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.



Haar measure of a subgroup
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.


