Let $G$ be a semisimple algebraic group over a number field $F$ with trivial center. Let $\mathfrak S \subset G(\mathbb A)$ be a Siegel domain (defined in terms of a given maximal split torus and minimal parabolic in $G(\mathbb A)$, for a reference see for example pg. 37 of Arthur's [introduction to the trace formula](http://www.claymath.org/library/cw/arthur/pdf/62.pdf)).


My question is, from the following two facts, is it possible to deduce the finiteness of the volume of $G(F) \backslash G(\mathbb A)$?


* $\operatorname{meas}(\mathfrak S) < \infty$ (follows from a standard computation from the definition of $\mathfrak S$)
* For sufficiently large $\mathfrak S$, we have $G(\mathbb A) = G(F) \mathfrak S$.


I have seen computations (which I think are) to the effect of


$$\int\limits\_{G(F) \backslash G(\mathbb A)} dg = \int\limits\_{G(F) \backslash G(F) \mathfrak S} dg = \int\limits\_{(G(F) \cap \mathfrak S) \backslash \mathfrak S} dg \leq \int\limits\_{\mathfrak S} dg < \infty$$


but was uncertain of the legality of this reasoning. For one thing, $\mathfrak S$ and $G(F) \cap \mathfrak S$ are not groups, and although one can transfer the right invariant measure on $G(F) \backslash G(\mathbb A)$ to one on $(G(F) \cap \mathfrak S) \backslash \mathfrak S$ via the natural bijection


$$(G(F) \cap \mathfrak S) \backslash \mathfrak S \rightarrow G(F) \backslash G(\mathbb A)$$


$$(G(F) \cap \mathfrak S)x \mapsto G(F)x$$


it is not obvious how this measure on $(G(F) \cap \mathfrak S) \backslash \mathfrak S$ compares to the on $G(\mathbb A)$ with which $\int\limits\_{\mathfrak S} dg$ is calculated.



Finiteness of the volume of $G(F) \backslash G(\mathbb A)$
$\DeclareMathOperator\intr{int}\DeclareMathOperator\meas{meas}$This is one—probably not the best—way of thinking about it. (I am used to the $p$-adic world, where one can think about (real-valued) measures much less subtly than in the real world.) I hope someone will come along and write something more elegant.


Since $G(F)$ is discrete, so that its Haar measure may be taken to be the counting measure, the way we know we have the correct measure on $G(F)\backslash G(\mathbb A)$ is that
$$
\int\_{G(F)\backslash G(\mathbb A)} \sum\_{h \in G(F)} f(h g)\,\mathrm d\dot g = \int\_{G(\mathbb A)} f(g)\mathrm dg
$$
for all $f \in \operatorname C\_c^\infty(G(\mathbb A))$. Now take $\mathfrak S$ and $\mathfrak S'$ such that $G(\mathbb A) = G(F)\mathfrak S'$ and $\mathfrak S' \subseteq \intr(\mathfrak S)$. Write $\mathfrak S'$ and $\mathfrak S$ as countable, increasing unions of compacts $\mathfrak S'\_k$ and $\mathfrak S\_k$ such that $\mathfrak S'\_k \subseteq \intr(\mathfrak S\_k)$ for all $k$; and, for each $k$, let $f\_k \in \operatorname C\_c^\infty(G(\mathbb A))$ be a non-negative function that is $1$ on $\mathfrak S'\_k$ and $0$ outside $\mathfrak S\_k$. Notice that, for each $k$ and each $g \in \mathfrak S'\_k$, we have that
$$
\sum\_{h \in G(F)} f\_k(h g) \ge f\_k(g) = 1,
$$
so that
\begin{multline\*}
\meas\_{\mathrm d\dot g}(G(F)\backslash G(\mathbb A))
= \lim\_{k \to \infty} \meas\_{\mathrm d\dot g}(G(F)\backslash G(F)\mathfrak S'\_k)
\le \lim\_{k \to \infty} \int\_{G(F)\backslash G(\mathbb A)} \sum\_{h \in G(F)} f\_k(h g)\,\mathrm d\dot g \\
= \lim\_{k \to \infty} \int\_{G(\mathbb A)} f\_k(g)\mathrm dg
\le \lim\_{k \to \infty} \meas\_{\mathrm dg}(\mathfrak S\_k)
= \meas\_{\mathrm dg}(\mathfrak S) < \infty.
\end{multline\*}


