Suppose $G$ is a (Hausdorff) compact group with normalised Haar measure $\mu$, and that $H\trianglelefteq G$ is a closed normal subgroup. Is it true that the pushforward of $\mu$ to $G/H$ is the normalised Haar measure of $G/H$?
That is, is it true that $$\nu(A)=\mu(\pi^{-1}(A))$$ for every Borel $A\subseteq G/H$, where $\nu$ is the normalised Haar measure on $G/H$?
EDIT: I understand the pushforward measure is a left-invariant Borel measure assigning finite measure to compact sets and positive measure for open sets; but for it to be a Haar measure it also needs to be regular, at least in order to use the uniqueness of Haar measures proved in Hewitt and Ross's book (which is the only reference I know).