Finiteness of the volume $Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)$ for $G = \operatorname{GL}_2$ I am reading Gelbart's lectures on the trace formula and am confused on how the Siegel domain is used to prove the finiteness of the volume of $Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)$ for $G = \operatorname{GL}_2$.  Here compact sets $C_1 \subset N(\mathbb A)$ and $C_2 \subset (\mathbb A^{\ast})^1$ are given, together with a real number $c$, and the Siegel set $\mathfrak S_c$ is defined to be the set of 
$$g = znh_tmk \in G(\mathbb A)$$
with $z \in Z(\mathbb A), n \in N(\mathbb A), h_t = \begin{pmatrix} e^t \\ & e^{-t} \end{pmatrix}$, $m = \begin{pmatrix} a \\ & 1 \end{pmatrix}$ with $a \in \mathbb A^{\ast}$, and $k \in K$, such that
$$n \in C_1, a \in C_2, t > c/2.$$
From the Iwasawa decompostion, we can say that 
$$\int\limits_{Z(\mathbb A)\backslash G(\mathbb A)} f(g)dg = \int\limits_K \int\limits_{N(\mathbb A)} \int\limits_{(\mathbb A^{\ast})^1} \int\limits_{\mathbb R} f(nh_tmk) dt dm dn dk.$$
Here is what Gelbart says about how we can prove the finiteness of the volume of $Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)$:

I don't understand the computation
$$\int\limits_{Z(\mathbb A) G(\mathbb Q) \backslash G(\mathbb A)} dg \leq \int\limits_{Z(\mathbb A) \backslash \mathfrak S_c} dg.$$
What is going on here?  Are we using the fact that
$$\int\limits_{Z(\mathbb A) \backslash G(\mathbb A)} dg = \int\limits_{\big(Z(\mathbb A) \backslash Z(\mathbb A)G(F)\big)\backslash \big(Z(\mathbb A) \backslash G(\mathbb A)\big)} \sum\limits_{\gamma \in Z(\mathbb A) \backslash Z(\mathbb A)G(F)} dg $$
with $\big(Z(\mathbb A) \backslash Z(\mathbb A)G(F)\big)\backslash \big(Z(\mathbb A) \backslash G(\mathbb A)\big) = Z(\mathbb A)G(F) \backslash G(\mathbb A)$?
 A: A special case of this was addressed in my later question here.
The main point is that if $G$ is a unimodular second countable, locally compact Hausdorff group, and $H$ a discrete subgroup of $G$, then for any Borel subset $B$ of $H \backslash G$, and any Borel set $A$ of $G$ whose image in $H \backslash G$ is $B$, there exists a Borel subset $X \subseteq A$ which projects bijectively onto $B$.  It follows that $\operatorname{meas}(X) = \operatorname{meas}(B)$ from the equation
$$\int\limits_G f(g) dg = \int\limits_{H \backslash G} \Bigg( \sum\limits_{h \in H} f(hg) \Bigg) dg$$
which holds for any positive measurable function $f$ (taking $f$ to be the characteristic function of $X$, the map $Hg \mapsto \sum\limits_{h \in H} f(hg)$ will be the characteristic function of $B$).
Now, let $G$ be a connected, reductive group over a number field $k$, and $\mathfrak S_c$ be a Siegel set for $G$, so that $G(\mathbb A) = G(k) \mathfrak S_c$.
If we consider the image $\overline{\mathfrak S_c}$ of $\mathfrak S_c$ in $Z(\mathbb A) \backslash G(\mathbb A)$, then this is easily seen to remain a Borel set, and it surjects to the quotient $(\mathbb Z(A) \backslash Z(\mathbb A)G(k) ) \backslash (Z(\mathbb A) \backslash G(\mathbb A)) = Z(\mathbb A)G(k) \backslash G(\mathbb A)$.
Letting $X$ be a Borel subset of $\overline{\mathfrak S_c}$ which maps bijectively onto $Z(\mathbb A)G(k) \backslash G(\mathbb A)$, we get
$$\operatorname{meas}(Z(\mathbb A)G(k) \backslash G(\mathbb A)) = \operatorname{meas}(X) \leq \operatorname{meas}(\overline{\mathfrak S_c}) < \infty.$$
