Banach Valued Integration and the Radon Nikodym Property

Form a finite common refinement. [given, L1] Suppose $s={\sum }_{j=1}^m{x}_j{\mathbf{1}}_{A_j}={\sum }_{k=1}^n{y}_k{\mathbf{1}}_{B_k}$ are two representations from [L1]. Because every displayed coefficient is nonzero, both unions ${\bigcup }_j{A}_j$ and ${\bigcup }_k{B}_k$ are the same set $\{\omega :s(\omega )\mathrm{\ne }0\}$. Hence the cells $C_{jk}=A_j\cap B_k$ with $1\le j\le m$ and $1\le k\le n$ partition every $A_j$ and every $B_k$. On every nonempty $C_{jk}$, pointwise equality gives $x_j=y_k$.

Compare the two integral sums. By finite additivity in [L2], step 1.1 gives

Prove linearity. [L1, step 2.1] For integrable $s,t$ and scalars $a,b$, refine their finite level partitions. On each refined cell $as+bt$ has coefficient $a{x}_j+b{y}_k$. Every cell on which this coefficient is nonzero lies in the union of the finite-measure supports of $s$ and $t$, so $as+bt$ is integrable. Applying step 2.1 and distributing the finite vector sum yields ${\int }_E(as+bt)=a{\int }_{E}s+b{\int }_{E}t$.

Prove null-insensitivity. [L2, step 3.1] If integrable simple functions $s,t$ agree off a null set $N$, refine their level partitions as above. A refined cell on which their coefficients differ is contained in $N$, hence has measure zero by [L2]. Its contribution to ${\int }_E(s-t)$ is zero, and linearity from step 3.1 gives ${\int }_{E}s={\int }_{E}t$.

Prove the norm inequality and conclude. [L1, L3, step 4.1] Write $s={\sum }_j{x}_j{\mathbf{1}}_{A_j}$ in its nonzero disjoint-level form. The triangle inequality in $X$, [L3], and the finite-measure support rule give

Strong measurability gives an essentially separable range. [given, L1, L4] Assume first that $f$ is strongly measurable, witnessed by $s_n$ and $N$ as in [L1]. The union of the finite ranges of the $s_n$ is countable. Its closed linear span $Y$ is separable by [L4], and every $f(\omega )$ with $\omega \mathrm{\notin }N$ is a norm limit of points of $Y$. Thus $f$ is essentially separably valued.

Strong measurability gives weak measurability. [given, L1, L2, L3] For $x^{\ast }\in X^{\ast }$, [L2] gives $x^{\ast }(s_n(\omega ))\to x^{\ast }(f(\omega ))$ off $N$. Each $x^{\ast }\circ s_n$ is scalar simple and measurable. A pointwise scalar limit is measurable off $N$, and [L3] makes its arbitrary values on subsets of $N$ measurable as well. Hence $f$ is weakly measurable.

Fix countable dense data for the reverse implication. [given, L4, choose] Conversely assume conditions 1 and 2. If $Y=\{0\}$, the constant zero simple functions converge to $f$ off $N$, so suppose $Y\mathrm{\ne }\{0\}$. By [L4] choose a sequence $(y_m)$ dense in $Y$ and a sequence $(z_k)$ dense in its unit sphere.

Construct a countable norming family. [A1, L2, L5, step 1.3] For each $k$, in the complex case define on the underlying real plane $\mathbb{C}{z}_k$ the norm-one real functional $u_k(a{z}_k)=\mathrm{Re}a$; in the real case use $u_k(a{z}_k)=a$ on $\mathbb{R}{z}_k$. Apply [L5] and [A1] to extend these simultaneously to real functionals $U_k$ on the underlying real space of $X$. In the complex case put $x_k^{\ast }(x)=U_k(x)-i{U}_k(ix)$; in the real case put $x_k^{\ast }=U_k$. Then $x_k^{\ast }\in X^{\ast }$, $\mathrm{\parallel }{x}_k^{\ast }\mathrm{\parallel }=1$, and $x_k^{\ast }(z_k)=1$. Consequently, for $y\in Y$,

Norm distances to fixed centres are measurable. [L3, step 2.1] For fixed $y\in Y$, step 2.1 and weak measurability give, off $N$, $\mathrm{\parallel }f-y\mathrm{\parallel }={\sup }_k\mathrm{\mid }{x}_k^{\ast }(f-y)\mathrm{\mid }$. The right side is the supremum of a countable family of measurable scalar functions. With any values assigned on $N$, [L3] therefore makes $\omega \mapsto \mathrm{\parallel }f(\omega )-y\mathrm{\parallel }$ measurable.

