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This is an exercise from a book:

Let $G$ be a locally compact group with Haar measure $\mu$.

  1. $\mu(\{e\})>0$ if and only if $G$ is discrete.
  2. $\mu(G)<\infty$ if and only if $G$ is compact.

I think I can solve the first part:

If $\mu(\{e\})>0$ then $\mu$ is a scalar multiple of the counting measure. Since $\mu$ is outer-regular, this means that $\{e\}$ is open.

I need help with the second part, here's what I've been trying.

Maybe I can try to use the fact that $\mu$ is inner-regular? I'm not sure how to. Other possible approaches: Since $\mu(G)<\infty$, every subgroup $H$ of $G$ must either be of measure zero or of finite index.

Gils
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    See Remark 2.8 and Proposition 2.9 of http://www.math.ethz.ch/~torniers/download/2014/haar_measures.pdf. – Martín-Blas Pérez Pinilla Feb 14 '14 at 19:06
  • Note part 1 requires the additional Hausdorff hypothesis, otherwise the singleton may even fail to lie in the Borel sigma-algebra (e.g. in the case the topology is the indiscrete). – pre-kidney Sep 02 '18 at 21:31

1 Answers1

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This is proved in NACHBIN, The Haar Integral, chapter II, Proposition 4. I'll rewrite his proof here:

Suppose $G$ is not compact. Choose $f:G\rightarrow\mathbb{R}$ continuous with compact support such that $0\leq f\leq 1$ and $\int_{G}fd\mu>0$. Let $K=\text{supp}(f)$, so $K$ is compact. Set $s_1=e$.

Suppose we already chose $s_1,\ldots,s_n\in G$ such that the sets $s_iK$ are parwise disjoint.

Notice that, since $\bigcup_{i=1}^n s_iK K^{-1}$ is (a finite union of compact sets, hence) compact, then there exists $s_{n+1}\in G\setminus\bigcup_{i=1}^n s_iKK^{-1}$. It is obvious that $s_{n+1}K$ is disjoint to all others $s_iK$.

Therefore, we obtain a sequence $s_1,s_2,\ldots$ such that $s_iK$ are pairwise disjoint. For every $n$, define the continuous function $f_n(t)=\sum_{i=1}^nf(s_i^{-1}t)$, so that $\text{supp}f_n=\bigcup_{i=1}^ns_iK$ and $f_n\leq 1$. Finally, $$0<n\int fd\mu=\int f_nd\mu\leq\mu(G),\qquad\text{for all }n=1,2,\ldots$$ so $\mu(G)=\infty$.Q.E.D.

Luiz Cordeiro
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  • A comment for myself and future people, I found it is better to write the last equation as $0<n\int_G fd\mu = \int_G f_nd\mu \leq \mu(G),\qquad\text{for all }n=1,2,\ldots$ – No One Sep 22 '20 at 00:28
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    It seems like we can simplify things a little bit, working only with the measure instead of the integral. All we need is that $K$ is a compact set with nonzero Haar measure; it would suffice to choose any compact set with nonempty interior, which by local compactness must exist. And for the last step, the $s_i K$ are disjoint sets with measure $\mu(K) > 0$, so $\mu(G) \ge \sum \mu(s_i K) = \sum \mu(K) = \infty \cdot \mu(K) = \infty$ by countable additivity. – Nate Eldredge Oct 19 '21 at 06:22