Is the wave function the Radon-Nikodym derivative of a complex measure? I read somewhere (latest version of a webcomic to be honest) that "superposition" means:

a complex linear combination of a $0$ state and a $1$ state... Quantum mechanics is just a certain generalization of probability... Quantum mechanics has probability amplitudes, and they can be positive or negative or even complex.

Question: Does this mean that quantum wave functions are the Radon-Nikodym derivatives of (normalized) complex measures? In particular, when the complex measure has no atoms (in the sense of measure theory), then is the wave function the density of a complex-valued measure?
In particular, that would seem to imply a certain analogy between the (time-dependent) Schrodinger equation, which would then involve the time derivative of the density of a complex-valued measure, and the Fokker-Planck equation, which involves the time derivative of the density of a (real-valued) probability measure. 
(Just checking Wikipedia's article about the Fokker-Planck equation now, and apparently there is a formal analogy between the two equations -- is the analogy what I've described above?)
Discussion: Wikipedia's article on probability amplitude mentions that the complex modulus of the probability amplitude is the Radon-Nikodym derivative of a probability measure, i.e. $$\int_X | \psi |^2 d\mu = 1\,, $$ but that is not my question -- what I am asking about is $\psi$ itself (i.e. not $|\psi|^2$), and whether $\psi$ is the "Radon-Nikodym derivative" of a complex measure.
The Wikipedia page for the Radon-Nikodym theorem says that the theorem generalizes even to complex-valued measures, and apprently all Hilbert spaces have the Radon-Nikodym property.
Also please note that when I say a "normalized" complex measure, I just mean a complex measure whose variation is a probability measure. So perhaps another way to state my question is:

Does the fact that $|\psi|^2$ is a Radon-Nikodym derivative of a probability measure imply that $\psi$ is the Radon-Nikodym derivative of a complex measure whose variation is the probability measure defined by $|\psi|^2$? Is it at least the Radon-Nikodym derivative of some complex measure?

Note: I decided to major in math, not physics, so although I know some basic Lagrangian and Hamiltonian mechanics, I am very ignorant of quantum mechanics, but do know a lot of relevant math for the subject (e.g. functional analysis), so I would prefer answers which more heavily emphasize mathematical formulation of the situation and not physical intuition.
Let me know if this should be migrated to Math.SE -- since what motivated the question was my reaction to an attempt to describe physical intuition, I imagine it might not be well-received there.
Related but different question: Probability amplitude in Layman's Terms
 A: The wavefunction $\psi$ of a nonrelativistic spinless particle is not to be interpreted as the Radon-Nikodym derivative of an absolutely continuous complex measure (i.e. a measure of the form $\text d \nu = \psi \text d^3 \mathbf x$). 
First of all, in general $\psi \notin L^1(\mathbb R ^3)$, so that mathematically $\text d \nu = \psi \text d^3 \mathbf x$ doesn't make any sense.
On the other hand, we require $\psi \in L^2(\mathbb R ^3)$, so that $\vert \psi \vert ^2$ could be interpreted as a probability density in position space. Indeed, if we know that the particle is in the state $\psi$, and we want to compute, say, the expectation value of its position $\mathbf x$, we can compute it as an ordinary mathematical expectation (here and henceforth I assume that every integral is well defined): $$\langle  \mathbf x \rangle =\intop \mathbf x \vert \psi (\mathbf x)\vert ^2\text d^3\mathbf x.$$
However, there are observables that are not of the form $F(\mathbf x)$. One of these is the momentum $\mathbf p = -i\hbar \frac{\partial}{\partial \mathbf x}$; to compute, it's expectation value, we need the whole $\psi$ (not just its modulus):$$\langle \mathbf p \rangle = \intop \psi^*(\mathbf x )( -i\hbar \frac{\partial}{\partial \mathbf x} )\psi(\mathbf x)\,\text d^3 \mathbf x.$$
We can perform a quantum Fourier transform $$\phi(\mathbf p )= \intop\frac{\text d ^3 \mathbf x}{(2\pi \hbar)^{\frac{3}{2}}}e^{-i\frac{\mathbf p \cdot \mathbf x}{\hbar}}\psi (\mathbf x)$$ 
so that the above expectation value becomes $$\langle \mathbf p \rangle = \intop \mathbf p \vert \phi(\mathbf p) \vert^2\text d^3 \mathbf p.$$
Note that, in this representation: $$\langle \mathbf x \rangle = \intop \phi ^*  (\mathbf p ) (i\hbar \frac{\partial}{\partial \mathbf p})\phi (\mathbf p ) \,\text d ^3 \mathbf p$$

Moral. The state $\psi$ is not a complex measure and the square $\vert \psi \vert ^2$ is not a probability measure. This is ultimately rooted in the fact that there are propositions, like "the particle has position $\mathbf x$ and momentum $\mathbf p$" which simply don't make sense in quantum mechanics. This is also related to quantum interference, but I think I'm going a bit off-topic. I suggest you to read Feynman, if you want to get a better idea of why a quantum state is described by a probability amplitude and not a probability distribution.
