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.


1 Answer 1

  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 :


  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.