Build finite-valued nearest-centre approximants. [L1, step 1.3, step 3.1] For each $n\ge 1$ and $\omega \mathrm{\notin }N$, choose the least $m\in \{1,\dots ,n\}$ minimizing $\mathrm{\parallel }f(\omega )-y_m\mathrm{\parallel }$; put $s_n(\omega )=y_m$ there and $s_n=0$ on $N$. The finitely many tie-broken Voronoi cells are measurable by step 3.1, so $s_n$ is a measurable simple function. Density of $(y_m)$ gives $\mathrm{\parallel }{s}_n(\omega )-f(\omega )\mathrm{\parallel }\to 0$ for every $\omega \mathrm{\notin }N$.

Steps 1.1--1.2 prove the forward implication, and step 4.1 supplies the simple approximants required by [L1] for the reverse implication. The only non-finite choice is [A1]: it supplies the Hahn--Banach extensions in step 2.1 (and hence also covers their countable simultaneous selection).

Prove necessity of scalar norm integrability. [given, L1, L2] Suppose first that $f$ is Bochner integrable and choose $(s_n)$ as in [L1]. For some $n$, $\int \mathrm{\parallel }f-s_n\mathrm{\parallel }<\mathrm{\infty }$, while integrability of the simple function gives $\int \mathrm{\parallel }{s}_n\mathrm{\parallel }<\mathrm{\infty }$. Since $\mathrm{\parallel }f\mathrm{\parallel }\le \mathrm{\parallel }f-s_n\mathrm{\parallel }+\mathrm{\parallel }{s}_n\mathrm{\parallel }$, [L2] gives $\int \mathrm{\parallel }f\mathrm{\parallel }<\mathrm{\infty }$.

Prove approximation independence. [given, L1] If $(s_n)$ and $(t_n)$ are any two defining approximations, the simple norm inequality gives $\mathrm{\parallel }\int s_n-\int t_n\mathrm{\parallel }\le \int \mathrm{\parallel }{s}_n-t_n\mathrm{\parallel }$, which is at most $\int \mathrm{\parallel }{s}_n-f\mathrm{\parallel }+\int \mathrm{\parallel }f-t_n\mathrm{\parallel }$. Both terms tend to zero, so the two norm limits coincide.

Construct dominated simple approximants for sufficiency. [given, L3, construct] Conversely assume $\int \mathrm{\parallel }f\mathrm{\parallel }<\mathrm{\infty }$. On the exceptional measurable null set of a strong approximation, replace both $f$ and every approximant by zero (and call the representative again $f$). Thus measurable simple functions $u_n$ converge pointwise to $f$; [L3] makes $\mathrm{\parallel }f\mathrm{\parallel }$ measurable. Define $s_n=u_n{\mathbf{1}}_{\{\mathrm{\parallel }{u}_n\mathrm{\parallel }\le 2\mathrm{\parallel }f\mathrm{\parallel }\}}$. Then $s_n$ is simple and measurable, $\mathrm{\parallel }{s}_n\mathrm{\parallel }\le 2\mathrm{\parallel }f\mathrm{\parallel }$, and $s_n\to f$ pointwise: when $f(\omega )\mathrm{\ne }0$, the inequality defining the retained part holds eventually, while at a zero of $f$ either retained values tend to zero or the replacement is zero.

Verify that the constructed simple functions are integrable. [L2, step 1.3] Each $s_n$ is integrable. Indeed, if a nonzero value $x$ occurs, its level set is contained in $\{2\mathrm{\parallel }f\mathrm{\parallel }\ge \mathrm{\parallel }x\mathrm{\parallel }\}$, whose measure is at most $2\int \mathrm{\parallel }f\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }<\mathrm{\infty }$ by [L2].

