# Measure of subsets of profinite groups

Let $$G$$ be an infinite profinite group, so $$G=\lim_{\longleftarrow}G/N$$ where $$N$$ runs through the open normal subgroups. I have two questions:

1. Is $$G$$ of Haar measure zero in the compact group $$\prod_NG/N$$?
2. What is the relation between the Haar measure of a subset $$E$$ of $$G$$ and the numbers $$\frac{|EN/N|}{|G/N}$$, the size of the image of $$E$$ in $$G/N$$?
• You probably want $G$ connected? Otherwise, just take a finite simple group. – abx Dec 23 '19 at 11:51
• @abx Profinite group is just a totally disconnected compact group. Sure I mean an infinite profinite group as I would edit. – Meisam Soleimani Malekan Dec 23 '19 at 11:57
• Yes, I think that 1 has a positive answer (it's a good exercise). – YCor Dec 23 '19 at 14:43
• For 2, there's an obvious inequality (for $E$ measurable, $\lambda(E)\le\inf_N|EN/N|/|G/N|$) which for $E$ dense of measure zero is clearly not an equality. – YCor Dec 23 '19 at 14:47
• @YCor I choose this title because of my first question. – Meisam Soleimani Malekan Dec 23 '19 at 15:01

1) The measure of a closed subgroup $$H$$ of a profinite group $$G$$ is $$\frac{1}{\vert G:H \vert}$$. So $$G$$ has measure zero in $$\prod G/N$$ if and only if it has infinite index. This way you should be able to show that $$G$$ always has measure zero in $$\prod G/N$$.
2) As Yves mentioned, you always have the inequality $$\mu(S) \le \inf \frac{\vert NS/N\vert}{\vert G/N \vert}$$ for any measurable subset $$S$$. In fact the right-hand side is the measure of the closure of $$S$$. If $$S$$ is closed it is an equality. If $$G$$ is the profinite completion of an abstract countable group $$\Gamma$$, then $$\mu(\Gamma)=0$$, but the closure of $$\Gamma$$ is $$G$$.