Trouble with definition of a lattice: Meaning of finite volume of $\Gamma\setminus G$ Let $G$ be a Lie group and $\Gamma$ be a discrete subgroup of $G$. Definition 1.3.5 of Dave Witte Morris's Introduction to Arithmetic Groups says the following:

We say that $\Gamma$ is a lattice in $G$ if $\Gamma\setminus G$ has finite volume (with respect to the Haar measure on $G$).

I am unable to understand this definition. What is meant by saying "$\Gamma\setminus G$ has finite volume with respect to the Haar measure on $G$?"
 A: A Haar measure on $G$ induces a canonical measure on the coset space $\Gamma\backslash G$.  There are various ways of defining it, but here's probably the simplest: a Haar measure on $G$ can be represented by a left invariant volume form $\omega$, and then you can push that volume form down to $\Gamma\backslash G$ along the quotient map (which is a local diffeomorphism).  The pushforward is well-defined since $\omega$ is left invariant, and in particular invariant under left multiplication by any element of $\Gamma$.
A: Let $\Gamma$ be a discrete subgroup of $G$, and let $\mu$ be a right Haar measure on $G$.  Up to scalar multiple, there is a unique Radon measure $\bar{\mu}$ on the quotient space $\Gamma \backslash G$ such that 
$$\bar{\mu}(Eg) = \bar{\mu}(E)$$
for all Borel subsets $E \subset \Gamma \backslash G$ and all $g \in G$.  If $f \in C_c(G)$ is a continuous and compactly supported complex valued function, then the averaged function
$$\dot f: \Gamma \backslash G \rightarrow \mathbb C$$
$$\dot f(\Gamma g) = \sum\limits_{\gamma \in \Gamma} f(\gamma g)$$
is well defined and lies in $C_c(\Gamma \backslash G)$. 
 The measure can be scaled so that (with a slight abuse of notation)
$$\int\limits_G f(g)d\mu(g) = \int\limits_{\Gamma \backslash G} \dot f(\Gamma g) d\bar{\mu}(g) = \int\limits_{\Gamma \backslash G} \Bigg(\sum\limits_{\gamma \in \Gamma} f(\gamma g) \Bigg)d\bar{\mu}(g)$$
for all $f \in C_c(G)$.  
The volume of $\Gamma \backslash G$ is the measure of $\Gamma \backslash G$ with respect to $\bar{\mu}$.  When $\Gamma$ is a lattice, the volume of $\Gamma \backslash G$ of course depends on the normalization of the measures involved.
If $\Gamma \backslash G$ is compact, then $\Gamma$ is automatically a lattice, e.g. $G = \mathbb R$ and $\Gamma = \mathbb Z$.  But there are other examples  where $\Gamma \backslash G$ has finite volume but is not compact, e.g. $G = \operatorname{SL}_2(\mathbb R)$ and $\Gamma = \operatorname{SL}_2(\mathbb Z)$.  
