At the start of the proof for Theorem 3.22, Folland says that $dv=d\lambda + fdm$ implies $d|v|=d|\lambda| + |f|dm$. I get why this is the case for positive and signed measures, but I'm not sure how to prove this for the complex measure case. Could anyone guide me in the right direction?
This is my attempt based on Eric Towers', comment:
$dv=d\lambda + fdm$ is such that $\lambda \perp m$ (ie. There exists a measurable set $E$ such that $\lambda(E)=m(E^c)=0$). We can decompose $v$ in this way: $\forall A$ measurable, $v(A)=v(A\cap E) + v(A\cap E^c)$. Moreover, $v(A\cap E)=\int_{A\cap E}fdm$ and $v(A\cap E^c)=\lambda(A\cap E^c)$. Note that $v_E(A)=v(A\cap E)$ and $v_{E^c}(A)=v(A\cap E^c)$ are measures. Since there exists a positive measure $u$ where $v<<u$ as in Folland chapter 3.3, we can represent $v=\int g du$ for some $g$. Then $v(A\cap E)=\int_{A\cap E} g du=\int_{A\cap E}fdm=v_E(A)$. Similarly, $v(A\cap E^c)=\int_{A\cap E^c} g du=\lambda(A\cap E^c)=\int_{A\cap E^c} \frac{d\lambda}{d|\lambda|}d|\lambda|=v_{E^c}(A)$. Therefore, $|v|(A\cap E)=\int_{A\cap E}|g|du=|v_{E}|(A)=\int_{A\cap E}|f|dm$ and similarly, $|v|(A\cap E^c)=\int_{A\cap E^c} |g| du=\int_{A\cap E^c} |\frac{d\lambda}{d|\lambda|}|d|\lambda|=|v_{E^c}|(A)$. But from Folland chapter 3.3, $|\frac{d\lambda}{|d\lambda|}|=1$ a.e. Thus, $|v|(A)=|v|(A\cap E) + |v|(A\cap E^c)=\int_{A\cap E^c}1d|\lambda| + \int_{A\cap E} |f|dm$. Recalling from above that $\lambda(E)=m(E^c)=0$, the result follows.