Are closed subgroups of positive measure open?

Question

Let $G$ be a locally compact Hausdorff group and $H \le G$ a closed subgroup with positive Haar measure. Is $H$ open then?

For Lie groups this should be true (since positive measure gives full dimension) but I'm wondering if the result can be generalized.

Answer 1

If $H$ is $\sigma$-finite, then yes. In that case, by inner regularity of the Haar measure for $\sigma$-finite Borel sets, there is a compact $K \subset H$ with $\mu(K) > 0$. Then consider the function

$$f = \chi_K \ast \chi_K \colon x \mapsto \int_{G} \chi_K(y^{-1}x)\chi_K(y)\,d\mu(y).$$

Since $\mu(K) < \infty$, $f$ is well-defined and continuous, and since $\mu(K) > 0$, $f$ doesn't vanish identically.

$$\int_{G} f(x)\,d\mu(x) = \mu(K)^2 > 0.$$

But $f(x) \neq 0$ implies that there is an $y \in K$ with $y^{-1} x \in K$, i.e. $x \in yK \subset K\cdot K \subset H$, so $f^{-1}(\mathbb{R}\setminus \{0\})$ is a nonempty open subset of $H$. Any subgroup with nonempty interior is open in a topological group.

If $H$ is not $\sigma$-finite, then $H$ need not be open. Let $\mathbb{R}_D$ be the group $(\mathbb{R}, +)$, endowed with the discrete topology, and let $G = S^1 \times \mathbb{R}_D$, where $S^1$ carries the standard topology. Then $G$ is locally compact, and the subgroup $H = \{1\} \times \mathbb{R}_D$ has positive measure ($\mu(H) = \infty$), yet it is not open.

Answer 2

I just found Steinhaus theorem. It states that for $A \subseteq G$ with $\mu(A)>0$ we have that $AA^{-1}$ is a neighbourhood of unity. So take $A:=H$. Then $HH^{-1}=H$ has non-empty interior, hence is open.