Let $G$ be a compact abelian group, $H$ be a closed subgroup of $G$.
Suppose that both $G,H$ equipped with the Borel $\sigma$-algebra and the Haar measure. Why is it true that $G$ is isomorphic to $G/H\times H$ as measure spaces? (i.e there exists a measurable map $\pi:G\rightarrow G/H\times H$ that is one to one and onto up to a set of measure zero).