Suppose that $\nu$ is a complex measure. Show that $L^1(\nu)=L^1(|\nu|)$.
This is my attempt to prove this assertion. To begin with, suppose that $\mu=|\nu_r|+|\nu_i|$. Then $\nu_r <<\mu$ and $\nu_i <<\mu$. So by Radon Nikodym Theorem, $$d\nu_r=\frac{d\nu_r}{d\mu}d\mu,d\nu_i=\frac{d\nu_i}{d\mu}d\mu$$ Therefore, $$d\nu=\left(\frac{d\nu_r}{d\mu}+i\frac{d\nu_i}{d\mu}\right)d\mu$$ By definition, we have $$d|\nu|=\left|\frac{d\nu_r}{d\mu}+i\frac{d\nu_i}{d\mu}\right|d\mu$$
Moreover, $\nu_r <<|\nu|,\nu_i <<|\nu|$, again using Radon Nikodym Theorem give us that $$d\nu_r=\frac{d\nu_r}{d|\nu|}d|\nu|,d\nu_i=\frac{d\nu_i}{d|\nu|}d|\nu| ,\text{ hence } d\nu=\left(\frac{d\nu_r}{d|\nu|}+i\frac{d\nu_i}{d|\nu|}\right)d|\nu| \text{ and } d|\nu|=\left|\frac{d\nu_r}{d|\nu|}+i\frac{d\nu_i}{d|\nu|}\right|d|\nu|=\left|\frac{d\nu_r}{d|\nu|}+i\frac{d\nu_i}{d|\nu|}\right|\left|\frac{d\nu_r}{d\mu}+i\frac{d\nu_i}{d\mu}\right|d\mu$$
By uniqueness of Radon Nikodym derivative we must have that $$\left|\frac{d\nu_r}{d\mu}+i\frac{d\nu_i}{d\mu}\right|=\left|\frac{d\nu_r}{d|\nu|}+i\frac{d\nu_i}{d|\nu|}\right|\left|\frac{d\nu_r}{d\mu}+i\frac{d\nu_i}{d\mu}\right| \text{ $\mu$-a.e and hence $|\nu|$-a.e. } $$ Let $h=\frac{d\nu_r}{d\mu}+i\frac{d\nu_i}{d\mu},E=\{x :h(x)=0\}$. $$ \text{ Then }|\nu|(E)=\int_E |h| d\mu=0 \text{ gives us that } \left|\frac{d\nu_r}{d|\nu|}+i\frac{d\nu_i}{d|\nu|}\right|=1 \text{ $|\nu|$-a.e from whence it follows that} $$
$$\left|\frac{d\nu_r}{d|\nu|}\right| \le 1,\left|\frac{d\nu_i}{d|\nu|}\right| \le 1 \text{ and } 1=\left|\frac{d\nu_r}{d|\nu|}+i\frac{d\nu_i}{d|\nu|}\right|\le \left|\frac{d\nu_r}{d|\nu|}\right|+\left|\frac{d\nu_i}{d|\nu|}\right| $$ Now suppose that $f \in L^1(|\nu|)$. Then $$ \int|f| d|\nu_r|=\int |f| \left|\frac{d\nu_r}{d|\nu|}\right|d|\nu| \le \int |f| d|\nu| \text{ and } \int|f| d|\nu_i|=\int |f| \left|\frac{d\nu_i}{d|\nu|}\right|d|\nu| \le \int |f| d|\nu|$$ give us that $f \in L^1(|\nu_r|)\cap L^1(|\nu_i|)$ which is equivalent to $f$ being in $L^1(\nu)$.
On the other hand, if $f$ were in $L^1(\nu)=L^1(\nu_r) \cap L^1(\nu_i)=L^1(|\nu_r|) \cap L^1(|\nu_i|)$, then we have $$\int |f| d|\nu| \le \int |f|\left(\left|\frac{d\nu_r}{d|\nu|}\right|+\left|\frac{d\nu_i}{d|\nu|}\right|\right) d|\nu|=\int |f| d|\nu_r|+\int |f| d|\nu_i| \lt \infty$$
It took me a lot of time to come up with the solution. It would be great help if someone could verify it.
Thanks for the help!!