Let $\nu$ is a complex measure then its total variation $|\nu|$ is
(1) determined by the property that if $d\nu= f d\mu$ where $\mu$ is a positive measure, then $d|\nu|=|f|d\mu$.
(2) $|\nu|=|\nu_r|+|\nu_i|=({\nu}_r^+ + {\nu}_r^-)+ ({\nu}_i^+ + {\nu}_i^-)$.
First definition is given in Real Analysis by Folland. I want to ask if the second definition is true
Thank you!
$|\nu|(E)$ is defined to be $\int_E|f|d\mu$ whenever $\nu(E)=\int_E fd\mu$. You can talk about the whether the definition is well-defined. It doesn't make sense to say that $|\nu|=|\nu_r|+|\nu_i|$ a priori. What you can say is that, if $\lambda=|\nu_r|+|\nu_i|$, then $\nu <<\lambda$. By Radon-Nikodym theorem, we have $\nu(E)=\int_Efd\lambda$ and therefore $|\nu|(E)=\int_E|f|d\lambda$.
For Welldefinedness: Suppose that $d\nu=f_1d\mu_1=f_2d\mu_2$, then take $\phi=\mu_1+\mu_2$. See that $\mu_1<<\phi$ and $\mu_2<<\phi$. Again by Radon Nikodym Theorem, we have $$\frac{d\nu}{d\phi}=\frac{d\nu}{d\mu_1}\frac{d\mu_1}{d\phi}=f_1\frac{d\mu_1}{d\phi} \implies d\nu=f_1\frac{d\mu_1}{d\phi}d\phi$$ Similar argument gives that $$ d\nu=f_2\frac{d\mu_2}{d\phi}d\phi$$
Thus, $$|f_1|d\mu_1=|f_1|\frac{d\mu_1}{d\phi}d\phi=|f_2|\frac{d\mu_2}{d\phi}d\phi=|f_2|d\mu_2$$
