My current task at hand is to prove that on a measurable space $(X,\mathcal A)$, for a complex measure $\nu$ and a $\sigma$-finite measure $\mu$ with $\nu\ll\mu$, there holds $$ \frac{d\left\lvert \nu\right\rvert}{d \mu}=\left\lvert\frac{d\nu}{d\mu}\right\rvert \quad \mu\text{-a.e.}, $$ where $\left\lvert \nu\right\rvert$ denotes the total variation of $\nu$. The inequality $\frac{d\left\lvert \nu\right\rvert}{d \mu}\leq\left\lvert\frac{d\nu}{d\mu}\right\rvert$ is quite straightforward for it is enough to prove that $$ \int_X\frac{d\left\lvert \nu\right\rvert}{d \mu}\,\mathrm d\mu \leq \int_X \left\lvert\frac{d\nu}{d\mu}\right\rvert\,\mathrm d\mu $$ and since LHS is equal to $\lvert\nu\rvert(X)$, we use the definition of the total variation, i.e. pick an arbitrary measurable finite partition $(A_i)_{i=1}^n$ of $X$ and show that $$\sum_i\left\lvert\nu(A_i)\right\rvert=\sum_i \left\lvert\int_{A_i} \frac{d\nu}{d\mu}\,\mathrm d\mu\right\rvert\leq\sum_i\int_{A_i}\left\lvert\frac{d\nu}{d\mu}\right\rvert\,\mathrm d\mu =\int_X \left\lvert\frac{d\nu}{d\mu}\right\rvert\,\mathrm d\mu.$$
The only remaining question I have up to this point is whether it's enough to consider $X$ or does this need to hold $\forall E\in\mathcal A$.
However, the inverse inequality is more tricky and I've problems with even beginning to tackle the problem. I've looked in some textbooks (e.g. Folland), but they usually state this as an exercise.