Real Analysis, Folland Problem 3.3.20 Complex Measures Related definitions - A complex measure on a measurable space $(X,M)$ is a map $\nu: M\rightarrow\mathbb{C}$ such that 
i.) $\nu(\emptyset) = 0;$
ii.) if $\{E_j\}$ is a sequence of disjoint sets in $M$, then $\nu\left(\bigcup_{1}^{\infty}E_j\right) = \sum_{1}^{\infty}\nu(E_j)$ where the series converges absolutely.
Problem 3.20 - If $\nu$ is a complex measure on $(X,M)$ and $\nu(X) = |\nu|(X)$, then $\nu = |\nu|$
Attempted proof: The total variation of $\nu$ is defined as wherever $d\nu = f d\mu$, $d|\nu| = |f|d\mu$. We also have $\nu(X) = |\nu|(X)$ By equality, $$\int_X f d\mu = \int_X |f|d\mu$$ hence by proposition 2.23b we have $$\int_X |f - f|d\mu = 0$$ thus $f = |f|\mu$-a.e. hence $\nu = |\nu|$.
 A: 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 $\nu=|\nu|$.
A: $$\nu(X)=|\nu|(X) \implies |\nu(X)|=|\nu|(X)$$
$$|\nu|(E)+|\nu|(E^c) = |\nu(E)+\nu(E^c)|\leq |\nu(E)|+|\nu(E^c)|$$
Since $|\nu|$ is finite
$$0\leq |\nu|(E^c)-|\nu(E^c)| \leq |\nu(E)| - |\nu|(E) \leq 0$$
From here, deduce that $|\nu| = \nu = \nu_{\text{real}}^{+}$.
Alternatively
Note that $d\nu = e^{i\theta}d|\nu|$ for some measurable function $\theta$. Then
\begin{align}
&\nu(X) = |\nu|(X)\\
&\implies \int (1-e^{i\theta})d|\nu| = 0\\
&\implies \int (1-\cos\theta)d|\nu| - i\int \sin\theta d|\nu| = 0\\
&\implies \cos\theta = 1 \text{ and } \sin\theta = 0 \text{ a.e.}
\end{align}
where the first implication follows because $|\nu|(X) < \infty$.
A: According to Folland, the total variation of $\nu$ is defined as whenever $d\nu = f\,d\mu, d|\nu| = |f|d\mu$. By equality,
$$ \int_S f\,d\mu = \int_S |f|d\mu$$
Then deduce that $f = |f|$ $\mu$-a.e. and the measures coincide.
