What is the correct definition of the cuspidal subspace of $L^2$? I have a few (semi-)related questions regarding certain Hilbert space representations of locally compact groups that come up in the theory of automorphic forms. 
Let $G$ be a unimodular locally compact Hausdorff group, $\Gamma$ a discrete (hence closed) subgroup of $G$, and $Z$ a closed subgroup of the center of $G$. The example of interest to me is:
(*) $G$ is the group of adelic points of a connected reductive $\mathbf{Q}$-group $\mathscr{G}$, $\Gamma=\mathscr{G}(\mathbf{Q})$, and $Z$ is the group of adelic points of the maximal split torus in the center of $\mathscr{G}$. 
I would be happy to know the answer to the following question just in case (*).
Is the group $Z\Gamma$ closed in $G$, and if so, is it unimodular?
The answer to the first part of the question would be yes if one of $Z$, $\Gamma$ were compact, but this isn't so in many cases of interest to me (e.g. $G=\mathrm{GL}_n$). I suspect the answer to both parts of the question are "yes" because of the following definition which appears in the literature (I will assume $Z\Gamma$ is closed and unimodular to make the definition, although I'm hoping, at least in the case of interest mentioned above, that this is unnecessary):
For a unitary character $\omega$ of $Z$, let $L^2(Z\Gamma\setminus G,\omega)$ be the space of equivalence classes (the equivalence being almost every equality relative to the unique-up-to-scaling $G$-invariant measure on $\Gamma\setminus G$) of Borel measurable $f:\Gamma\setminus G\rightarrow\mathbf{C}$ with the following properties: for each $z\in Z$, $f(\Gamma zg)=\omega(z)f(\Gamma g)$ for almost every $\Gamma g\in\Gamma\setminus G$, $\vert f\vert$ is Borel measurable on $Z\Gamma\setminus G$ (this makes sense because of the unitary of $\omega$), and $\int_{Z\Gamma\setminus G}\vert f\vert^2<\infty$ (the integral being taken with respect to the unique-up-to-scaling $G$-invariant measure on $Z\Gamma\setminus G$). Assuming I've got this definition right, it will be a Hilbert space with the natural inner product.
My second question concerns the definition of the so-called cuspidal subspace $L_0^2(Z\Gamma/G,\omega)$, and I'm only interested in the case of (*) with $\mathscr{G}=\mathrm{GL}_n$, so $G=\mathrm{GL}_n(\mathbf{A}_\mathbf{Q})$, $\Gamma=\mathrm{GL}_n(\mathbf{Q})$, and $Z=\mathbf{A}_\mathbf{Q}^\times$ is the center of $G$. The (apparently) standard definition of $L_0^2(Z\Gamma\setminus G,\omega)$ is: the subspace of $L^2(Z\Gamma\setminus G,\omega)$ consisting of equivalence classes containing a function $f$ such that $\int_{\mathbf{A}_\mathbf{Q}/\mathbf{Q}}f\big(\big(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\big)g\big)dx=0$ for almost every $g\in G$ ($dx$ means Haar measure on $\mathbf{A}_\mathbf{Q}/\mathbf{Q}$). There are some things to check to make precise sense of this definition, but the important thing is that it should be a closed subspace of the full $L^2$-space. This is taken for granted in, e.g., Bump's book on automorphic representations and Godement's notes on Jacquet-Langlands. The closedness doesn't seem completely obvious to me, and I worry about it because some other authors, notably Lang in his book on $\mathrm{SL}_2$, define the cuspidal subspace as the Hilbert space completion of the space of bounded continuous functions on $\Gamma\setminus G$ with the appropriate properties. I would hope these definitions yield the same space but it isn't clear to me.
Assuming the definition I've given of $L_0^2(Z\Gamma\setminus G,\omega)$ is technically correct (i.e. I haven't left out any conditions on the functions under consideration), is it closed in $L^2(Z\Gamma\setminus G,\omega)$?
 A: First, yes, $Z\Gamma$ is closed in $G$. In any particular example this is pretty easy to prove.
Second, for groups bigger than $GL_2$, even the imprecise definition of "cuspform" needs to refer to all conjugacy classes of (unipotent radicals of) rational parabolics. For $GL_n$, these are specified by (ordered) partitions $n_1+\ldots+n_\ell=n$ of $n$, specifying the upper-block-triangular matrices with diagonal blocks of sizes $n_j$, respectively. Their unipotent radicals are upper-triangular with identity matrices in those blocks. Then the imprecise definition is that all the integrals over $N_k\backslash N_\mathbb A$ vanish.
You're right that there is something fishy here, since it doesn't really make sense to integrate $L^2$ functions over smaller subsets, and since these supposed integrals are definitely not continuous functionals on $L^2$, we cannot conclude in any obvious way that the kernels are closed... :)
(Lang's definition dodges the imprecision, but doesn't go far enough...)
For this reason (and also for proving automorphic Plancherel) it is better to say that cuspforms are the orthogonal complement in $L^2$ to all pseudo-Eisenstein series with smooth, compactly-supported data. In effect, this is an integrated form of the naive/imprecise version often seen. The compact support (and continuity) of the data for the pseudo-Eisenstein series produces continuous, compactly-supported (mod the center) functions on $G_k\backslash G_\mathbb A$, so certainly in $L^2$, so giving sensible integrals against $L^2$ functions. Thus, the kernels are closed, and the simultaneous kernel is an intersection of closeds, so is closed.
A: If you look at article on the definition of automorphic forms/reps. in Corvalis
(by Borel and Jacquet, I think, and in vol. I, again I think) I believe everything is precise and correct.
In particular, "cuspidal" will be defined for automorphic forms (which are necessarily real analytic, so that restricting and integrating over unipotent radicals of parabolics is meaningful).  Since cuspidal automorphic forms are
necessarily $L^2$, one could then close them up in $L^2$ to get $L^2_0$; this is an alternative approach.  
The cuspidal automorphic forms will be the literal direct sum of irred. admissible
reps. (the cuspidal automorphic reps.), each with finite multiplicity; their closure (i.e. $L^2_0$) will be the Hilbert space direct sum of appropriate completions of these irreps. (which will then be topologically irred., 
i.e. have no proper invariant closed subspaces). One can get back from the Hilbert space setting to the more algebraic setting of admissible reps. by passing to $K$-finite vectors. 
