Suppose that $\nu$ is a complex measure on $(X,\mathcal M)$ and $\nu(X)=|\nu|(X)$. Then show that $\nu=|\nu|$
My Attempt:
Suppose that $d\nu=fd\mu$, where $\mu=|\nu_r|+|\nu_i|$. Then $d|\nu|=|f|d\mu$. For every measurable set $E$, we have $$\int_{E}f d\mu + \int_{E^c}f d\mu =\int_{E}|f| d\mu + \int_{E^c}|f| d\mu$$
Clearly $$\int_X Im(f) d\mu=0$$.
Now $$ \int_E (f-|f|) d\mu=\int_{E^c} (|f|-f) d\mu$$ Thus, writing $f=\text{Re}(f)+i\text{ Im}(f)$, we have $$\int_E(\text{Re(f)}-|f|)d\mu + i \int_E \text{Im}(f) d\mu=\int_{E^c}(|f|-\text{Re}(f)) d\mu + i \int_{E^c} \text{-Im}(f) d\mu$$
Comparing the real and imaginary parts we have $$0 \ge \int_E(\text{Re(f)}-|f|)d\mu = \int_{E^c}(|f|-\text{Re}(f)) d\mu \ge 0$$
Thus, we have $$ \int_E(\text{Re(f)}-|f|)d\mu=0=\int_{E^c}(|f|-\text{Re}(f)) d\mu$$ So for any measurable set $E$, we have $\sqrt{{\text{Re}(\nu(E))}^2+{\text{Im}(\nu(E))}^2} =|\nu(E)| \le |\nu|(E)=\text{Re}(\nu(E))$ which implies that $\text{Im}(\nu(E))=0$. Thus, $\nu(E)=\text{Re}(\nu(E))=|\nu|(E)$.
Is this alright?? Thanks for the help!!