the action of $\mathbb{Q}$ to the original $\mathbb{R}$, using the bijection. So, the original non-measurable will be the result of the same Vitali construction, but for $\mathbb{Q}$ acting in that way. – plop Aug 16 '22 at 13:49

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Your conjecture is currently far too imprecisely stated to be disproved, but certainly there are naturally-occurring types of nonmeasurable set (that is, the sets themselves might not be natural, but their key properties are) which are quite different from Vitali sets.

For example, there are Bernstein sets. A set $A$ is Bernstein iff both $A$ and $A^c$ (= the complement of $A$) have nonempty intersection with every perfect (= nonempty closed without isolated points) set. Bernstein sets need not be Vitali. There are also Hamel bases (= bases for $\mathbb{R}$ as a vector space over $\mathbb{Q}$); no Hamel basis can be Vitali, and non-null Hamel bases must be non-measurable (there do exist null, so a fortiori measurable, Hamel bases). And there are many more types of pathological set, such as Luzin sets and Sierpinski sets. See these notes of Beriashvili.

Basically, the sheer variety of interesting types of non-measurable sets out there strongly suggests to me that there is no sense in which all instances of non-measurability must be related to Vitali sets.

Noah Schweber
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  • This is exactly what I was looking for - great answer - thank you! –  Aug 16 '22 at 05:24
  • Btw: tried to make my conjecture refutable — which I think you have in your answer –  Aug 16 '22 at 12:19