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 $A_1A_2:=\{a_1a_2|\ a_1\in A_1,a_2\in A_2\}$?

  • 3
    $\begingroup$ 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 $\endgroup$ Apr 9, 2016 at 12:42
  • $\begingroup$ 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. $\endgroup$ Apr 10, 2016 at 23:03
  • $\begingroup$ I'd like to know whether $\mu_H(A_1A_2)>0$. $\endgroup$
    – Luke
    Apr 11, 2016 at 10:07

1 Answer 1


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.

  • $\begingroup$ This argument isn't right. The restriction of $\mu_H$ could be the zero measure (and frequently is, I think). $\endgroup$
    – HJRW
    Apr 10, 2016 at 6:35
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    $\begingroup$ @HJRW Could you elaborate a bit ? $\endgroup$ Apr 10, 2016 at 8:06
  • $\begingroup$ Well, the claim that there's a unique invariant measure is clearly false, since we can scale. There should be lots of examples in which the Haar measure of a double-coset is zero. For instance, the double coset of two pro-cyclic subgroups of a profinite free group should have infinitely many disjoint translates, and hence have zero Haar measure. $\endgroup$
    – HJRW
    Apr 10, 2016 at 12:42
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    $\begingroup$ @HJRW, edited as promised. It came to my mind: maybe you missed reading the condition $\mu_H(H_1H_2)>0$ given in thecquestion. $\endgroup$
    – Uri Bader
    Apr 10, 2016 at 19:46
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    $\begingroup$ @user89334, sorry, you're quite right, I did miss that hypothesis. I withdraw my objection! (As Paul Garrett notes above, the first incarnation of the question wasn't especially clear.) $\endgroup$
    – HJRW
    Apr 11, 2016 at 9:02

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