TITLE: Need help to understand Lebesgue-Radon-Nikodym Theorem
QUESTION [8 upvotes]: On Folland's Real Analysis book page $90$, the Lebesgue-Radon-Nikodym Theorem is given as
Let $\nu$ be a $\sigma$-finite signed measure and $\mu$ a $\sigma$-finite positive measure on $(X,\mathcal{M})$. There exists unique $\sigma$-finite signed measure $\lambda,\rho$ on $(X,\mathcal{M})$ such that $\lambda\perp \mu$, $\rho\ll\mu$, and $\nu=\lambda+\rho$. Moreover, there is an extended $\mu$-integrable function $f: X\to\mathbb{R}$ such that $d\rho=fd\mu$, and any two functions are equal $\mu$-a.e.
As my understanding, this is a theorem of existence but seems not providing a method to construct $\lambda,\rho,f$. Probably I don't quite get the proof of the theorem.
To clarify my question, can anyone provide a method or recipe to compute the absolutely continuous part $\rho$ of a measure $\mu$ on the real line
? I think this will definitely help me to understand the theorem in practice.

REPLY [6 votes]: The proof of this theorem in Folland provides quite some insight into the construction of 
$$
  \mathrm d\nu = f\mathrm d\mu + \mathrm d\lambda
$$
where $\lambda\perp \mu$. First of all, the function $f$ is constructed using the class 
$$
  \mathscr F = \left\{f:X\to [0,\infty]:\int_E f\mathrm d\mu\leq \nu(E)\text{ for all }E\in \mathscr M\right\}.
$$
Secondly, one defines $a := \sup\{\int _X f\mathrm d\mu|f\in \mathscr F\}$ and as it follows from the definition of $a$, there exists a sequence $(f_n)_{n\in \Bbb N} \subset \mathscr F$ such that 
$$
  \int_X f_n\mathrm d\mu \to a
$$
The theorem claims that $f := \sup_nf_n$ is indeed such a function that 
$$
  \lambda := \int f\mathrm d\mu - \nu\perp\mu.
$$
As a result, the existence of the function $f$ is proved by constructing it. However, I often saw the statement that "there is no known constructive proof of Radon-Nikodym Theorem", which I guess means that in practice in general it's hard to get the explicit expression of $f$ (provided one can define what does the "explicit expression" mean).