A finite complex measure absolutely continuous with respect to a sigma-finite positive measure has an integrable complex density

By [L1] and [L2], the signed measures $\mathrm{Re}\nu$ and $\mathrm{Im}\nu$ are absolutely continuous with respect to $\mu$. By [L5], partition sums for their total variations are bounded by those of $\mathrm{\mid }\nu \mathrm{\mid }$, so both have finite total variation. Choose an increasing exhaustion $(X_n)$ with $\mu (X_n)<\mathrm{\infty }$; it is then common to $\mu$ and both signed measures. Applying [L3], choose real-valued $u,v\in L^1(\mu )$ such that $\mathrm{Re}\nu (E)={\int }_{E}ud\mu ,\mathrm{Im}\nu (E)={\int }_{E}vd\mu (E\in \mathcal{A})\mathrm{.}$

Put $h:=u+iv$. Then [L4] gives $h\in L^1(\mu )$, and for every measurable $E$ one has $\nu (E)=\mathrm{Re}\nu (E)+i\mathrm{Im}\nu (E)={\int }_{E}ud\mu +i{\int }_{E}vd\mu ={\int }_{E}hd\mu \mathrm{.}$ If $h^{\mathrm{\prime }}$ is another such density, then its real and imaginary parts give alternative signed densities for $\mathrm{Re}\nu$ and $\mathrm{Im}\nu$, so the uniqueness part of [L3] makes $h^{\mathrm{\prime }}=h$ $\mu$-almost everywhere.