# Non-measurable sets on groups from translation invariance

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)?

• It's not really a construction, and it's hard to call some specific set "the" Vitali set.
– YCor
Jan 21, 2021 at 0:42
• @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?
– aras
Jan 22, 2021 at 16:36

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 $$0 < \lambda(K) \leq \lambda( ( \Lambda \cap K K^{-1} ) V ) \leq \lambda(K K^{-1} K) < \infty.$$

• How often does such a pair $(K, \Lambda)$ exist? Jan 22, 2021 at 22:22
• 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 $K K^{-1}$. Jan 22, 2021 at 22:55
• Why does $\Lambda$ generated by a countable subset of $K K^{-1}$ intersect $K K^{-1}$ only countably often? Jan 22, 2021 at 23:13
• @LSpice because a countable subset generates a countable subgroup (in every group).
– YCor
Jan 23, 2021 at 1:29
• @YCor, thanks. I was thinking of topological generation and so forgot that this was just a bare not-necessarily-closed group. Jan 24, 2021 at 1:46

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.

• Why does $G$ support a countably additive left-invariant probability measure? Jan 21, 2021 at 13:29
• @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 $\mu$. Jan 21, 2021 at 14:13
• What is known about for which $G$ we can get such a countably additive left-invariant probability measure?
– aras
Jan 22, 2021 at 16:51
• @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. Jan 24, 2021 at 8:16
• 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
– BCLC
Feb 2, 2021 at 1:28

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.

• 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?
– YCor
Jan 21, 2021 at 0:45
• Here we use measurable for a measure. (For example, Lebesuge measurable.) I added some info. Jan 21, 2021 at 2:23
• 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? Jan 21, 2021 at 2:37
• @LSpice i guess the answer is in my comment, although Gerald Elgar didn't clearly confirm.
– YCor
Jan 21, 2021 at 6:49
• $\mu$-measurable in the sense of Caratheodory. When $\mu$ is a sigma-finite Borel measure this could mean the completion of the Borel sets for $\mu$. Jan 21, 2021 at 12:27