Real Analysis, Folland Problem 3.3.20 Complex Measures

Solution 1:

To prove that $\nu=|\nu|$, let $E\subset X$ be any measurable set. So $$\nu(E)+\nu(E^c) = \nu(X)=|\nu|(X) = |\nu|(E)+|\nu|(E^c).$$ Considering $d\nu=f\mu$, we have $$\int_E fd\mu+\int_{E^c} fd\mu = \int_E |f|d\mu+\int_{E^c} |f|d\mu.$$ Then $$\int_E f-|f|d\mu =\int_{E^c} |f|-fd\mu.$$ Now, write $f=f_r+if_i$, where $f_r$ and $f_i$ are real functions. We have $$\int_E f_r-|f|d\mu + i \int_E f_id\mu = \int_{E^c} |f|-f_rd\mu - i \int_{E^c} f_id\mu.$$ Comparing the real part, $$\int_E f_r-|f|d\mu = \int_{E^c} |f|-f_rd\mu.$$ Noticing that $f_r\le |f|$, we have $$0\ge\int_E f_r-|f|d\mu = \int_{E^c} |f|-f_rd\mu\ge0.$$ and then $$\int_E f_r-|f|d\mu=0.$$

As it holds for every measurable set $E$, then $|f|=f_r$ $\mu$-a.e. which means that $f$ is a real and positive function almost everywhere, so $$|\nu|(E) = \int_E |f|d\mu = \int_E fd\mu = \nu(E)$$ and therefore $\nu=|\nu|$.