I'm wondering if the following statement holds:
Let $(X, \mathfrak{B})$ be a measurable space, and $\mu, \nu$ be complex measures. Assume $\nu << \mu$. Is there a complex valued measurable function $f \colon X\to \mathbb{C}$, which satisfies $$\forall B\in \mathfrak{B}; \int_B f d\mu = \nu(B)$$
If so, I want to call $f = \frac{d\nu}{d\mu}$ the Radon-Nikodym derivative. My textbook says it holds (only) when $\mu$ is $\sigma$-finite positive measure. Any help will be appriciated.
P.S. Here complex measure means the function $\mu \colon \mathfrak{A}\to \mathbb{C}$ which satisfies $\mu(\varnothing)=0$ and $\sigma$-additivity in $\mathbb{C}$ (with "Euclidean topological convergence").
Recall that $|\mu|$, the total variation of $\mu$ is a finite measure, $\mu << |\mu|$ and $$|\frac{d\mu}{d|\mu|}| = 1 \text{(i.e. invertible in $\mathbb{C}$, $|\mu|$-a.e.)}$$ $$\int_A f d\mu = \int_A f \frac{d\mu}{d|\mu|}d|\mu|$$ holds when $f$ is $|\mu|$-integrable.
Now let $$g=\frac{d\nu}{d|\mu|}(\frac{d\mu}{d|\mu|})^{-1}$$ and calculate the integration of $g$. We see that $$\int_A gd\mu = \int_A \frac{d\nu}{d|\mu|}(\frac{d\mu}{d|\mu|})^{-1}\frac{d\mu}{d|\mu|}d|\mu| = \nu(A)$$ so $g$ is the Radon-Nikodym derivative.
Note that $g$ is unique up to $|\mu|$-a.e., since if $g$ and $h$ are both Radon-Nikodym derivative, $\int_A (g-f)\frac{d\mu}{d|\mu|}d|\mu| = 0$ for all $A$, then $(g-f)\frac{d\mu}{d|\mu|} = 0$, so $g-f=0$, $|\mu|$-a.e. .