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

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    $\begingroup$ 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 $\endgroup$ – Venkataramana Apr 9 '16 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. $\endgroup$ – paul garrett Apr 10 '16 at 23:03
  • $\begingroup$ I'd like to know whether $\mu_H(A_1A_2)>0$. $\endgroup$ – Luke Apr 11 '16 at 10:07
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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.

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  • $\begingroup$ This argument isn't right. The restriction of $\mu_H$ could be the zero measure (and frequently is, I think). $\endgroup$ – HJRW Apr 10 '16 at 6:35
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    $\begingroup$ @HJRW Could you elaborate a bit ? $\endgroup$ – Duchamp Gérard H. E. Apr 10 '16 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 '16 at 12:42
  • $\begingroup$ @HJRW, it indeed should be written "unique up to scale" instead of "unique". Thanks for this correction. I didn't elaborate, as this is a nearly trivial extension, but one could (and should) actually replace any reference to measure (in the question and in the answer) with measure class. In this form the answer applies as well to any locally compact group $H$. I guess this is why I haven't write "up to scale" in the first case. Anyway, I will update. $\endgroup$ – Uri Bader Apr 10 '16 at 19:41
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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 '16 at 19:46

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