This is a question from Richard Bass, Real Analysis for Graduate Students.
If $\mu$ is a signed measure on $(X,A)$ and $|\mu|$ is the total variation measure, prove that there exists a real-valued function $f$ that is measurable with respect to $A$ such that $|f| = 1 \text{ a.e.}$ with respect to $\mu$ and $d\mu = f d|\mu|$.
This is what I have so far:
"Since $\mu$ is a signed measure, by the Hahn Decomposition Theorem, we know that $X=P \cup N$, where $P$ is a positive set and $N$ is a negative set, such that they are disjoint. Now define $f=1$ on $P$, and $f=-1$ on $N$. Then, it is clear that $|f|=1 \text{ a.e.}$"
However, I am having trouble proving that $d\mu = f d|\mu|$. Any help would be greatly appreciated.
