# Some question on haar measure for sumsets of closed subsets of profinite groups

Let $H$ be a profinite group with the haar measure $\mu_H$. Let $H_1$ and $H_2$ be closed subgroups of $H$. $H_1$ and $H_2$ have their own haar measures $\mu_{H_1}$ and $\mu_{H_2}$ respectively.

Suppose $\mu_H(H_1H_2)>0$, where $H_1H_2:=\{h_1h_2|\ h_1\in H_1, h_2\in H_2\}$.

My question is as follows : For two closed subsets $A_1\subset H_1$ and $A_2\subset H_2$ with $\mu_{H_1}(A_1)>0$ and $\mu_{H_2}(A_2)>0$, is it the case that $\mu_{H}(A_1A_2)>0$, where $A_1A_2:=\{a_1a_2|\ a_1\in A_1,a_2\in A_2\}$?

• what is the question? Do you wish to know if $\mu _H(A_1A_2)>0$? In that case please edit the question. This question is indeed answered below – Venkataramana Apr 9 '16 at 12:42
• As suggested by @Venkataramana, I significantly edited the question in a direction that makes it ask the only question I could see that might make sense. And then the answer below is a good answer. – paul garrett Apr 10 '16 at 23:03
• I'd like to know whether $\mu_H(A_1A_2)>0$. – Luke Apr 11 '16 at 10:07

Yes. Consider $H_1H_2$ as a subset of $H$. This is a $H_1\times H_2$ homogeneous space (for left-times-right action). On such a space there is a unique (up to scale) $H_1\times H_2$-invariant measure. But in our case we see obvious two ones: the restriction of $\mu_H$ and the push foreward of $\mu_{H_1}\times \mu_{H_2}$. Thus these two conicide, and the answer follows.
• This argument isn't right. The restriction of $\mu_H$ could be the zero measure (and frequently is, I think). – HJRW Apr 10 '16 at 6:35
• @HJRW, edited as promised. It came to my mind: maybe you missed reading the condition $\mu_H(H_1H_2)>0$ given in thecquestion. – Uri Bader Apr 10 '16 at 19:46