Reflexive spaces have the Radon--Nikodym property

Discharge the choice hypotheses of the reflexivity suppliers. By [L1], AC supplies Countable Choice and every instance of the relative Hahn--Banach principle used below.

Reduce to one closed separable subspace. Let $Y\subseteq X$ be an arbitrary closed separable linear subspace. By [L2], $Y$ is a reflexive Banach space. Thus its canonical map $J_Y:Y\to Y^{\ast \ast }$ is onto by [L5] and is an isometry by [L2].

Exhibit the bidual as a separable dual space. Choose an at most countable norm-dense subset $D\subseteq Y$. The image $J_Y[D]$ is at most countable and is dense in $Y^{\ast \ast }$: if $\mathrm{\Phi }=J_{Y}y$ and $d\in D$ approximates $y$, then $\mathrm{\parallel }\mathrm{\Phi }-J_{Y}d\mathrm{\parallel }=\mathrm{\parallel }y-d\mathrm{\parallel }$. Hence $Y^{\ast \ast }=(Y^{\ast }{)}^{\ast }$ is a norm-separable dual Banach space. Moreover $J_Y$ is a bounded linear bijection with bounded inverse, indeed an isometry.

Transfer RNP from the bidual back to the subspace. The separable-dual theorem [L3] gives RNP to $Y^{\ast \ast }$. Isomorphism invariance along $J_Y$ then gives RNP to $Y$.

Apply separable determination. The closed separable subspace $Y$ was arbitrary, so every closed separable subspace of $X$ has RNP. The reverse implication in [L4] therefore gives RNP to $X$.

Record the scope and degenerate cases. [A1, step 1.1, step 3.1, step 5.1] If $X=\{0\}$, its sole closed subspace, bidual, vector measures, and densities are zero, so the same proof applies. A zero subspace has the singleton dense set and its canonical map is the zero bijection. The argument works in both scalar fields because [L2] and [L3] do. AC is used exactly to supply relative Hahn--Banach and Countable Choice in steps 1.1--3.1 and through the two RNP suppliers [L3]--[L4]. No dual-reflexivity theorem or unstated canonical-map isometry is used.