Lebesgue-Radon-Nikodym Theorem shows that:
If $\mathbb{M}$ is a $\sigma$-algbra on the set X. $\mu,\lambda$ are a $\sigma$-finite positive measure and $\sigma$-finite signed measure on $\mathbb{M}$ respectively. then $\lambda$ has a decomposition.
I wander why we do not consider the case when $\mu$ is not positive.
I think maybe it is difficult to define integral with signed measure or complex-valued measure,so we just do not take the case when $\mu$ is not positive. But I am not clear about this.
Any hint will be greatly appreciated!
Consider the total variation $|\mu|$ of $\mu$. Then $|\mu|$ is $\sigma$-finite and positive. Moreover $\mu \ll |\mu|$. Then, you get a function $\mu' \in L^1(|\mu|)$ (in fact $|\mu'|=1$ $|\mu|$-a.e.), such that $$\mu(E) = \int_E \mu' \, \mathrm{d}|\mu|.$$
Now, you can apply the Lebesgue-Radon-Nikodým theorem to $\lambda$ and $|\mu|$. In particular, there is $h \in L^1(|\mu|)$, such that $$\lambda_a(E) = \int_E h \, \mathrm{d}|\mu|.$$ This could also be written as $$\lambda_a(E) = \int_E \frac{h}{\mu'} \, \mathrm{d}\mu$$ if the integration w.r.t. the complex measure $\mu$ is defined accordingly. In particular, $h / \mu'$ could be considered a Radon-Nikodým derivative of $\lambda_a$ w.r.t. $\mu$.