Proving that $\overline{\int f \, d\mu}=\int \bar{f} \, d\bar{\mu}$ I need to show that $\overline{\int f \, d\mu}=\int \bar{f} \, d\bar{\mu}$, where $\mu$ is a complex measure. The integral with respect to a complex measure is defined by $\int f \, d\mu = \int f \frac{d\mu}{d|\mu|} \, d|\mu|$, where $|\mu|$ is the total variation of $\mu$ and $\frac{d\mu}{d|\mu|}$ is the Radon-Nykodym derivative of $\mu$ with respect to $|\mu|$. 
I don't know how to get the $\bar{\mu}$ from $\overline{\int f \, d\mu}=\Re{\int f \, d\mu}-i\Im{\int f \, d\mu}$. I'm very confused. Any help please? 
 A: Let us, for ease of notation, write
$$h := \frac{d\mu}{d\lvert\mu\rvert}.$$
Then we have
\begin{align}
\overline{\int f\,d\mu} &= \overline{\int f\cdot h\, d\lvert\mu\rvert}\\
&= \int \overline{f\cdot h}\, d\lvert\mu\rvert\\
&= \int \overline{f}\cdot \overline{h}\,d\lvert\mu\rvert,
\end{align}
since integration with respect to a positive (or real signed) measure commutes with complex conjugation.
Let us briefly verify that: Suppose $\nu$ is a positive measure, and $g = u+iv$ a complex valued integrable function. By definition, $\int g\,d\nu = \int u+iv\,d\nu = \int u\,d\nu + i\int v\,d\nu$. Since $\overline{g} = u-iv$, we thus have $\int \overline{g}\,d\nu = \int u-iv\,d\nu = \int u\,d\nu - i \int v\,d\nu = \overline{\int g\,d\nu}$.
On the other hand, we have
$$\int \overline{f} \,d\overline{\mu} = \int \overline{f} \frac{d\overline{\mu}}{d\lvert\mu\rvert}\, d\lvert\mu\rvert$$
where $\frac{d\overline{\mu}}{d\lvert\mu\rvert}$ is the Radon-Nykodým derivative of $\overline{\mu}$ with respect to $\lvert\mu\rvert$.
So it remains to see that
$$\overline{h} = \frac{d\overline{\mu}}{d\lvert\mu\rvert}.$$
Consider any $\lvert\mu\rvert$-measurable $A$. Then
$$\overline{\mu}(A) = \overline{\mu(A)} = \overline{\int_A h\,d\lvert\mu\rvert} = \int_A \overline{h}\,d\lvert\mu\rvert.\tag{$\ast$}$$
But $(\ast)$ is (since it holds for all measurable $A$) the defining property of the Radon-Nykodým derivative of $\overline{\mu}$ with respect to $\lvert\mu\rvert$. Thus we are done.
