Unique Haar measure on quotient is pushforward

Question

Let $G$ be an abelian locally compact Hausdorff group with discrete subgroup $H$. Let $\mu$ be a Haar measure on $G$ and $\lambda$ the usual counting measure on $H$. Then we obtain a unique Haar measure on $G/H$ satisfying $$\int_{G/H}\left(\int_{H}f(gh)\mathrm{d}\lambda(h)\right)\mathrm{d}\nu(gH)=\int_{G}f(g)\mathrm{d}\mu(g).$$ Since another natural way to obtain a Haar measure on $G/H$ is by considering the pushforward measure $\nu:=\mu\circ f^{-1}$, where $f$ is the quotient map, I wondered whether the two measures obtained coincide. What I tried is just showing that the pushforward measure satisfies the integral formula, but I failed doing so.

By definition of the pushforward measure we have for each measurable $M\subset G/H$, and corresponding characteristic function $\chi_{M}$, that $$\int_{G/H}\chi_{M}(gH)\mathrm{d}\nu(gH)=\int_{G}\chi_{f^{-1}(M)}(g)\mathrm{d}\mu(g).$$ This already looks a lot like the desired integral formula, nevertheless we would like to implement the measure on H in an appropiate way. But I don’t see how, or if this is even possible. Or is there any additional condition needed, for instance $G/H$ being compact? Any help would be appreciated!

Answer 1

I guess I found my own counterexample, which maybe is the most natural one to consider at first. Namely, pick $G=\mathbb{R}$ and $H=\mathbb{Z}$. In this case the quotient is endowed with the natural Lebesgue measure induced by the lebesgue measure on $[0,1)$ (by the pushforward of the canonical isomorphism). This clearly doesn’t equal the pushforward measure of the quotient map between $H$ and $G$.