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What is the connection between the normalized Haar measure of a compact group and the normalized Haar measure of one of its compact subgroups?

I am trying to solve the following problem:

Given $G$ a compact group with normalized measure $\mu$ and $\{H_n\}$ an increasing sequence of compact subgroups of $G$ with normalized Haar measures $\mu_k$ such that $\bigcup H_n$ is dense in $G$. Prove that $\mu_k$ converges in the weak star topology to $\mu$.

[edit] The problem is indeed an exercise, as you can see from my comments, but I don't know why this is so relevant. I asked a question which could enlighten me in order to solve the given problem, and I think that the given question about Haar measures is not so trivial, since no one gave an answer until now.

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    $\begingroup$ Is this an exercise? $\endgroup$
    – Andreas Thom
    Commented May 23, 2011 at 11:03
  • $\begingroup$ Yes, it is in the chapter on Haar Measure in Katznelson, Harmonic Analysis, but the theory presented there I couldn't find anything that would help me solve this. $\endgroup$
    – Beni Bogosel
    Commented May 23, 2011 at 11:35
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    $\begingroup$ @Beni: the question about whether this is an exercise is extremely relevant, whether you realize it or not. At the very least, by pinpointing where the problem can be found, you give users of this site an idea of what you are expected to use to solve the exercise in question, which can give a useful hint of where to start, but also is especially important when there may be completely different approaches to solving a problem. So kindly get off your high horse, thanks. $\endgroup$
    – Thierry Zell
    Commented May 23, 2011 at 18:24
  • $\begingroup$ @Beni: Something to keep in mind is the following. Most of the users here on MathOverflow don't like being told "... Prove that ...", in the "exercise" style. (Independent of whether an exercise is trivial or very difficult.) At the very least, many of us (certainly I speak for myself) prefer questions, like you give in your first paragraph. I would be happy for you to tell me that what you're looking for is an answer to your first question, and your motivation is towards being able to solve the exercise you quote. But you do not make this clear in the post. $\endgroup$
    – Theo Johnson-Freyd
    Commented May 24, 2011 at 0:55
  • $\begingroup$ I was going to ask just that. The answer given uses compactness of the weak star topology, and the first question is still not answered: what is the connection between the Haar measure on the group and the Haar measure on one of its compact subgroups, if there is any? $\endgroup$
    – Beni Bogosel
    Commented May 24, 2011 at 6:18

1 Answer 1

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Each of the Haar measures on $H_k$ defines a $H_k$-invariant probability measure $\mu_k$ on $G$, which is supported on $H_k \subset G$. Let now $\mu$ be any limit point in the weak-$*$-topology, then $\mu$ is a probability measure on $G$, which is invariant under $H_k$ for all $k \in \mathbb N$. Since the union of the $H_k$ is dense in $G$, $\mu$ is $G$-invariant. By uniqueness of the Haar measure, it must equal the Haar measure of $G$.

By the Banach-Alaoglu theorem, the space of probability measures is compact. Since the above argument applies to any weak-$*$-limit point of the sequence, the sequence $\lbrace \mu_k \rbrace_{k \in \mathbb N}$ converges to $\mu$.

Concerning your comment, the problem is indeed not difficult and I think that the level of the question is below the average on MathOverflow.

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  • $\begingroup$ Thank you. It was indeed simple, but I didn't encountered many such arguments before. $\endgroup$
    – Beni Bogosel
    Commented May 23, 2011 at 18:12
  • $\begingroup$ Do you have an answer to the initial question? Is there a connection between the Haar measure of the big group and the Haar measure of a compact subgroup? $\endgroup$
    – Beni Bogosel
    Commented May 25, 2011 at 19:16

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