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"](https://www.jstor.org/stable/3219308?seq=4#metadata_info_tab_contents) 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](http://tamuz.caltech.edu/teaching/ma144a/lectures.pdf).


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](https://en.wikipedia.org/wiki/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)?



Non-measurable sets on groups from translation invariance
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
$$
0 < \lambda(K) \leq \lambda( ( \Lambda \cap K K^{-1} ) V ) \leq \lambda(K K^{-1} K) < \infty.
$$


