Inequality between integrals with complex measures.

Question

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*}

Answer 1

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.

Answer 2

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*}