# Is the Haar measure of the boundary of an open subset of the $k$-torus zero?

I have just recently started reading the basics of topological groups and Haar measures, and have become really curious about when the boundary of a non-empty open subset $$U$$ of a compact abelian group $$G$$, has zero Haar measure. Here and in the following, by "Haar measure", I will mean the normalized Haar measure of a compact abelian group.

After some searching, I found these posts (When does the boundary have measure zero? and Conditions that ensure that the boundary of an open set has measure zero) which give very general sufficient conditions for subsets of $$\mathbb R^n$$ to have boundaries of Lebesgue measure zero, however I couldn't find any analogues for Haar measures.

I would like to restrict $$G$$ to the $$n$$-torus $$\mathbb T^n$$, so here's my

Semi-general question: Consider the $$n$$-torus $$\mathbb T^n = \mathbb R^n/\mathbb Z^n$$ which is a topological group, and a closed subgroup $$H$$ (which can be assumed to be connected if necessary). Consider the Haar measure $$\mu_H$$ on $$H$$ and an open set $$U \subset \mathbb T^n$$. What minimal additional conditions on $$U$$ might we need (if any) for the $$H$$-Haar measure of the boundary $$\partial(U \cap H)$$ of the open subset $$U \cap H \subset H$$ to be zero, that is to have $$\mu_H(\partial (U \cap H)) = 0$$?

If $$H$$ were $$\mathbb T^n$$ itself, then we know that $$\mu_H$$ is simply the $$n$$-dimensional Lebesgue measure and the question reduces to one already answered in the linked posts. If $$H$$ were an open subgroup, then (if I understand correctly) the Haar measure of $$\mathbb T^n$$ would also restrict to the normalized Haar measure of $$H$$, so that $$\mu_H$$ would just be the restriction of the $$n$$-dimensional Lebesgue measure, and once again the question is answered. But of course, $$\mathbb T^n$$ being connected has no non-empty proper open subgroups, so the last statement doesn't really add anything in my context. So what about the general setting, namely when $$H$$ is a proper non-empty closed subgroup (again, assumed connected if necessary)?

If nothing can be said in the above setting (even with some minimal additional conditions), the specific $$H$$ and $$U$$ I would like to know about are the following: consider any real numbers $$a_1, \cdots, a_n$$ and any complex numbers $$c_1, \cdots, c_n$$. My $$H$$ is the closure of the subgroup $$\{(e^{2 \pi i a_1 t}, \cdots, e^{2 \pi i a_n t}): t \in \mathbb R\} \subset \mathbb T^n$$, and $$U \subset \mathbb T^n$$ is the set of all $$(z_1, \cdots, z_n) \in \mathbb T^n$$, for which the complex number $$c_1z_1 + \cdots + c_n z_n$$ has strictly positive real part. Can we say in this case that $$\partial(U \cap H)$$ has $$H$$-Haar measure zero?

Note 1: I do not know if this will be relevant, but in the above example, $$U$$ can be written as the intersection of $$\mathbb T^n$$ with the set $$V := \{(z_1, \cdots, z_n) \in \mathbb C^n: \Re \sum_{j=1}^n c_j z_j > 0\} \subset \mathbb C^n$$, so that $$U \cap H = V \cap H$$. Here, $$V$$ is both open and convex in $$\mathbb C^n$$, hence its boundary has zero Lebesgue measure.

Note 2: It might seem that the role of $$U$$ in my semi-general question is kind of superficial since we are really working with the open subset $$U \cap H$$ of $$H$$, but it is in line with my last example.

• Do you know about Cantor sets of positive Lebesgue measure? Commented May 31, 2022 at 21:21
• @Moishe Kohan Wikipedia seems to give the example of a Smith–Volterra–Cantor set, but that is of course not convex. Also, could you please elaborate on what you have in mind? (In particular, how does it relate to minimal additional conditions I wish to find on open convex $U \subset \mathbb T^n$ for $U \cap H$ to have $H$-Haar measure zero?) Commented May 31, 2022 at 21:28
• If you assume that your open subset is convex, then, of course, its boundary will have measure zero. Commented May 31, 2022 at 21:37
• @Moishe Kohan Is this true for the Haar measure on the closed subgroup $H$ too? Could you please give a reference (or the reasoning if it is short)? Sorry I am a newbie with topological groups .... Commented May 31, 2022 at 21:43
• Every closed connected subgroup will be another torus, so it does not really matter, you can simply work with $R^n$ and Lebesgue measure. Commented May 31, 2022 at 22:07