Obtain convergence in $L^1$. [L1, L4, step 1.3, step 2.1] The pointwise convergence in step 1.3 and $\mathrm{\parallel }f-s_n\mathrm{\parallel }\le 3\mathrm{\parallel }f\mathrm{\parallel }$ allow [L4] to be applied. Hence $\int \mathrm{\parallel }f-s_n\mathrm{\parallel }\to 0$. Together with step 2.1 this is exactly the approximation required in [L1].

Step 1.1 proves necessity, step 3.1 proves sufficiency, and step 1.2 proves that the resulting integral is approximation-independent. The zero function, the empty measure space, and a single simple function are included by taking the constant zero or constant simple approximation.

Restrict a defining simple approximation to $E$. [given, L1, choose] Choose integrable simple $s_n$ with $\int \mathrm{\parallel }f-s_n\mathrm{\parallel }\to 0$. Replacing each by ${\mathbf{1}}_E{s}_n$ shows from [L1] that ${\mathbf{1}}_{E}f$ is Bochner integrable and that its integral is the norm limit of ${\int }_E{s}_n$.

Bound each simple integral. [L2, L3, step 1.1] The triangle inequality and [L2] give $\mathrm{\parallel }{\int }_E{s}_n\mathrm{\parallel }\le {\int }_E\mathrm{\parallel }{s}_n\mathrm{\parallel }$. Since $\mathrm{\parallel }{s}_n\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }+\mathrm{\parallel }{s}_n-f\mathrm{\parallel }$, [L3] yields $\mathrm{\parallel }{\int }_E{s}_n\mathrm{\parallel }\le {\int }_E\mathrm{\parallel }f\mathrm{\parallel }+{\int }_E\mathrm{\parallel }{s}_n-f\mathrm{\parallel }$.

Pass to the limit and conclude. [step 1.1, step 2.1] Let $n\to \mathrm{\infty }$ in step 2.1. Norm continuity and step 1.1 identify the left limit with $\mathrm{\parallel }{\int }_{E}f\mathrm{\parallel }$, while the last term tends to zero. This proves the inequality, including $E=\mathrm{\varnothing }$ and $f=0$.

Select simultaneous strong-measurability witnesses. [given, A1, L2, choose] Use [A1] exactly once to choose, for every $n$, a simple approximation sequence witnessing the strong measurability in [L2]. Unite their exceptional null sets with those from convergence and domination; countable additivity makes the union null. Modify all functions and approximants to be zero there. We now have pointwise convergence everywhere and a doubly indexed family of simple approximants.

Obtain a common countable range and measurable distances. [L3, step 1.1] Let $D$ be $\{0\}$ together with all values of all selected simple approximants. It is countable. For each $n$, the range of $f_n$ lies in $\bar{D}$, hence the pointwise limit $f$ also takes values in $\bar{D}$. For $y\in D$, the functions $\mathrm{\parallel }{f}_n-y\mathrm{\parallel }$ are measurable by applying [L3] to the simple approximants, and $\mathrm{\parallel }f-y\mathrm{\parallel }$ is measurable by applying [L3] once more to $f_n\to f$.

Build finite-valued approximants to the limit. [L2, step 2.1, construct] Enumerate $D$ with repetitions as $(y_j)$. For each $m$, assign $s_m(\omega )$ to be the least-indexed nearest point to $f(\omega )$ among $y_1,\dots ,y_m$. The finitely many tie-broken Voronoi cells are measurable by step 2.1, so $s_m$ is simple; density gives $s_m(\omega )\to f(\omega )$. Thus $f$ is strongly measurable.

Apply the scalar dominated-convergence theorem. [L1, L4, step 3.1] Passing to the pointwise limit in $\mathrm{\parallel }{f}_n\mathrm{\parallel }\le g$ gives $\mathrm{\parallel }f\mathrm{\parallel }\le g$. By [L1], $f$ and every $f_n$ are Bochner integrable. Moreover $\mathrm{\parallel }{f}_n-f\mathrm{\parallel }\le 2g$, so [L4] gives $\int \mathrm{\parallel }{f}_n-f\mathrm{\parallel }\to 0$.

