A Bochner density defines an absolutely continuous vector measure

Establish scalar control and absolute continuity. By [L4], $\mathrm{\parallel }f\mathrm{\parallel }$ is integrable. Put $\rho (E)={\int }_E\mathrm{\parallel }f\mathrm{\parallel }$. By [L3], $\rho$ is a finite positive measure. By [L2], $\mathrm{\parallel }{\nu }_f(E)\mathrm{\parallel }\le \rho (E)$, so ${\nu }_f\ll \mu$. For every finite partition $(E_j)$ of $E$, summing the same inequality gives ${\sum }_j\mathrm{\parallel }{\nu }_f(E_j)\mathrm{\parallel }\le \rho (E)$; hence $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)\le \rho (E)$ by [L1].

Fix a simple approximation for the reverse variation bound. Choose integrable simple $s_n$ with $\int \mathrm{\parallel }f-s_n\mathrm{\parallel }\to 0$ as supplied by [L4]. For fixed $E$, partition $E$ into the nonzero level sets of $s_n$ and the remaining zero cell.

Prove norm countable additivity without a new choice. For disjoint $(E_k)$ with union $E$, finite additivity follows from simple approximation and [L4]. Moreover $\mathrm{\parallel }{\nu }_f(E)-{\sum }_{k=1}^N{\nu }_f(E_k)\mathrm{\parallel }=\mathrm{\parallel }{\nu }_f(E\setminus {\bigcup }_{k=1}^N{E}_k)\mathrm{\parallel }\le \rho (E\setminus {\bigcup }_{k=1}^N{E}_k)\to 0$ by countable additivity of the finite measure $\rho$. Thus ${\nu }_f$ is norm-countably additive.

Prove the reverse variation inequality. On the partition from step 1.2, [L2] and [L4] give $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)\ge {\int }_E\mathrm{\parallel }{s}_n\mathrm{\parallel }-{\int }_E\mathrm{\parallel }f-s_n\mathrm{\parallel }$. The pointwise inequality $\mathrm{\parallel }{s}_n\mathrm{\parallel }\ge \mathrm{\parallel }f\mathrm{\parallel }-\mathrm{\parallel }f-s_n\mathrm{\parallel }$ then yields $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)\ge \rho (E)-2{\int }_E\mathrm{\parallel }f-s_n\mathrm{\parallel }$. Letting $n\to \mathrm{\infty }$ proves $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)\ge \rho (E)$.

Combine the bounds and close all cases. [L1, L5, step 1.1, step 2.1, step 2.2] Steps 1.1 and 2.2 give $\mathrm{\mid }{\nu }_f\mathrm{\mid }=\rho$; step 2.1 gives the required vector measure. In particular variation is finite (consistently with [L5]). For $E=\mathrm{\varnothing }$, $f=0$, or a one-level simple density, the equality reduces respectively to $0=0$, $0=0$, or $\mathrm{\parallel }\mu (E\cap A)x\mathrm{\parallel }=\mu (E\cap A)\mathrm{\parallel }x\mathrm{\parallel }$.