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The most well-known construction of a non-measurable set is the Vitali set. The idea behind Vitali sets is to split up the space (such as $[0,1]$) into equal-sized copies (guaranteed by translation invariance), by looking at something like $\mathbb{R}/\mathbb{Q}$. This same idea is used in "Visualizing a Nonmeasurable Set" to construct non-measurable sets on the torus.

Another construction I know about is on the probability space of infinitely many coin tosses $\{ 0, 1 \}^\mathbb{N}$. In this case, instead of modding out by $\mathbb{Q}$, you can mod out by "switching the outcome of finitely many coins". This approach is taken in these probability lecture notes.

All of these constructions seem closely related: each time, we have a way to decompose our set into "translation-invariant" copies. My question is how these sorts of constructions of non-measurable sets generalize. In the Wikipedia article on Haar measure I read:

Unless $G$ is a discrete group, it is impossible to define a countably additive left-invariant regular measure on all subsets of $G$, assuming the axiom of choice, according to the theory of non-measurable sets.

This seems very close to an answer to my question, but the Wikipedia article doesn't elaborate here. So, how does the construction of non-measurable sets on a non-discrete group $G$ work? Is the general intuition similar to the examples I've described here? What happens for discrete groups (could this be related to amenable groups, which I know you can put measures on)?

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  • $\begingroup$ It's not really a construction, and it's hard to call some specific set "the" Vitali set. $\endgroup$
    – YCor
    Jan 21 at 0:42
  • $\begingroup$ @YCor This is a fair point. I suppose my language there was chosen more for brevity than precision. By "the" I was intending to convey "the most well-known kind of Vitali set". Your objection to the word "construction" is interesting, do you just mean that it's not morally a construction since it depends on AC? $\endgroup$
    – aras
    Jan 22 at 16:36
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The proof for the reals can be generalized to any non-discrete locally compact group $G$. We let $K \subset G$ be any compact set with positive Haar measure $\lambda(K) > 0$ (e.g., $K = [0, 1]$ when $G = \mathbb R$), and we let $\Lambda < G$ be any subgroup such that $\Lambda \cap KK^{-1}$ is countably infinite (e.g., $\Lambda = \mathbb Q$ when $G = \mathbb R$). We define an equivalence relation on $G$ by the $\Lambda$-orbits coming from left multiplication and we let $V \subset K$ be a set containing exactly one representative of each equivalence class that intersects non-trivially with $\Lambda K$.

We then have $K \subset (\Lambda \cap K K^{-1}) V \subset K K^{-1} K$. The first inclusion here follows from the fact that if $k \in K$, then we may find $t \in \Lambda$ such that $tk \in V \subset K$. It then follows that $t = (tk) k^{-1} \in K K^{-1}$ and hence $k \in (\Lambda \cap K K^{-1})V$.

If we were able to extend the Haar measure to a countably additive left-invariant measure defined on all subsets of $G$, then we would have $\lambda(( \Lambda \cap K K^{-1} ) V) = \sum_{t \in \Lambda \cap K K^{-1}} \lambda(V) \in \{ 0, \infty \}$, but this would then contradict the inequalities $$ 0 < \lambda(K) \leq \lambda( ( \Lambda \cap K K^{-1} ) V ) \leq \lambda(K K^{-1} K) < \infty. $$

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  • $\begingroup$ How often does such a pair $(K, \Lambda)$ exist? $\endgroup$
    – LSpice
    Jan 22 at 22:22
  • $\begingroup$ Such pairs exist for all non-discrete locally compact groups. Haar measure is always finite on compact sets, and since Haar measure is inner-regular a set $K$ with the properties above exists. Since $G$ is not discrete the set $K$ must be infinite and you can take $\Lambda$ to be any subgroup generated by a countably infinite subset of $K K^{-1}$. $\endgroup$ Jan 22 at 22:55
  • $\begingroup$ Why does $\Lambda$ generated by a countable subset of $K K^{-1}$ intersect $K K^{-1}$ only countably often? $\endgroup$
    – LSpice
    Jan 22 at 23:13
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    $\begingroup$ @LSpice because a countable subset generates a countable subgroup (in every group). $\endgroup$
    – YCor
    Jan 23 at 1:29
  • $\begingroup$ @YCor, thanks. I was thinking of topological generation and so forgot that this was just a bare not-necessarily-closed group. $\endgroup$
    – LSpice
    Jan 24 at 1:46
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For any infinite group $G$ we can easily construct a Vitali subset $V$ of $G$. Indeed, pick an arbitrary countable infinite subgroup $H$ of $G$ and let $V$ be a subset of $G$ which intersects each right coset $Hg$ of $H$ is exactly one element. Then $G$ is a disjoint union of a countable infinite family $\{hV:h\in H\}$ consisting of left translation copies of the set $V$. Let $\mu$ be any countably additive left-invariant probability measure on $G$. Suppose for the sake of contradiction that $H$ is measurable with respect to $\mu$. If $\mu(H)=0$ then $\mu(G)=0$, a contradiction. If $\mu(H)>0$ then $\mu(G)$ is infinite, a contradiction.

