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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.

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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 $G$.

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  • $\begingroup$ If that is true (it seems like a nontrivial statement, but I'm willing to believe) the story ends here. $\endgroup$
    – Xandi Tuni
    Commented May 13, 2010 at 12:54
  • $\begingroup$ 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... $\endgroup$
    – Matthew Daws
    Commented May 13, 2010 at 14:19
  • $\begingroup$ 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. $\endgroup$
    – coudy
    Commented May 28, 2010 at 9:11
  • $\begingroup$ For future reference, this is known as Steinhaus's Theorem. The wikipedia page gives some links to elementary proofs: in particular this holds for any locally compact group. $\endgroup$
    – Matthew Daws
    Commented Mar 12, 2020 at 12:43

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