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Is it true that the Haar measure $\mu$ on a compact group $G$ is always completely positive, i.e. every nonempty open set has positive measure? I think I have a very simple proof of it, but honestly, the fact I tried Googling this fact and couldn’t find any mention of it, so I’m second-guessing my argument, which is as follows:

Let $G$ be a compact group, and $U \subseteq G$ a nonempty open subset of $G$. Then $\mathscr{U} = \{ g U : g \in G \}$ is an open cover of $G$, so there exist $g_1, \ldots, g_n \in G$ such that $G = \bigcup_{j = 1}^n g_j U$. Then $1 = \mu(G) \leq \sum_{j = 1}^n \mu (g_j U) = n \cdot \mu(U)$. Thus $\mu(U) \geq \frac{1}{n} >0$.

AJY
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1 Answers1

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The Haar measure on an arbitrary locally compact (not just compact) Hausdorff group is positive on all nonempty open subsets. This is because Haar measure on a locally compact Hausdorff group always satisfies two regularity conditions: (i) for each Borel set $A$, $\mu(A) = \inf_{U \supset A} \mu(U)$ where $U$ runs over open subsets that contain $A$ and (ii) for each open set $A$, $\mu(A) = \sup_{K \subset A} \mu(K)$ where $K$ runs over compact subsets of $A$.

Remark. You can't assume (ii) holds for all Borel sets. For example, let $G = S^1 \times \mathbf R_d$, where $S^1$ has its standard compact topology, $\mathbf R_d$ is the real line with the discrete topology, and $G$ has the product topology. This is a locally compact Hausdorff since $S^1$ and $\mathbf R_d$ are both locally compact Hausdorff. The product topology on $G$ is non-discrete and makes $G$ look like an infinitely long vertical cylinder whose horizontal slices are "topologically independent" circles. Haar measure on $G$ does not satisfy (ii) for the closed set $A = \{1\} \times [0,1]$, which is not an open set: compact subsets of $A$ are finite and thus have Haar measure $0$ (one-point sets in a nondiscrete group have Haar measure 0) and you should show open sets containing $A$ have infinite measure, so property (i) tells us $\mu(A) = \infty$ and thus (ii) isn't true for this $A$.

Returning to the general situation, how do (i) and (ii) show Haar measure is positive on all open subsets? I'll give the argument for left Haar measure (for right Haar measure replace my right translates below by left translates). Since Haar measure is not identically $0$ on the group, there is a Borel set $A_0$ such that $\mu(A_0) > 0$ (perhaps $\mu(A_0) = \infty$). By (i), $\mu(V_0) > 0$ for some open set $V_0$ containing $A$. By (ii), $\mu(K_0) > 0$ for some compact set $K_0$ contained in $V_0$. For an arbitrary nonempty open set $U$ in the group, its left translates $gU$ as $g$ runs over $G$ are an open cover of $G$, so finitely many of them cover $K_0$: $K_0 \subset \bigcup_{i=1}^n g_iU$. Then $\mu(K_0) \leq \sum_{i=1}^n \mu(g_i U) = n\mu(U)$. Thus positivity of $\mu(K_0)$ implies positivity of $\mu(U)$.

KCd
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