For signed and complex measures, absolute continuity is equivalent for the measure, its Jordan or real-imaginary parts, and its total variation

Suppose first that $\nu$ is signed. If $\nu \ll \mu$ and $\mu (E)=0$, then every measurable $F\subseteq E$ is $\mu$-null, hence $\nu (F)=0$; [L1] therefore gives ${\nu }^{+}(E)=0$ and ${\nu }^{-}(E)=0$. Conversely, if ${\nu }^{+}\ll \mu$ and ${\nu }^{-}\ll \mu$, then $\nu (E)={\nu }^{+}(E)-{\nu }^{-}(E)=0$ on every $\mu$-null set, so $\nu \ll \mu$. Thus $\nu \ll \mu$ iff ${\nu }^{+}\ll \mu$ and ${\nu }^{-}\ll \mu$.

Suppose now that $\nu$ is complex. If $\nu \ll \mu$ and $\mu (E)=0$, then $\nu (E)=0$, so both $\mathrm{Re}\nu (E)$ and $\mathrm{Im}\nu (E)$ vanish; hence $\mathrm{Re}\nu \ll \mu$ and $\mathrm{Im}\nu \ll \mu$. Conversely, if both real and imaginary parts are absolutely continuous, then [L3] gives $\nu (E)=0$ on every $\mu$-null set, so $\nu \ll \mu$.

Still in the signed case, if $\nu \ll \mu$, then step 1.1 and [L1] give $\mathrm{\mid }\nu \mathrm{\mid }(E)={\nu }^{+}(E)+{\nu }^{-}(E)=0$ on every $\mu$-null set, so $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$. Conversely, if $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$, then [L2] gives $\mathrm{\mid }\nu (E)\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(E)=0$ on every $\mu$-null set, hence $\nu \ll \mu$. This completes clause 1.

For a complex measure, $\nu \ll \mu$ implies $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$ because if $\mu (E)=0$, then every piece of every measurable partition of $E$ is $\mu$-null, so every partition sum in [L2] is $0$ and therefore $\mathrm{\mid }\nu \mathrm{\mid }(E)=0$. Conversely, if $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$, then [L2] again gives $\mathrm{\mid }\nu (E)\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(E)=0$ on every $\mu$-null set, hence $\nu \ll \mu$. Together with step 1.2 this proves clause 2.

Steps 1.1 and 2.1 prove clause 1, and steps 1.2 and 2.2 prove clause 2.