Is the map $G\to G/\Gamma$ 'locally measure preserving?' Let $G$ be a Lie group with Haar measure $\mu$ and $\Gamma$ be a discrete subgroup of $G$. Assume that $X:=G/\Gamma$ admits a nonzero $G$-invariant Radon measure $\nu$.
Recall that the map $p:G\to G/\Gamma$ is a normal covering map.

Question. Is it true that if $U$ is any evenly covered neighborhood of $X$ and $V$ is any slice of $p^{-1}(U)$, then $\mu(V)=\nu(U)$?

I thought of constructing a $G$-invariant measure on $G/\Gamma$ satifying the property mentioned in the above question, so that we'd be done by using the uniqueness (up to a constant multiple) of $\nu$.
The natural thing to do would be to define a measure $\theta$ on $G/\Gamma$ as follows: If $U$ is an evenly covered open set in $X$, then we define $\theta(U)$ as the Haar measure of any slice of $p^{-1}(U)$.
(It is clear that $\mu(V)$ is independent of the choice of the slice $V$ by the fact that $p$ is a normal covering.)
For open sets which can be covered by finitely many evenly covered open sets we may use inclusion exclusion to assign measure to them (the well-definedness is an issure here).
It is not clear how we can go to arbitrary borel sets from here.
 A: I think this should be true. The $G$-invariant Radon measure $\nu$ satisfies some uniqueness properties. In particular (after scaling appropriately) it satisfies the following Fubini identity for $f\in C_c(G)$:
$$\int\limits_{G}f d\mu = \int\limits_{G/\Gamma}P(f) d\nu
$$
Where $P:C_c(G) \to C_c(G/\Gamma)$ is the operator 'integration along fibers' (with respect to a Haar measure on $\Gamma$).
You can see Folland's book on harmonic analysis (theorem $2.49$ and before) for a complete explanation.
In this case, it is simply $P(f)(g\Gamma):= \sum\limits_{\gamma\in \Gamma} f(g\gamma)$.
Now assuming for simplicity that $V \subset G$ has compact closure, we get from the above formula that 
$$\mu(V) = \int\limits_{G/\Gamma} P(1_V) d\nu
$$
And $P(1_V)(g\Gamma) = \sum 1_V(g\gamma)= |\lbrace \gamma \in \Gamma : g \in V\gamma^{-1}\rbrace | = |\lbrace \gamma \in \Gamma : \gamma \in g^{-1}V |$. Since you assumed $V$ was a slice of some stack over $U$, this number is either $0$ or $1$.
It is non-zero if and only if $g\Gamma\in \pi(V)=U$.
Using this in the above equation, we get
$$\mu(V) = \nu(U).
$$
Hope this isn't nonsense.
