2
$\begingroup$

I want to know why if $\mu$ is a haar measure on a compact $G$ and $\mu(A)=\mu(G)$ then $A$ is dense in $G$. This fact is mentioned in the wikipedia page, but I couldn't find a proof for it.

$\endgroup$

4 Answers 4

2
$\begingroup$

In fact, every subset of full Haar measure must be dense. This follows from the following statement, that you can also find on the wikipedia page:

Claim If $U$ is a nonempty open set in the locally compact group $G$, then the left Haar measure satisfies $\mu(U)>0$.

Proof: We will use the fact that Haar measure is inner regular, so there must be some compact set $K \subset G$ with $\mu(K)>0$. Given a nonempty open set $U \subset G$, fix some $u\in U$, and note that the sets $\{gu^{-1}U: g \in K\}$ form an open cover of $K$. There is a finite subcover $\{g_ju^{-1}U: 1 \le j \le m\}$. If $\mu(U)=0$ then this finite subcover, and the left-invariance of Haar measure, would yield $\mu(K)=0$, a contradiction.

$\endgroup$
2
$\begingroup$

I will assume that $\mu (G \setminus A)=0$. Then $\mu (G\setminus \overline A)=0$ and $G\setminus \overline A$ is open. Since Haar measure has full support this implies that $ G\setminus \overline A=\emptyset$. Hence, $ \overline A =G$.

[If $K$ is compact and $U$ is a non-empty open set then $K \subset \bigcup_x (x+U)$ and there is a finite subcover. If $\mu (U)=0$ then translation invariance gives $\mu (K)=0$. This implies that $\mu $ is the $0$ measure. Hence, $\mu (U)>0$ for any nonempty open set $U$].

$\endgroup$
2
  • $\begingroup$ Could you explain why does a haar measure has full support? $\endgroup$
    – Saviour
    May 24 at 23:49
  • $\begingroup$ If $K$ is compact and $U$ is open then $K \subset \bigcup_x (x+U)$ and there is a finite subcover,. If $\mu (U)=0$ then translation invariance gives $\mu (K)=0$. This implies that $\mu $ is the $0$ measure. $\endgroup$ May 25 at 0:00
1
$\begingroup$

False as stated. Let $G$ be $\mathbb Z^2$ with the discrete topology. Haar measure is counting measure. Let $A = \left\{(x,0) \mid x \in \mathbb Z\right\}$. Then $A$ is closed, $\mu(A) = +\infty = \mu(G)$. But $A$ is not dense in $G$.


To get the correct statement: replace $\mu(A) = \mu(G)$ by $\mu(G \setminus A) = 0$.

$\endgroup$
2
  • $\begingroup$ I corrected the question adding the condition for G to be compact, which should force the measure to be finite. thanks! $\endgroup$
    – Saviour
    May 25 at 0:21
  • $\begingroup$ Good observation! I have edited my answer accordingly. $\endgroup$ May 25 at 6:31
0
$\begingroup$

Suppose that $A$is not dense in $G$. If $\mu(G) < \infty$, there is an open subset $U$ of $G$ with $\mu(U) = 0$. A compactness argument can be used to show that $\mu(G) = 0$.

I am not sure about the case of $\mu(G) =\infty$.

$\endgroup$

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.