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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.

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  • $\begingroup$ I had an unwarranted $\sigma$-finiteness assumption in the first version of my answer. That's fixed now. $\endgroup$ – Daniel Fischer Dec 12 '16 at 14:27
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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.

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  • $\begingroup$ I'd say $\mu(H)=0$ in your counterexample. I'd define the Haar measure in $G$ to be $\mu(M) := \sum_{r \in \mathbb R} \mu'( \pi_1 (M \cap (S^1 \times \{r\})) )$ where $\mu'$ is the Haar measure of $S_1$. This measure is inner regular (but not outer regular). For some groups (like this one) there are two Haar measures an inner regular and an outer regular. $\endgroup$ – principal-ideal-domain Dec 12 '16 at 14:50
  • $\begingroup$ That is the case for e.g. open sets. But by the outer regularity of the Haar measure, it doesn't hold for all sets. An open set $U$ containing $H$ always has infinite measure, since for every $r\in \mathbb{R}$ there is an open arc $U_r$ containing $1$ in $S^1$ such that $U_r\times \{r\} \subset U$. Since $\mu'(U_r) > 0$, $\sum_{r\in\mathbb{R}} \mu'(U_r)$ contains uncountably many strictly positive terms, hence that sum is $+\infty$. So $\mu(U) = +\infty$ for all open $U\supset H$, and by outer regularity, $\mu(H) = \inf \{\mu(U) : U \supset H,\,U\text{ open}\} = +\infty$. $\endgroup$ – Daniel Fischer Dec 12 '16 at 14:57
  • $\begingroup$ If you take an inner-regular measure, the part for the $\sigma$-finite case should apply. But e.g. the wikipedia definition requires outer regularity for a Haar measure. $\endgroup$ – Daniel Fischer Dec 12 '16 at 14:59
  • $\begingroup$ Yeah, I know that some authors require outer regularity. But I've also seen some who want the Haar measure to be inner regular. But since there are representation theorems for both cases (to make the positive linear form $C_C(G,\mathbb R) \to \mathbb R$ into a measure), it should be fine. By the way: What do you think about my answer to the question (added a few minutes ago)? Do they assume the Haar measure to be inner regular in the Steinhaus theorem? $\endgroup$ – principal-ideal-domain Dec 13 '16 at 9:53
  • $\begingroup$ Since without inner regularity the assertion doesn't follow, that must be assumed. Perhaps they consider the Baire measure, rather than the Borel measure, then regularity is automatic. Since $\{1\} \times \mathbb{R}_D$ is not a Baire set, it wouldn't be considered in that case. $\endgroup$ – Daniel Fischer Dec 13 '16 at 10:29
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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.

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