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).
Best Answer
The answer is yes: if you write $\pi^{-1}(A)=AH=HA$ (since the subgroup $H$ is normal in $G$), then you realize that for any element $gH=Hg\in G/H$ you have $$gH.A=gH.AH=g.HA=g.\pi^{-1}(A).$$
Since the measure $\mu$ is invariant under the multiplication by $g$, so is the pushforward measure: $$ \nu(gH.A)=\mu(\pi^{-1}(gH.A))=\mu(gHA)=\mu(AH)=\mu(\pi^{-1}(A))=\nu(A). $$
Moreover the pushforward measure is clearly non-trivial ($\nu(G/H)=\mu(G)\neq 0$), and Radon: (EDIT) by the definition of the quotient topology, the compact subsets of $G/H$ are of the form $\pi(K)=KH$, with $K\subset G$ compact. This implies that the measure $\nu=\pi_*\mu$ is inner regular.
Concerning the reference that you are asking, I really love the book by Halmos on Measure Theory. Another place where I've first learned about integration on locally compact groups is the appendices of the book by Bekka, de la Harpe & Valette about Kazhdan's property (T) (it is freely available online).