Why is the Haar measure times the volume of the eigenvalue simplex considered a good measure of Hilbert space volume? In particular, why do we need both of these to find the volume? And should I be thinking of it as an actual volume or not?
This Hilbert space volume is talked about in this paper. It says

There exist the unique, uniform measure $\nu$ on $P$ induced by the Haar measure on the group of unitary matrices $U(N)$. Integration over the set $P$ thus amounts to an integration of the corresponding angles and phases in N-dimensional complex space that determine the families of orthonormal projectors. 

And 

Since the simplex $\Delta$ is a subset of the ($N-1$)-dimensional hyperplane, there exist a natural measure on $\Delta$ which is deﬁned as a usual normalized Lebesgue measure $\mathcal{L}_{N-1}$ on $R^{N-1}$.

The volume of the Hilbert space is then defined to be $\mu=\nu \times \mathcal{L}_{N-1}$, but why?
 A: 1) First of all, $\mu$ is not a measure on the $N$-dimensional Hilbert space $H\cong\mathbb{C}^N $, but on the space $C$ of density operators $\rho:H\to H$, i.e., positive operators $\rho\in B(H)\equiv {\cal L}(H,H)$ with trace $\mathrm{tr}(\rho)=1$. Let us below sketch an argument for the choice of the measure $\mu$.
2) A positive operator $\rho$ can always be diagonalized 
$$U^{-1}\rho U=  \mathrm{diag}(\lambda_1, \ldots,  \lambda_N) ,$$
by an unitary operator $U\in U(H)\cong U(N)$. 
3) The positivity yields 
$$\lambda_1, \ldots,  \lambda_N\geq 0,$$
and the trace condition yields 
$$1=\mathrm{tr}(\rho)=\sum_{i=1}^N  \lambda_i.$$
4) It is not hard to show that if the eigenvalues $\lambda_1, \ldots,  \lambda_N$ are all different (which is the generic situation; more precisely, below we shall say that this is true almost everywhere wrt. the Lebesgue measure $d^N\lambda$), then any other unitary operator $V\in U(H)$, that diagonalizes $\rho$, must be related to $U$ as
$$ VU^{-1}=  P_{\pi} \ \mathrm{diag}(e^{i\varphi_1}, \ldots, e^{i\varphi_N}),\qquad \pi\in S_{N},\qquad \varphi_1, \ldots, \varphi_N \in \mathbb{R}, $$
where  $P_{\pi}:H\to H$ is a permutation operator corresponding to a permutation $ \pi\in S_{N}$. So the unitary operator $U\in U(H)$ is unique modulo phase factors and permutations of the eigenvalues 
$$(\lambda_1, \ldots,  \lambda_N)\qquad\rightarrow \qquad(\lambda_{\pi(1)}, \ldots,  \lambda_{\pi(N)}),\qquad \pi\in S_{N}.$$ 
We can thus generically identify a density operator $\rho$ with a unitary operator $U$ and a set of eigenvalues $(\lambda_1, \ldots,  \lambda_N)$ modulo the above mentioned relations.
5) Note that there is not a preferred orthonormal basis in $H\cong\mathbb{C}^N$. Therefore it is natural to use the Haar measure $\nu$ on $U(H)\cong U(N)$. Hence, a natural class of measures $\mu$ on the space $C$ of density operators $\rho$ is,
$$\mu=\nu \times  d^N\lambda\ f(\lambda_1, \ldots,  \lambda_N) \ \delta(1-\sum_{i=1}^N \lambda_i)\ \prod_{j=1}^n \theta(\lambda_j),$$
where $\delta$ is the Dirac delta distribution; $\theta$ is the Heaviside step function;  $d^N\lambda$ is the $N$-dimensional Lebesgue measure on the space $\mathbb{R}^N$ of real eigenvalues, and $f: \mathrm{R}^n\to [0,\infty[ $ is a totally symmetric 
$$f(\lambda_{\pi(1)}, \ldots,  \lambda_{\pi(N)})=f(\lambda_1, \ldots,  \lambda_N),\qquad \pi\in S_{N},$$ 
measurable function, that is integrable 
$$\int d^N\lambda \ f(\lambda_1, \ldots,  \lambda_N) \ \delta(1-\sum_{i=1}^N \lambda_i)\prod_{j=1}^N \theta(\lambda_j) ~<~ \infty$$
on the simplex 
$$\Delta_{N-1}=\left\{\vec{\lambda}\in \mathbb{R}^N \mid \lambda_1\geq 0 , \ldots,  \lambda_N\geq 0; \sum_{i=1}^N  \lambda_i=1\right\},$$ 
which is a compact set in $\mathbb{R}^N$.
6) Note that the Haar measure $\nu$ on the compact group $U(H)$ is finite, $\nu(U(H))<\infty$, and the volume of the simplex 
$$ \mathrm{Vol}(\Delta_{N-1})=
\int d^N\lambda  \ \delta(1-\sum_{i=1}^N \lambda_i)\prod_{j=1}^N \theta(\lambda_j)~<~ \infty,$$ 
and hence the total volume $\mu(C)<\infty$ is finite. Therefore, one may normalize the measure by dividing with $\mu(C)$. 
7) A choice $f(\vec{\lambda})=1$ is just a convenient choice, but not necessary. Another natural choice, from the perspective of the Gaussian unitary ensemble (GUE), is the square of the Vandermonde determinant $f(\vec{\lambda})=\Delta^2(\vec{\lambda})$.
