(complex measures) $d\nu=d\lambda +f\,dm \Rightarrow d|\nu|=d|\lambda| +|f|\,dm$ and $f\in L^1(\nu)\Rightarrow f\in L^1(|\nu|)$ Two simple exercises on complex measures I don't know how to solve, from Folland's Real Analysis. I find it difficult to manipulate the definitions in computations. For context, if $\nu$ is a complex measure on $(X,\mathcal{M})$ then $|\nu|$ is defined as $d|\nu|=|g|\,d\mu$ where positive measure $\mu$ and $g\in L^1(\mu)$ are such that $\nu \ll \mu$ and $\nu = g\,d\mu$ (and in the textbook it's shown that this definition does not depend on the choice of $\mu$).

1) Let $\nu$ be a reguar complex Borel measure on $\mathbb{R}^n$, $m$ denote   Lebesgue measure, and
  let $d\nu = d\lambda + f\,dm$ be its Lebesgue-Radon-Nikodym
  representation. Then ("it is easily verified")
  $d|\nu|=d|\lambda|+|f|\,dm$.
2) If $f\in L^1(\nu)$ then $f\in L^1(|\nu|)$. I was able to prove the
  reverse implication, but not this one.

The definition of $|\nu|$ in the case of real signed measures (as $|\nu|=\nu^+ + \nu^-$) makes sense to me but I can't see intuitively into the definion of $|\nu|$ for complex measures as given in the textbook. Do you have any advice on this?
Thank you!
 A: 1) You may first try to prove the easy inequality: $|\nu_1+\nu_2|\le|\nu_1|+|\nu_2|$ (See Proposition 3.14 in Folland's book for a proof). Since in our case the two measures are mutually singular (this means, in Folland's proof, that the supports of $f_1$ and $f_2$ are disjoint), the equality holds.
2) It is easy to prove this when $\nu$ is a real-valued measure. In the complex measure case, we may write $\nu=\nu_r+i\,\nu_i$, where $\nu_r$ and $\nu_i$ are real-valued measures. Define $\mu=|\nu_r|+|\nu_i|$. Then $d\nu_r=h\,d\mu$ and $d\nu_i=k\,d\mu$ for some real-valued functions $h,k\in L^1(\mu)$. We also have $\nu\ll\mu$. So $\frac{d\nu}{d\mu}=h+i\,k$. Let $g=h+i\,k$. Let's use $L^1(d\nu)$ to denote $L^1(\nu)$. Then $$L^1(d\nu)=L^1(h\,d\mu)\cap L^1(k\,d\mu)=L^1(|h|\,d\mu)\cap L^1(|k|\,d\mu)=L^1(|g|d\mu)=L^1(d|\nu|)$$ 
The first and the last equalities are just definitions. The second equality is the real measure case, and the third one is a result of the trivial inequality $\max\{|h|, |k|\}\le|g|\le|h|+|k|$.
A: Remark

By construction it is guaranteed:
  $$\lambda\geq0:\quad|\mu(E)|\leq\lambda(E)\implies|\mu|(E)\leq\lambda(E)$$
  (This minimality property is of no use here!)

But it has a consequence worth mentioning: $\mu=\nu+\nu'\implies|\mu|\leq|\nu|+|\nu'|$
Setting
Given the Lebesgue measure $\lambda$.
Consider a complex measure $\mu$.
Part 1
By Lebesgue-Radon-Nikodym one has:
$$\mathrm{d}\mu=h\mathrm{d}\lambda+\mathrm{d}\mu_\perp$$
By singularity it holds:
$$\mathrm{d}\mu\restriction_S=h\mathrm{d}\lambda\restriction_S\quad\mathrm{d}\mu\restriction_{S^\complement}=\mathrm{d}\mu_\perp\restriction_{S^\complement}$$
So one gets:
$$\mathrm{d}|\mu|\restriction_S=|h|\mathrm{d}\lambda\restriction_S\quad\mathrm{d}|\mu|\restriction_{S^\complement}=\mathrm{d}|\mu_\perp|\restriction_{S^\complement}$$
Concluding:
$$\mathrm{d}|\mu|=|h|\mathrm{d}\lambda+\mathrm{d}|\mu_\perp|$$
Part 2
One has by definition due to Rudin:
$$f\in\mathcal{L}(\mu):\iff f\in\mathcal{L}(|\mu|)$$
For the equivalence of definitions see: Complex Measures: Integrability
