Dentable average ranges give vector-measure densities

Normalize by the variation measure. Put $\rho =\mathrm{\mid }\nu \mathrm{\mid }$. By [L3], $\rho$ is finite, and $\nu \ll \mu$ implies $\rho \ll \mu$ by refining subsets of a $\mu$-null set. If $\rho (\mathrm{\Omega })=0$, $\nu =0$ has the zero density. Otherwise [L4], using [A1], gives $w\ge 0$ with $\rho =wd\mu$. We first construct a density with respect to $\rho$, for which $\mathrm{\mid }\nu \mathrm{\mid }\le \rho$ holds tautologically.

Pass dentability to every average range. For $A$ with $\rho (A)>0$, put $x_A=\nu (A)\mathrm{/}\rho (A)$ and $D_A=\{x_B:B\subseteq A,\rho (B)>0\}$. Then $\mathrm{\parallel }{x}_B\mathrm{\parallel }\le 1$. The closed convex hull $C_A$ is nonempty, bounded, and dentable by hypothesis. A small slice of $C_A$ meets $D_A$, because its defining supremum over $C_A$ equals the supremum over $D_A$; its intersection with $D_A$ has no larger diameter. Thus every $D_A$ is dentable.

Find a positive subset with a small average range. Fix $\epsilon >0$ and positive $A$. If every positive $B\subseteq A$ had $\mathrm{diam}{D}_B>2\epsilon$, then for each such $B$ and every $x\in X$ some positive $C\subseteq B$ would satisfy $\mathrm{\parallel }x-x_C\mathrm{\parallel }>\epsilon$. Using [A1] and a maximal-disjoint-family argument, choose disjoint positive $B_j\subseteq B$ with $\mathrm{\parallel }{x}_B-x_{B_j}\mathrm{\parallel }>\epsilon$ which exhaust $B$ modulo $\rho$. Norm countable additivity gives $x_B={\sum }_j\rho (B_j)x_{B_j}\mathrm{/}\rho (B)$, so $x_B$ lies in the closed convex hull of points of $D_A$ more than $\epsilon$ away. A slice of $D_A$ of diameter less than $\epsilon$ cannot contain both $x_B$ and a member of this convex combination above the same slice threshold, contradicting step 1.2. Hence some positive $B\subseteq A$ has $\mathrm{diam}{D}_B\le 2\epsilon$.

Exhaust the space by good pieces and estimate the error. Choose a maximal disjoint family $(A_j)$ of positive sets with $\mathrm{diam}{D}_{A_j}\le 2\epsilon$. Step 2.1 forces it to cover $\mathrm{\Omega }$ modulo $\rho$; finiteness of $\rho$ makes the family countable. Put $g_{\epsilon }={\sum }_j{\mathbf{1}}_{A_j}{x}_{A_j}$. Its finite partial sums show strong measurability, and $\mathrm{\parallel }{g}_{\epsilon }\mathrm{\parallel }\le 1$, so [L6] gives Bochner integrability. For measurable $E$, $\nu (E)-{\int }_E{g}_{\epsilon }d\rho$ is the sum over $j$ of $\rho (E\cap A_j)(x_{E\cap A_j}-x_{A_j})$ (zero intersections omitted). Consequently every finite partition of $E$ gives the variation estimate $\mathrm{\mid }\nu -g_{\epsilon }\rho \mathrm{\mid }(E)\le 2\epsilon \rho (E)$.

Produce an $L^1(\rho ;X)$-Cauchy sequence. Use [A1] to choose $g_n=g_{2^{-n}}$ for all $n$. By [L5] and the triangle inequality for variation, $\int \mathrm{\parallel }{g}_n-g_{n-1}\mathrm{\parallel }d\rho \le \mathrm{\mid }\nu -g_n\rho \mathrm{\mid }(\mathrm{\Omega })+\mathrm{\mid }\nu -g_{n-1}\rho \mathrm{\mid }(\mathrm{\Omega })\le 6\cdot 2^{-n}\rho (\mathrm{\Omega })$. Thus the series of $L^1$ differences is summable.

Construct and identify the normalized density. By [L6], monotone convergence applied to ${\sum }_{n=1}^N\mathrm{\parallel }{g}_n-g_{n-1}\mathrm{\parallel }$ shows that the norm series is finite a.e. Hence, by completeness of $X$, $g_0+{\sum }_{n\ge 1}(g_n-g_{n-1})$ converges a.e. to a strongly measurable $h$, and the same tail estimate gives $g_n\to h$ in $L^1(\rho ;X)$. The criterion in [L6] makes $h$ Bochner integrable. For every $E$, step 3.1 and the norm integral inequality give $\mathrm{\parallel }\nu (E)-{\int }_{E}hd\rho \mathrm{\parallel }\le 2^{1-n}\rho (E)+{\int }_E\mathrm{\parallel }{g}_n-h\mathrm{\parallel }d\rho \to 0$. Thus $\nu (E)={\int }_{E}hd\rho$.

Transfer the density back to the original control measure. Set $f=wh$ (and $f=0$ where $w=0$). Products of scalar and vector simple approximants show that $f$ is strongly measurable. By [L4], $\int \mathrm{\parallel }f\mathrm{\parallel }d\mu =\int \mathrm{\parallel }h\mathrm{\parallel }d\rho <\mathrm{\infty }$, so [L6] makes $f$ Bochner integrable. If $s_n\to h$ in $L^1(\rho ;X)$ are simple, [L7] and [L4] give ${\int }_{E}w{s}_{n}d\mu ={\int }_E{s}_{n}d\rho$ level by level, while [L4] identifies the two $L^1$ errors. Passing to the limit yields ${\int }_{E}fd\mu ={\int }_{E}hd\rho =\nu (E)$.

Conclude RNP and record the choice cost. [A1, L1, step 6.1] The construction applies to arbitrary data in [L1], so $X$ has RNP. The exact non-finite uses of [A1] are scalar Radon--Nikodym in step 1.1, maximal disjoint families in steps 2.1 and 3.1, and simultaneous selection of the sequence $(g_n)$ in step 4.1. Empty and zero-variation cases were settled in step 1.1; one-piece exhaustions are included in step 3.1.