Complex Measures: Variation I highly doubt a proof of mine for complex measures...

First, a complex measure can be decomposed into positive measures:
$$\mu=(\Re\mu)_+-(\Re\mu)_-+i(\Im\mu)_+-i(\Im\mu)_-=:\sum_{\alpha=0\ldots3}i^\alpha\mu_\alpha$$

Next, each of these has a Radon-Nikodym derivative:
$$\mu_\alpha(E)=\int_Eu_\alpha d|\mu|$$
giving rise to the complex Radon-Nikodym derivative:
$$\mu(E)=\sum_{\alpha=0\ldots3}i^\alpha\mu_\alpha=\int_E\sum_{\alpha=0\ldots3}i^\alpha u_\alpha d|\mu|=:\int_Eud|\mu|$$

Finally, the Radon-Nikodym derivative decomposes too:
$$u=(\Re u)_+-(\Re u)_-+i(\Im u)_+-i(\Im u)_-=:\sum_{\alpha=0\ldots3}i^\alpha u'_\alpha$$


  How to check that a fortiori: $u'_\alpha=u_\alpha$


(Put in words, wether the variation of measure agrees with the variation of derivative.)
Meanwhile, I got it by some lemma...

It respects real and imaginary variation by linearity:
$$\Re\int h\mathrm{d}\lambda=\int\Re h\mathrm{d}\lambda$$
$$\Im\int h\mathrm{d}\lambda=\int\Im h\mathrm{d}\lambda$$
as well as positive and negative variation by variation of integrals:
$$\left(\int h\mathrm{d}\lambda\right)_\pm=\int h_\pm\mathrm{d}\lambda$$
(For a rigorous check see: Integral Measures: Variation)

Therefore one has for the real and similarly imaginary part:
$$\int_Eu_\alpha\mathrm{d}|\mu|=(\Re\mu)_\pm(E)=\int_E(\Re u)_\pm\mathrm{d}|\mu|=\int_Eu'_\alpha\mathrm{d}|\mu|\quad(\alpha=0,2)$$

Concluding that they agree pointwise: $u'_\alpha=u_\alpha\mod|\mu|$
Meanwhile, I got it by some lemma...

It respects real and imaginary variation by linearity:
$$\Re\int h\mathrm{d}\lambda=\int\Re h\mathrm{d}\lambda$$
$$\Im\int h\mathrm{d}\lambda=\int\Im h\mathrm{d}\lambda$$
as well as positive and negative variation by variation of integrals:
$$\left(\int h\mathrm{d}\lambda\right)_\pm=\int h_\pm\mathrm{d}\lambda$$
(For a rigorous check see: Integral Measures: Variation)

Therefore one has for the real and similarly imaginary part:
$$\int_Eu_\alpha\mathrm{d}|\mu|=(\Re\mu)_\pm(E)=\int_E(\Re u)_\pm\mathrm{d}|\mu|=\int_Eu'_\alpha\mathrm{d}|\mu|\quad(\alpha=0,2)$$

Concluding that they agree pointwise: $u'_\alpha=u_\alpha\mod|\mu|$