Inequality between integrals with complex measures. I am given $\mu:\mathfrak{A}\to\mathbb{C}$ a complex measure. 
I am asked to show that there exists a function  $g:X\to\mathbb{C}$ such that $|g|=1$ and $\mu = g \odot|\mu|$.
I think that it's easy to show the first part but I'm stuck with the following:
If $f\in\mathscr L_1(X,\mu)$ then
$f\in\mathscr L_1(X, |\mu |)$ and that
   \begin{align*}
      \left| \int_X f\,d\mu \right|
      \le  \int_X |f|\,d |\mu |,
   \end{align*}
   where   \begin{align*}
      \int_X f\,d\mu 
      :=
      \int_X f\;d\, Re(\mu)^+
      - \int_X f\;d\, Re(\mu)^-
      + i \int_X f\;d\, Im(\mu)^+
      -i \int_X f\;d\, Im(\mu)^-.
   \end{align*}
 A: Let $\int_X f \, d\mu =re^{it}$ with $r \geq  0$ and $t$ real.Then $r=\int_X e^{-it} f \, d\mu=\int_X e^{-it} fg \, d|\mu|$ from linearity of the integral. Hence $r =\int_X \Re (e^{-it} fg)  \, d|\mu|$ and this gives $ r \leq \int_X |e^{-it} fg| \, d|\mu|=\int_X | f| \, d|\mu|$. Since the left side is $|\int_X f \, d\mu|$ we are done.
A: For $f=\chi_{A}$, then 
\begin{align*}
\int_{X}\chi_{A}d\mu=(\text{Re}\mu)^{+}(A)-(\text{Re}\mu)^{-}(A)+i(\text{Im}\mu)^{+}(A)-i(\text{Im}\mu)^{-}(A),
\end{align*}
then 
\begin{align*}
\left|\int_{X}\chi_{A}d\mu\right|&=\left(\left((\text{Re}\mu)^{+}(A)-(\text{Re}\mu)^{-}(A)\right)^{2}+\left((\text{Im}\mu)^{+}(A)-(\text{Im}\mu)^{-}(A)\right)^{2}\right)^{1/2}\\
&=\left(\left((\text{Re}\mu)(A)\right)^{2}+\left((\text{Im}\mu)(A)\right)^{2}\right)^{1/2}\\
&=|\mu(A)|\\
&\leq|\mu|(A)\\
&=\int_{X}\chi_{A}d|\mu|,
\end{align*}
the rest use the density of simple functions in $L^{1}$.
Note that 
\begin{align*}
&|\mu|(A)\\
&=\sup\left\{\sum_{k=1}^{n}|\mu(E_{k})|: E_{1},...,E_{n}~\text{are pairwise disjoint sets in }\mathfrak{A}~\text{such that } A=\bigcup_{k=1}^{n}E_{k}\right\}.
\end{align*}
