Complex Measures: Integrability Problem
On the one hand, a complex measure decomposes into:
$$\mu=\Re_+\mu-\Re_-\mu+i\Im_+\mu-i\Im_-\mu=:\sum_{\alpha=0\ldots3}i^\alpha\mu_\alpha$$
This gives rise to the integrability condition:
$$f\in L(\mu)\iff f\in L(\mu_\alpha)\quad(\alpha=1,\ldots,3)$$
On the other hand, a complex measure admits a derivative:
$$\mu(E)=\int_Eu\mathrm{d}|\mu|\quad(|u|=1)$$
This gives rise to the integrability condition:
$$f\in L(\mu):\iff fu\in L(|\mu|)\iff f\in L(|\mu|)$$

So the question arises wether these approaches coincide:
  $$f\in L(|\mu|)\iff f\in L(\mu_\alpha)\quad(\alpha=1,\ldots,3)$$

Attempt
So far by construction it is:
$$\mu_\alpha(E)=\int_Eu_\alpha\mathrm{d}|\mu|\quad(0\leq u_\alpha\leq1)$$
which gives for positive functions:
$$\int|f|\mathrm{d}\mu_\alpha=\int|f|u_\alpha\mathrm{d}|\mu|\leq\int |f|\mathrm{d}|\mu|$$
That proves the inclusion:
$$f\in L(|\mu|)\implies f\in L(\mu_\alpha)\quad(\alpha=1,\ldots,3)$$
But what about the converse?
Related
For a formal treatment see: Complex Measures: Integration
For a related recapitulation see: Complex Measures: Variation
For a similar problem see: Complex Functions: Integrability 
For a general problem see: Radon-Nikodym: Integrability?
 A: If we have a signed measure $\mu$ on a measurable space $(\Omega,\Sigma)$, then $\mu$ may be allowed to assume values of $\pm \infty$, but not both. A complex measure $\mu$ is not allowed to assume any type of infinite value. For either type of measure $\mu$ which is not allowed to assume infinite values, the variation measure $|\mu|$ is a finite measure, where
$$
             |\mu|E = \sup_{\pi}\sum_{S \in \pi}|\mu S|
$$
is taken over all finite partitions $\pi$ of measurable subsets of $E$. Because we want to deal with a complex measure, then we assume without loss of generality that all measures related to this discussion are finite.
If $\mu$ is a signed measure, then we can decompose $\mu$ into positive meausres $\mu_{+}$, $\mu_{-}$ according to the (Hahn) Jordan decomposition. This decomposition is the same as a variation measure approach:
$$
                    |\mu|=\mu_{+}+\mu_{-},\;\;\; \mu = \mu_{+}-\mu_{-}.
$$
This extends to the complex case as well
$$
                \mu = (\Re\mu)_{+}-(\Re\mu)_{-}+i\{(\Im\mu)_{+}-(\Im\mu)_{-}\}
$$
with
$$
            |\mu| \le |\Re\mu| + |\Im\mu| = |(\Re\mu)_{-}|+|(\Re\mu)_{+}|+|(\Im\mu)_{-}|+|(\Im\mu)_{+}|.
$$
But we also have
$$
                       |\Re\mu| \le |\mu|,\;\;\; |\Im\mu| \le |\mu|.
$$
So integrability of a function $f$ with respect to $|\mu|$ is equivalent to simultaneous integrability with respect to $|(\Re\mu)_{-}|,|(\Re\mu)_{+}|,|(\Im\mu)_{-}|,|(\Im\mu)_{+}|$.
A: 
In fact, the very very heart of the whole story here are the two inequalities:
  $$|\mu(E)|\leq|\mu|(E)<\infty\quad(E\in\Sigma)$$

For positive measures given as derivative:
$$\kappa(E)=\int_Eh\mathrm{d}\lambda\quad(h\geq0)$$
the positive integrals agree:
$$\int f\mathrm{d}\kappa=\int fh\mathrm{d}\lambda\quad(f\geq0)$$
(Note, these are allowed to be infinite!)
Now, consider the representation by the Radon-Nikodym derivative:
$$\mu_\alpha=\int_Eu_\alpha\mathrm{d}|\mu|\quad(u_\alpha\geq0,|u|=1)$$
Exploiting the bounds:
$$u_\alpha\leq|u|$$
$$|u|\leq\sum_{\alpha=0\ldots3}u_\alpha$$
one has the estimates:
$$\int|f|\mathrm{d}\mu_\alpha=\int|f|\cdot u_\alpha\mathrm{d}|\mu|\leq\int|f|\cdot|u|\mathrm{d}|\mu|=\int|f|\mathrm{d}|\mu|$$
$$\int|f|\mathrm{d}|\mu|=\int|f|\cdot|u|\mathrm{d}|\mu|\leq\sum_{\alpha=0\ldots3}\int|f|\cdot u_\alpha\mathrm{d}|\mu|=\sum_{\alpha=0\ldots3}\int|f|\mathrm{d}\mu_\alpha$$
and therefore:
$$f\in L(|\mu|)\iff f\in L(\mu_\alpha)\quad(\alpha=0\ldots3)$$
