Closed subgroup of a locally compact Hausdorff group whose Haar measure is non-zero.

Question

Let $G$ be a locally compact Hausdorff group, $H$ its closed subgroup. To avoid pathologies, we assume the underlying topological space of $G$ has a countable base. Let $\mu$ be a Haar measure on $G$.

Is the following asserion true?

If $\mu(H) \gt 0$, then $H$ is open.

It seems to be false, but I was unable to find a counter-example. For exanple, if $G$ is a Lie group, the assertion seems to be true.

Answer 1

It seems the following.

If the measure $\mu$ is $\sigma$-finite, that is the group $G$ can be represented as a countable union of sets with finite measure then $|G:H|$ is countable. By Theorem 2.3 from [HC], every locally compact Hausdorff space is Baire. So the group $H=\overline{H}$ contains a non-empty open subset $U$ of the group $G$. Then $H=H+U$ is an open subgroup of the group $G$.

References

[HC] R. C. Haworth, R. C. McCoy, Baire spaces, Warszawa, Panstwowe Wydawnictwo Naukowe, 1977.

Answer 2

This is a corollary to the following: If $A, B$ have positive measure, then $AB = \{ab : a \in A, b \in B \}$ has non empty interior. See the following post.