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).

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.

open(in $P_0(\mathbb A)$), then $\mathfrak S$ is open (in $G(\mathbb A)$), and that makes measures behave much more nicely. (EDIT: Hmm, we can't take it to be compactandopen. This requires a bit more care, but should still work.) $\endgroup$ – LSpice Jan 30 at 22:47