there is a measurable function $f$ on $X$ such that $|{f(x)}|=1$ for a.a $x \in X$ and $\nu(E)=\int_Efd|{\nu}|$ for any $E \in \mathfrak{M}$ any hints on this problem: 
 Let $\nu$ be a finite signed measure on a measure space $(X, \mathfrak{M})$ and let $|{\nu}|$ be its total variation, prove that there is a measurable function $f$ on $X$ such that $|{f(x)}|=1$ for a.a $x \in X$ and $\nu(E)=\int_Efd|{\nu}|$ for any $E \in \mathfrak{M}$
I know that we need to use the Radon-Nikodym theorem, but I don't know how to start!
Any help is greatly appreciated. Thanx in advance. 
 A: Since $\nu$ is absolutely continuous with respect to $|\nu|$, there exists by (Radon-Nikodym) $f\in L^{1}(X,\mathfrak{M},|\nu|)$ such that $d\nu = fd|\nu|$.  Equivalently, $\nu(E) = \int_{E}fd|\nu|$ for all $E\in\mathfrak{M}$.
To verify the other condition, suppose there were an set $U$ of non-zero measure such that $|f(x)|\neq 1$ for all $x\in U$.  
Without loss of generality (use measurability of $f$ here), assume that either $|f(x)| > 1$ on $U$ or $|f(x)| < 1$ on $U$.
Then write $U = U_{1}\cup U_{2}$, where $U_{1}$ is a positive set and $U_{2}$ is a negative set.  One of these must have positive $\nu$-measure $0$.
Assume here that $\nu(U_{1})\neq 0$ (the argument should be similar for when $\nu(U_{2}) \neq 0$.
Compute $|\nu|(U_{1})$ by definition and derive a contradiction.
A: Is it not enough to take $A, B \subset X$ such that $\nu^+(E) = \nu(E\cap A)$ and $\nu^-(E) = -\nu(E\cap B)$ and then to define $f(x) = 1$ if $x\in A$ and $f(x) = -1$ if $x \in B$?
We also need that $A \cap B = \emptyset$ (of measure 0 might suffice) and $X = A \cup B$ (less a set of measure 0 might suffice). These are guaranteed by Hahn Decomp. 
