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.
Best Answer
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.