Remark that the above construction of $V$ works even for amenable groups, because a left-invariant probability measure required by amenability is required to be finitely-additive.

Also remark that the main result of the paper [BGR] easily implies that if $G$ is a meager Hausdorff (para)topological group then the set $V$ can be constructed to be nowhere dense in $G$.

References

[BGR] Taras Banakh, Igor Guran, Alex Ravsky, Characterizing meager paratopological groups, Applied general topology 12:1 (2011) 27–33.

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    $\begingroup$ Why does $G$ support a countably additive left-invariant probability measure? $\endgroup$
    – LSpice
    Jan 21 at 13:29
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    $\begingroup$ @LSpice I do not claim that $G$ supports such measure. But if $G$ supports such measure $\mu$ then $H$ is not measurable with respect to $\mu$. $\endgroup$ Jan 21 at 14:13
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    $\begingroup$ What is known about for which $G$ we can get such a countably additive left-invariant probability measure? $\endgroup$
    – aras
    Jan 22 at 16:51
  • $\begingroup$ @aras The characterization of such $G$ depends on which its subsets are required to be measurable. My answer implies that all subsets of $G$ are measurable iff $G$ is finite. If $G$ is countably infinite then a one-point subset $\{x\}$ of $G$ cannot be measurable, because each of assumptions $\mu(\{x\})=0$ and $\mu(\{x\})>0$ easily implies a contradiction. $\endgroup$ Jan 24 at 8:16
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    $\begingroup$ ah, i see you too were suspended by the maths se mods/admins. math.stackexchange.com/users/71850/alex-ravsky anyhoo thanks for editing the post on which i set a bounty math.stackexchange.com/posts/2057393/revisions $\endgroup$
    – BCLC
    Feb 2 at 1:28
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Not an answer (does not use translation invariance)

Another non-measurable set, which may generalize more easily, is the Bernstein set ... That is, a set $E$ such that for every uncountable closed set A, we have $A \cap E \ne \varnothing$ and $A \setminus E \ne \varnothing$ .

[With AC] we can prove that any uncountable Polish space admits a Bernstein set (indeed, $\mathfrak c$ disjoint Bernstein sets). If $\mu$ is any atomless Borel measure on an uncountable Polish space, then a Bernstein set is not $\mu$-measurable.

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  • $\begingroup$ Isn't "measurable" relative to a $\sigma$-algebra rather than a measure? which $\sigma$-algebra do you mean? the completion of the Borel $\sigma$-algebra with respect to every $\sigma$-finite (finite?) Borel measure? $\endgroup$
    – YCor
    Jan 21 at 0:45
  • $\begingroup$ Here we use measurable for a measure. (For example, Lebesuge measurable.) I added some info. $\endgroup$ Jan 21 at 2:23
  • $\begingroup$ What does "measurable for a measure" mean, if not belonging to the $\sigma$-algebra on which the measure is defined? Since you refer to a Borel measure, does that mean that measurability is with respect to the Borel $\sigma$-algebra? $\endgroup$
    – LSpice
    Jan 21 at 2:37
  • $\begingroup$ @LSpice i guess the answer is in my comment, although Gerald Elgar didn't clearly confirm. $\endgroup$
    – YCor
    Jan 21 at 6:49
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    $\begingroup$ $\mu$-measurable in the sense of Caratheodory. When $\mu$ is a sigma-finite Borel measure this could mean the completion of the Borel sets for $\mu$. $\endgroup$ Jan 21 at 12:27

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