Let $G = \operatorname{GL}_n(\mathbb Q_p)$, and let $Z$ be the center of $G$. Let $\nu: Z \rightarrow \mathbb C^{\ast}$ be a continuous, unitary character of $Z$. Then $L^2(G/Z,\nu)$ is defined to be the space of measurable functions $f: G \rightarrow \mathbb C$ such that $f(zg) = \nu(z)f(g)$ for all $z \in Z$ and $g \in G$ for which
$$\int\limits_{G/Z} |f(g)|^2 \space dg < \infty$$
Here $|f(g)|$ is well defined on $G/Z$. Then $L^2(G/Z,\nu)$ is a Hilbert space with inner product
$$\langle f,h\rangle = \int\limits_{G/Z} f(g) \overline{h(g)} \space dg$$
I'm a bit puzzled with this definition of requiring $f$ to be measurable and also requiring $f(zg) = \nu(z)f(g)$ for all $g \in G$ and $z \in Z$. Usually with $L^2$ spaces, you take a complete measure and identify those functions which are equal almost everywhere. In particular, the individual values $f(g)$ that a given function $f$ takes are not so important. But here the individual values $f(g)$ are important, since we want $|f|$ to be well defined as a function on $G/Z$.
Is the definition I've stated above really the correct one? I would think in order for $L^2(G/Z,\nu)$ to be a Hilbert space, rather than a pre Hilbert space, one would need to be less restrictive in the definition. Maybe something like requiring that for almost all $(g,z) \in G \times Z$, we have $f(zg) = \nu(z)f(g)$. Then $f$ is equal almost everywhere to a function $h$ for which $|h|$ is well defined on $G/Z$. I would appreciate if anyone could shed some light on this definition.