Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a compact connected topological group and let $H$ be a subgroup of $G$. Suppose that $H$ is measurable with respect to the normalised Haar measure $\mu$ on $G$. Do we necessarily have $\mu(H)=0$ or $\mu(H)=1$?

Maybe this is well--known, I ask it just out of curiosity. The question is related to this one: If you provide a measurable subgroup $H$ of $\mathbb R/\mathbb Z$ of measure not 0 or 1, then the characteristic function of $H$ violates the conjecture stated there.

share|improve this question

1 Answer 1

up vote 10 down vote accepted

Don't we still have this: if $A$ is measurable of positive measure, then $A A^{-1}$ contains a neighborhood of the identity...? So: a measurable subgroup of positive measure itself contains a neighborhood of the identity, and thus by connectedness is all of $G$.

share|improve this answer
    
If that is true (it seems like a nontrivial statement, but I'm willing to believe) the story ends here. –  Xandi Tuni May 13 '10 at 12:54
    
I'm certain that this is in Hewitt+Ross volume 1, but I don't have a copy handy to check. I'm not sure of a more modern textbook reference... –  Matthew Daws May 13 '10 at 14:19
    
This is true for any locally compact second countable group. This is proven e.g. in the appendices at the end of Zimmer book "Ergodic theory and semi-simple Lie groups". The proof is not very different from the proof for the group R of real numbers, that you can find in many textbooks. –  user6129 May 28 '10 at 9:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.