Separable dual spaces have the Radon--Nikodym property

Obtain the needed separability and choice interfaces. By [L1], AC supplies Countable Choice and proves every instance of the relative Hahn--Banach principle. Thus [L2] applies and makes $Y$ norm separable. Adjoin zero to chosen countable dense subsets of $Y$ and $X$ so that [L8] enumerates both even in the zero-space case.

Fix a vector measure and dominate it by one scalar density. Let $(\mathrm{\Omega },\mathcal{A},\mu )$ be a finite measure space and let $\nu :\mathcal{A}\to Y^{\ast }$ have bounded variation with $\nu \ll \mu$. If $\mu (E)=0$, every cell in a finite partition of $E$ has zero $\nu$-value, so $\mathrm{\mid }\nu \mathrm{\mid }(E)=0$. Hence $\mathrm{\mid }\nu \mathrm{\mid }\ll \mu$. By [L3] it is a finite positive measure, and [L4] gives an integrable real density $g$ with $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_{E}gd\mu$. Positivity, tested on $\{g\le -1\mathrm{/}m\}$, permits replacing $g$ on a null set so that $g\ge 0$ everywhere.

Choose a countable linear test space and all its scalar densities. Let ${\mathbb{K}}_0=\mathbb{Q}$ in the real case and ${\mathbb{K}}_0=\mathbb{Q}+i\mathbb{Q}$ in the complex case. The ${\mathbb{K}}_0$-linear span $D$ of a countable dense subset of $Y$ is countable and norm dense by [L8]. For $d\in D$ define the finite signed or complex measure ${\nu }_d(E)=\nu (E)(d)$. Its partition sums satisfy $\mathrm{\mid }{\nu }_d\mathrm{\mid }(E)\le \mathrm{\parallel }d\mathrm{\parallel }\mathrm{\mid }\nu \mathrm{\mid }(E)$, and ${\nu }_d\ll \mu$. Apply [L4], using AC to choose simultaneously for all $d\in D$, measurable $h_d\in L^1(\mu )$ such that ${\nu }_d(E)={\int }_E{h}_{d}d\mu$.

Make the scalar representatives pointwise linear and bounded. Uniqueness of scalar densities says, for every $a,b\in {\mathbb{K}}_0$ and $d,e\in D$, that $h_{ad+be}=a{h}_d+b{h}_e$ almost everywhere. There are only countably many such relations. Moreover [L5] and the variation estimate in step 2.1 give

Extend the pointwise functionals to $Y^{\ast }$. For every remaining $\omega$, continuity and density of $D$ extend $d\mapsto h_d(\omega )$ uniquely to a scalar-linear functional $f(\omega )\in Y^{\ast }$ with $\mathrm{\parallel }f(\omega )\mathrm{\parallel }\le g(\omega )$. In the complex case, $\mathbb{Q}+i\mathbb{Q}$-linearity and continuity give full complex linearity. Set $f=0$ on the common null set. Then $f(\omega )(d)=h_d(\omega )$ for all $d\in D$ off that set.

Prove strong measurability rather than merely coordinate measurability. Fix $y\in Y$. Using an enumeration of $D$, for every integer $m\ge 1$ take the least indexed $d_m\in D$ with $\mathrm{\parallel }{d}_m-y\mathrm{\parallel }<1\mathrm{/}m$. Step 3.1 gives $h_{d_m}(\omega )\to f(\omega )(y)$ off the common null set, so [L6] makes every coordinate $\omega \mapsto f(\omega )(y)$ measurable. Let $(u_j{)}_{j\ge 1}$ enumerate a countable dense subset of the unit ball of $Y$ obtained from $D$ by rational rescaling. For each $x^{\ast }\in X$,

Integrate the extension and identify the vector measure. The bound $\mathrm{\parallel }f\mathrm{\parallel }\le g$ and [L7] make $f$ Bochner integrable. For $d\in D$, boundedness of evaluation at $d$, commutation in [L7], and step 2.1 give

Conclude RNP and close the degenerate cases. [A1, L3, step 1.1, step 6.1] The measure space and $\nu$ were arbitrary, so step 6.1 proves the RNP condition in [L3]. If $Y^{\ast }=\{0\}$, every scalar measure and every density above is zero; if $\mu (\mathrm{\Omega })=0$ or $\nu =0$, take $g=f=0$. A one-point dense set and a one-element rational span are covered by the same construction. AC is used for Hahn--Banach and Countable Choice in step 1.1, scalar RN and simultaneous representatives in steps 1.2--2.1, and the common countable family of a.e. relations; no stronger unstated choice is used.