I'm taking a topics course and trying to fill in details from a lecture. I'm hoping someone can help fill things in or point to a resource where these basics are covered.
We have a lattice $\Lambda < G$, where $G$ is a locally compact group, so by definition there is finite measure fundamental domain $F$ (ie, $G= \bigcup_{\lambda \in \Lambda}F \lambda$). How can we refine this $F$ so that $G = \bigsqcup_{\lambda \in \Lambda}F \lambda$? My professor says this statement is true 'up to null sets', but it seems like below I will really need the fact that all of $G$ is covered, and the union really is disjoint below.
Suppose now that we have this disjoint union property. We can identify $F$ with $G/\Lambda$, since the covering of $G$ by translates of $F$ says every coset is represented in $F$, and the disjointness says there is a unique representative. Define an action of $G$ on $G/\Lambda = F$ by taking $x \in F$, noticing that $gx \in F \lambda$ for some unique (by disjointness) $\lambda \in \Lambda$, so that $gx\lambda^{-1} \in F$. Write $g \cdot x $ to distinguish this from regular multiplication in $G$. How would I be able to do this if we relaxed the disjoint union/covering to being 'only up to null sets'?
Next, let $\pi:G \longrightarrow G/\Lambda$ be the projection. Supposedly there should always exist a measurable section $\sigma: G/\Lambda \longrightarrow G$. Given that, we could just take $F = \sigma(G/\Lambda)$. These sections and the $F$ above are clearly equivalent. But how do we get them?
Finally, I should be able to show $\sigma(g\cdot x)^{-1}g\sigma(x) \in \Lambda$ for any $g,x$ by using $\pi$ is $G$ equivariant, but I don't quite see it.