Let $\mu$ be a signed measure. Then it has the Jordan decomposition $\mu=\mu^{+}-\mu{-}$. The total variation is defined as $|\mu|=\mu^{+}+\mu{-}$. I am now wondering how to compute an integral $\int f d|\mu|$? I have seen in some other questions here, that it may be defined as $$\int f d|\mu| = \int f d\mu^{+} +\int f d\mu^{-}.$$ I am wondering why this makes sense? Can it also be constructed somehow via simple functions just like the usual Lebesgue integral? Could it also be defined differently?
Are there any books or lecture notes that treat these things in detail?
Edit: So after some further research I found some solutions to Follands book here
https://math24.files.wordpress.com/2013/02/ch3-folland.pdf
There, the author constructs the above integral via simple functions from scratch as follows: Let $\phi = \sum_{i=1}^n a_i \chi_{E_i}$ be a simple function. Then: \begin{align} \int \phi d|\nu|&=\sum_{i=1}^n a_i |\nu|(E_i)=\sum_{i=1}^n a_i (\nu^+(E_i) + \nu^-(E_i))=\sum_{i=1}^n a_i \nu^+(E_i) + \sum_{i=1}^n a_i\nu^-(E_i)\\ &= \int \phi d\nu^+ + \int \phi d\nu^- \end{align}
Now the integral $\int f d|\mu|$ can be computed as follows: \begin{align} \int f d|\mu|=\sup\bigg\{\int \phi d\nu^+ + \int \phi d\nu^- \bigg| \phi \text{ simple and }\phi\leq f\bigg\} \leq \int f d\mu^{+} +\int f d\mu^{-}. \end{align}
So is it really correct that in the line there is an smaller or equal? Or can I get an equal there somehow?
You have $\nu = |\mu| = \mu^{+} + \mu^{-}$ by definition, which is a sum of mutually singular positive measures. Let $P \perp N$ be a Hahn decomposition for $\mu$, where $P$ and $N$ are disjoint positive and negative sets respectively, so $\mu^{+}(N) = 0$ and $\mu^{-}(P) = 0$.
In other word, $\chi_{P} d |\mu| = d \mu^{+}$ and $\chi_{N} d |\mu| = d \mu^{-}$.
Clearly $f\chi_{P} + f\chi_{N} = f$, and $f^{+} = f^{+} \chi_{P} + f^{+} \chi_{N}$, $f^{-} = f^{-} \chi_{P} + f^{-} \chi_{N}$.
By definition of integral, $\int f d |\mu| = \int f^{+} d |\mu| - \int f^{-} d |\mu| = (\int f^{+} \chi_{P} d|\mu| + f^{+} \chi_{N} d|\mu|) - (\int f^{-} \chi_{P} d|\mu| + f^{-} \chi_{N} d|\mu|) = (\int f^{+} d \mu^{+} + \int f^{+} d \mu^{-}) - (\int f^{-} d \mu^{+} + \int f^{-} d \mu^{-}) = \int f d \mu^{+} + \int f d \mu^{-}$.
$\textbf{Edit}$: I will outline an approach using simple functions that will work.
Assume you have proved this for simple functions.
Write $f = f^{+} - f^{-}$, then $\int f d |\mu| = \int f^{+} d |\mu| - \int f^{-} d |\mu|$.
Let $\phi_n^{+} \uparrow f^{+}$ and $\phi_n^{-} \uparrow f^{-}$ be simple functions monotonically increasing to $f^{+}$ and $f^{-}$ respectively.
Then $ \lim_{n} \int \phi_n^{+} d |\mu| = \int f^{+} d |\mu|$ by the monotone convergence theorem or by definition of integral.
So $\lim_{n} \int \phi_n^{+} d |\mu| = \lim_{n} \int \phi_n^{+} d \mu^{+} + \int \phi_n^{-} d \mu^{-} = \int f^{+} d \mu^{+} + \int f^{+} d \mu^{-}$ again by the monotone convergence theorem or the definition of integral and the hypothesis that the conclusion holds for simple functions.
Do the same thing with $\phi_n^{-}$ and $f^{-}$, and put the results together to get your desired conclusion.
$\textbf{Edit 2}$ : The below response should resolve the matter that troubles you discussed in the comments (namely on splitting the supremum).
Let $ f \geq 0$ and assume you have the desired conclusion for simple functions. Denote simple functions by $\mathcal{E}$.
Then $\int f d |\mu| = \sup \{ \int \phi d |\mu| : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} = \sup \{ \int \phi d \mu^{+} + \int \phi d \mu^{-} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \}$. Clearly $\sup \{ \int \phi d \mu^{+} + \int \phi d \mu^{-} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} \leq \sup \{ \int \phi d \mu^{+} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} + \sup \{ \int \phi d \mu^{-} : 0 \leq \phi \leq f , \phi \in \mathcal{E} \} = \int f d\mu^{+} + \int f d\mu^{-}$.
To show the other side, take $0 \leq \phi,\psi \in \mathcal{E} \leq f $ with $\int \phi d \mu^{+} $ and $\int \psi d \mu^{-}$ both within $\frac{\epsilon}{2}$ away from $\int f d \mu^{+}$ and $\int f d \mu^{-}$ respectively.
Then consider $0 \leq \phi \chi_{P} + \psi \chi_{N} \in \mathcal{E} \leq f $ where $P,N$ are as in my post (they form a Hahn decomposition for $\mu$).
Clearly $\int (\phi \chi_{P} + \psi \chi_{N}) d |\mu| = \int \phi d \mu^{+} + \int \psi d \mu^{-} $, and since $(\phi \chi_{P} + \psi \chi_{N})$ is simple, $\int (\phi \chi_{P} + \psi \chi_{N}) d |\mu| \leq \int f d |\mu|$.
But since $\int \phi d \mu^{+} $ and $\int \psi d \mu^{-}$ both within $\frac{\epsilon}{2}$ away from $\int f d \mu^{+}$ and $\int f d \mu^{-}$ respectively, we have that $\int f d\mu^{+} + \int f d\mu^{-} - \epsilon \leq \int f d|\mu|$.
Let $\epsilon \downarrow 0$ to recover $\int f d\mu^{+} + \int f d\mu^{-} \leq \int f d |\mu|$.
This gives the other side.