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I have been reading on Haar measure recently.

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

  1. $\mu(\{e\})>0$ then $G$ is discrete.
  2. $\mu(G)<\infty$ then $G$ is compact.
  3. we know that every locally compact Hausdorff group admits a Haar measure, is the same true for monoids(semigroup with identity e)? If not, is there any counterexample? Is there a class of semigroups that admits a Haar measure?

I just saw the "Finite Haar Measure if and only if Compact" by Gils.

About the first part, I can't understand what Gils said, 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.

About the second part, I'd like a proof without any integrals.

About the third part, My question on Haar Measure on Locally Compact monoids hasn't been answered yet. You can answer this question on that page. I am willing to accept the answer.

Any help will be appreciated.

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1 Answer 1

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  1. Suppose $\alpha := \mu(\{e\}) > 0$, then $\mu(\{x\}) = \alpha$ for all $x\in G$. Now if $\epsilon := \alpha/2 > 0, \exists U$ open such that $e\in U$ and $$ \mu(U) < \mu(\{e\}) + \epsilon $$ Conclude that $U = \{e\}$ must hold.

  2. Suppose $G$ is not compact, then by local compactness, choose a neighbourhood $U$ of $e$ such that $\overline{U}$ is compact. Clearly, $\exists g_1 \in G\setminus U$. Now, $$ U\cup g_1U $$ has compact closure, so $\exists g_2 \in G\setminus (U\cup g_1U)$. Thus proceeding, we obtain a sequence $(g_n)$ such that $$ g_n \in G\setminus \left( \bigcup_{i=0}^{n-1} g_i(U)\right) $$ where $g_0 = e$. Now choose an open set $V$ such that $e\in V$ and $VV^{-1} \subset U$, then the sets $$ \{g_n V: n\in \mathbb{N}\} $$ are mutually disjoint. Since $V$ is open, $\mu(V) > 0$ and this would contradict the fact that $\mu(G) < \infty$.

  3. See this :

https://mathoverflow.net/questions/180019/haar-measure-on-locally-compact-semigroups

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  • $\begingroup$ Thank you very much! Bnd for the third question, I just found that the counterexample is not a semigroup without identity e, I want an example about a semigroup with identity e... $\endgroup$
    – David Chan
    May 20, 2015 at 6:02
  • $\begingroup$ You can append an identity, and then $S = \{e, a, b\}$, $aS = \{e, a\}$ and $bS = \{e, b\}$. So $\mu(S) = \mu(aS)$, $0=\mu(\{b\}) = \mu(\{a\}) = \mu(\{e\})$ which is fails to be positive on non empty open sets, regardless of the topology. $\endgroup$
    – user24142
    May 20, 2015 at 6:17

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