Haar measure on isomorphic subgroups Let $G$ be an abelian locally compact Hausdorff group endowed with a chosen Haar measure $\mu$. Moreover, let $H,N$ be subgroups of $G$ where $N$ is discrete. Then we have the isomorphic quotient groups
$$\frac{(H+N)}{N}\cong\frac{H}{H\cap N}.$$ If we put the natural counting measure on the discrete subgroups, then we get a unique induced measure on the quotient. Now I wonder whether the groups in the isomorphism have the same volume under the canonical measures? Or maybe a better formulation is, whether the pushforward measure induced by the canonical measure on the left hand side coincides with the canonical measure on the right hand side?
Edit: We assume that the restricted measure on $H$ is nonzero, and therefore a Haar measure. On $N$ we denote the measure by $\lambda$, and on $H\cap N$ the restricted measure by $\lambda_{H}$. On $G$ we have the measure $\mu$, and on the quotient $G/N$ the measure $\nu_{N}$. The unique induced measure on $G/(H\cap N)$ is denoted by $\nu_{H}$. We want to show that $\nu_{N}((H+N)/N) = \nu_{H}(H/(H\cap N))$ (if it is true).
I guess that if this it true, one needs the quotient integral formula. But I dont exactly see how to obtain it here.
$$\nu_{N}((H+N)/N)=\int_{G/N}\chi_{(H+N)/N}(x)\mathrm{d}\nu_{N}(x)$$
$$\nu_{H}((H/(H\cap N))=\int_{G/(H\cap N)}\chi_{H/(H\cap N)}(x)\mathrm{d}\nu_{H}(x),$$
where the integrands are the characteristic funtions. The reason why I think the quotient integral formula is needed, is that the measure on $H\cap N$ is just the restriction of the measure on $N$. And therefore you might get a nice equality when you can write the integral
$$\int_{G/N}\chi_{(H+N)/N}(x)\mathrm{d}\nu_{N}(x) = \int_{G/N}\int_{N}f(yn)\mathrm{d}\lambda\mathrm{d}\nu_{N}(yN),$$
for some integrable $f$.
 A: Maybe not exactly what you're wanting, but a true thing that I can tell you. :)
For example, given a unimodular topological group $G$ and unimodular subgroup $H$ (discrete, for example), there is a small theorem that asserts that any two out of the three (positive, regular, Borel) measures on $G$, $H$, and $H\backslash G$ uniquely determine a measure on the third, so that the natural identity holds, for $f\in C^o_c(G)$,
$$
\int_G f \;=\; \int_{H\backslash G} \Big(\int_H f\Big)
$$
If one wants, one can insert more notation to be explicit, but nothing would really be gained. In fact, I like this decluttered version...
EDIT: To compare the adelic quotient $\mathbb A/K$ and purely archimedean quotient $\mathbb A_\infty/\mathfrak o$, use the fact that $\mathbb A_\infty\times \prod_v\mathfrak o_v$ is open in $\mathbb A$, so $\mathbb A_\infty/\mathfrak o\times \prod_v \mathfrak o_v$ is open in the adelic quotient. In fact, it is in bijection with it. Thus, specifying the (Haar) measure on $\prod_v \mathfrak o_v$ completely determines the comparison of the adelic quotient's measure and the archimedean quotient's measure.
