Why does the finiteness of the Haar measure of Siegel set implies that $SL(n,\mathbb Z)$ is a lattice? Let $G=\operatorname{SL}(n,\mathbb R)$ and $\Gamma =\operatorname{SL}(n, \mathbb Z)$. Suppose we have already shown that $G=\Sigma_{t,u} \Gamma$ for some subset $\Sigma_{t,u}$ of $G$, called Siegel set. Let $\mu$ be a Haar measure on $G$. Suppose also we have shown that $\mu(\Sigma_{t,u})$ is finite.
I wonder how to show that there exists a $G$-invariant measure on $G/\Gamma$ such that $\mu(G/\Gamma)$ is finite.
First obstacle: Let $\pi\colon G\to G/\Gamma$ be the canonical projection. I am not sure if the pushforward measure $\pi_*\mu$ on $G/\Gamma$ is $G$-invariant or not. Since it is unclear to me whether
$$\mu(\pi(gU))=\mu(g\pi(U))$$
or not.
Second obstacle: I don't see $\mu(\Sigma_{t,u})$ being finite implies $\pi_*\mu(G/\Gamma)$ being finite.
 A: I was also confused on this point (when studying the analogue of your question for Siegel domains of adele groups).  See my question and the helpful technical answers here.
So first of all, $G = \operatorname{SL}_2(\mathbb R)$ and its discrete subgroup $\Gamma = \operatorname{SL}_2(\mathbb Z)$ are both unimodular.  Fixing a Haar measure $dg$ on $G$, the general theory of topological groups guarantees you a unique left $G$-invariant Radon measure $d \dot g$ on $G/\Gamma$ such that
$$\int\limits_G f(g)dg = \int\limits_{G/\Gamma} \Big[\sum\limits_{\gamma \in \Gamma} f(g\gamma) \Big] d \dot g$$
for every $f \in \mathscr C_c(G)$ (continuous and compactly supported complex valued functions on $G$).  This measure is not the pushforward measure (since, the measure of $G$ being infinite, it would be impossible for the measure of $G/\Gamma$ to be finite with the pushforward measure).
If you're unfamiliar with this type of "quotient measure" construction on coset spaces, consider the standard example of the additive group $\mathbb R$ and its discrete subgroup $\mathbb Z$: the invariant Radon measure on the quotient space $\mathbb R/\mathbb Z = S^1$ is not the pushforward measure.
Anyway, to answer your question: why the finiteness of the measure of a Siegel set $\Sigma$ with respect to $dg$ implies the finiteness of the measure of $G/\Gamma$ with respect to $d \dot g$.  Standard topological/measure theoretic arguments tell you that since your Siegel set $\Sigma$ is a Borel set, there is a Borel subset $B \subset \Sigma$ with the property that:

*

*$B$ and $\Sigma$ have the same image in $G/\Gamma$.


*$B$ is mapped bijectively onto its image in $G/\Gamma$.


*The measure of $B$ with respect to $dg$ is equal to the measure of its image in $G/\Gamma$ with respect to $d \dot g$.
The fact that $\Sigma$ is a Siegel set tells you that the image of $\Sigma$ (and $B$) is all of $G/\Gamma$.  Therefore,
$$ d\dot g(G/\Gamma) = dg(B) \leq dg(\Sigma) < \infty.$$
