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For a given compact (lie) group there exists a unique haar propability measure $\mu_G$. Is it true that any open dense subset $U\subseteq G$ has full measure, i.e. $\mu_G(U) = 1$?

I know the statement is wrong for locally compact groups, see e.g. the euclidean case.

Since the case $|G| < \infty$ is trivial we can rely on the fact that singletons have zero measure. I was trying to prove by some compactness + approximation argument, that reduces $\mu_G(G\setminus U)$ to some $n\cdot \mu_G(\{g\})$, which implies that the complement must be a zero set, but could not really get it/finish...

The book Compact Lie Groups - M. Sepanski seems to use this fact in the proof of Theorem 7.34 and 7.16 but gives no precise information/reasoning, this is where i stumbled onto this hypothesis.

Tina
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1 Answers1

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I believe the following is a counterexample to the claim. Consider the unit circle $\mathbb T=\{e^{i\theta}:\theta\in\mathbb R\}$ in $\mathbb C$, which is clearly a compact Lie group, and its normalized Haar measure $m$ is the normalized arclength measure (i.e. the pushforward of Lebesgue measure on the unit interval). Let $\{\zeta_j\}_{j\in\mathbb N}$ be an enumeration of all points in $\mathbb T$ such that $\arg\zeta_j\in\mathbb Q$, where $\arg$ denotes the principal argument with values in $[0,2\pi)$. Note that $\{\zeta_j:j\in\mathbb N\}$ is dense in $\mathbb T$. For $j\in\mathbb N$, let

$$I_j=\left\{e^{i(\arg\zeta_j+\theta)}:\theta\in\left(-\frac{2\pi}{2^{j+3}},\frac{2\pi}{2^{j+3}}\right)\right\}.$$

Observe that each $I_j$ is an open subset of $\mathbb T$ with

$$m(I_j)=\frac{1}{2^{j+3}}-\left(-\frac{1}{2^{j+3}}\right)=\frac{1}{2^{j+2}}.$$

In particular, if we set

$$U=\bigcup_{j\in\mathbb N}I_j,$$

then $U$ is an open dense subset of $\mathbb T$ with

$$m(U)\leq\sum_{j=0}^\infty m(I_j)=\frac{1}{4}\underbrace{\sum_{j=0}^\infty\frac{1}{2^j}}_{=2}=\frac{1}{2},$$

so in particular $U$ does not have full measure. Of course by a similar argument you can make the measure of $U$ as small as you want.

Lorago
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