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.

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 $\mu(G) =\infty$.
