A subgroup of full measure is dense given a haar measure 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.
 A: 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 $U$].
A: 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.
A: 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 $\mu(G \setminus A) = 0$.
A: 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 $\mu(G) =\infty$.
