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The Wikipedia article on measurable groups says "let $(G, \circ)$ be a group and further let $\mathcal{G}$ be a $\sigma$-algebra on the set $G$." I am confused because they assume a sigma-algebra exists. But I don't know how to define sigma-algebras without well defined union and complementation operators. Consider a symmetry group, a finite set of functions like rotation and reflection. How do we define union and complement of these geometric operations?? Maybe not all groups have associated sigma algebras and the Wikipedia page just needs to be clarified.

I cannot find any readable references on this, please help. Thank you.

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    $\begingroup$ Isn't just on the set $G$ with no structure? $\endgroup$ – Randall Feb 13 '19 at 17:33
  • $\begingroup$ @Randall. Maybe you're right. I'm not sure what "good" the set $R_{120} \cup R_{240}$, where $R_{\theta}$ denotes rotation, does us, but I'll read on and maybe this will become clearer to me. Thanks. $\endgroup$ – RMurphy Feb 13 '19 at 17:35
  • $\begingroup$ The definition is on $G$ relative to a chosen and fixed $\sigma$-algebra on the set $G$ of elements. Once that is fixed you can then ask if all this data makes a "measurable group." A single group $G$ may have several different $\sigma$-algebras, some measurable, some not. $\endgroup$ – Randall Feb 13 '19 at 17:36
  • $\begingroup$ Also, $R_{120} \cup R_{240}$ isn't a "set." $\endgroup$ – Randall Feb 13 '19 at 17:37
  • $\begingroup$ @Randall. Thanks. I meant $\{ R_{120} \} \cup \{ R_{240} \}$, surely that's a set, right :)? $\endgroup$ – RMurphy Feb 13 '19 at 17:55
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A $\sigma$-algebra on a set $X$ is a collection $\mathcal{A}$ of subsets of $X$ which is closed under countable unions and complements. So, when the article refers to a $\sigma$-algebra on $G$, it means a $\sigma$-algebra on the set $X=G$. The elements of this $\sigma$-algebra are subsets of $G$ with their usual Boolean operations, not elements of $G$.

I would add that this statement in no way at all asserts that such a $\sigma$-algebra exists for arbitrary $G$ (though it turns out that it does). It is just saying that given a $\sigma$-algebra $\mathcal{G}$ on $G$, then the triple $(G,\circ,\mathcal{G})$ is called a measurable group if certain conditions hold.

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