Why is it possible to normalize the Haar measure on the quotient?

Question

I just asked a question which is related to the one I'm about to ask, but I realized my question can be reduced to the following: let $G$ be a locally compact abelian group with Haar measure $\mu$, and $H$ a discrete subgroup of $G$. Then $G/H$ is locally compact with Haar measure $\bar{\mu}$.

I believe it should be possible to normalize $\bar{\mu}$ to make it have the following property: if $W \subseteq G$ is any measurable set and the projection $\pi: G \rightarrow G/H$ maps $W$ injectively into the quotient, then $\mu(W) = \bar{\mu}(\pi(W))$.

Answer 1

This is possible. Suppose that $\mu(W)<\infty$ and that $\pi(W)$ is measurable. Then $\mu(W)$ is the supremum of all compact sets $K$ contained in $W$, as $\mu$ is a Radon measure. The same holds for $\pi(W)$. For a compact set $K$ the measure $\mu(K)$ is the infimum of all integrals $\int_Gf(x)d\mu(x)$ where $f\ge 0$ is continuous of compact support. (This is part of the proof of Riesz's representation theorem.) So the question boils down to the following: Can $\bar\mu$ be normalized such that $$ \int_Gg(x)d\mu(x)=\int_{G/H}\sum_{h\in H}g(xh)d\bar\mu(x). $$ Now the sum map $C_c(G)\to C_c(G/H)$ is surjective hence one can take the left hand side as a definition for a positive linear functional on $C_c(G)$ which by Riesz gives a unique Radon measure.