# 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.

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.

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 $$U$$].

• Could you explain why does a haar measure has full support? May 24 at 23:49
• 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. May 25 at 0:00

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 $$\mu(G \setminus A) = 0$$.

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

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 $$\mu(G) =\infty$$.