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

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.

• Linked: Steinhaus-like theorems (see, here, here, here) imply that if $H$ is a closed subgroup of a locally compact topological group and Haar measure of the group $H$ is positive then $H$ is open. – Alex Ravsky Jun 24 '15 at 5:10

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.

• "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." Would you explain why $|G:H|$ is countable? – Makoto Kato May 3 '15 at 0:54
• As I already wrote, “it seems”, but I did not check this, so I try to do it now. Let $G=\bigcup\{ H_\alpha:\alpha\in A\}$ be a representation of a group $G$ as a union of a disjoint family of left cosets of the form $H_\alpha=h_\alpha H$ for some element $h_\alpha\in G$. Note that $A=|G:H|$. The translation invariance of Haar measure $\mu$ implies that each coset $H_\alpha$ is $\mu$-measurable. Let $G=\bigcup G_n$ be a representation of a group $G$ as a countable union of a family $\{G_n\}$ with finite measure $\mu$. – Alex Ravsky May 3 '15 at 3:41
• Then for each $n$ and each $\alpha$ the set $G_n\cap H_\alpha$ is $\mu$-measurable and has finite measure $\mu$. The additivity of the measure $\mu$ implies that for each $n$ and each $\varepsilon>0$ the set $\{\alpha\in A:\mu(G_n\cap H_\alpha)>\varepsilon\}$ is finite. Therefore the set $\{(n,\alpha): \mu(G_n\cap H_\alpha)>0\}$ is countable. So if the set $A$ is uncountable then there exists an index $\alpha\in A$ such that $\mu(G_n\cap H_\alpha) =0$ for each $n$. Here we have to use $\sigma$-additivity of the measure $\mu$, which implies $\mu(H_\alpha)=\mu(H)=0$, a contradiction. – Alex Ravsky May 3 '15 at 3:41

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.

• I think this is correct, but unfortunately I can only accept one answer. – Makoto Kato May 8 '15 at 2:25