Pass from $L^1$ convergence to integral convergence. [L1, L5, L6, step 4.1] Combining simple approximations to $f_n$ and $f$, [L6] and approximation independence from [L1] show that $\int (f_n-f)=\int f_n-\int f$. Apply [L5]: $\mathrm{\parallel }\int f_n-\int f\mathrm{\parallel }\le \int \mathrm{\parallel }{f}_n-f\mathrm{\parallel }\to 0$ by step 4.1. If the measure space is empty or $g=0$ a.e., every integral is zero; no separate endpoint convention is needed.

Restrict a defining approximation to the measurable set. Choose integrable simple $s_n$ with $\int \mathrm{\parallel }f-s_n\mathrm{\parallel }\to 0$. Then ${\mathbf{1}}_{E}T{s}_n$ is an integrable $Y$-valued simple function: every nonzero level is a finite union of level sets of ${\mathbf{1}}_E{s}_n$.

Prove Bochner integrability after applying $T$. By [L1], $\int \mathrm{\parallel }{\mathbf{1}}_{E}Tf-{\mathbf{1}}_{E}T{s}_n\mathrm{\parallel }\le C{\int }_E\mathrm{\parallel }f-s_n\mathrm{\parallel }\to 0$. Thus [L2] makes ${\mathbf{1}}_{E}Tf$ Bochner integrable.

Commute $T$ with the defining limit. [L1, L2, L3, step 1.1, step 2.1] For each simple $s_n$, finite linearity in [L3] gives $T({\int }_E{s}_n)={\int }_{E}T{s}_n$. Boundedness makes $T$ norm-continuous, so taking limits in this equality and using [L2] proves the displayed identity. If $T=0$, $f=0$, or $E=\mathrm{\varnothing }$, both sides are explicitly zero.

Prove finite additivity of variation. For disjoint $A,B$, joining finite partitions of $A$ and $B$ shows $\mathrm{\mid }\nu \mathrm{\mid }(A)+\mathrm{\mid }\nu \mathrm{\mid }(B)\le \mathrm{\mid }\nu \mathrm{\mid }(A\cup B)$ (use partitions within $\epsilon$ of each supremum). Conversely, intersect any finite partition of $A\cup B$ with $A$ and $B$; finite additivity of $\nu$ and the triangle inequality show that its norm sum is at most $\mathrm{\mid }\nu \mathrm{\mid }(A)+\mathrm{\mid }\nu \mathrm{\mid }(B)$. Taking the supremum gives equality.

Prove the functional domination estimate. For every finite partition $(A_j)$ of $E$, [L2] gives ${\sum }_j\mathrm{\mid }{x}^{\ast }(\nu (A_j))\mathrm{\mid }\le \mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }{\sum }_j\mathrm{\parallel }\nu (A_j)\mathrm{\parallel }$. Taking suprema as in [L1] proves the displayed inequality, including $x^{\ast }=0$ and $E=\mathrm{\varnothing }$.

Prove countable additivity. Let $E={⨆}_{n\ge 1}{E}_n$. Finite additivity gives ${\sum }_{n=1}^N\mathrm{\mid }\nu \mathrm{\mid }(E_n)\le \mathrm{\mid }\nu \mathrm{\mid }(E)$, hence ${\sum }_n\mathrm{\mid }\nu \mathrm{\mid }(E_n)\le \mathrm{\mid }\nu \mathrm{\mid }(E)$. For the reverse inequality, take any finite partition $(A_j)$ of $E$. Norm countable additivity gives $\nu (A_j)={\sum }_n\nu (A_j\cap E_n)$, so ${\sum }_j\mathrm{\parallel }\nu (A_j)\mathrm{\parallel }\le {\sum }_n{\sum }_j\mathrm{\parallel }\nu (A_j\cap E_n)\mathrm{\parallel }\le {\sum }_n\mathrm{\mid }\nu \mathrm{\mid }(E_n)$. Taking the supremum over $(A_j)$ proves the reverse inequality.

Conclude finiteness and all boundary cases. [L1, step 1.2, step 2.1] The empty partition gives $\mathrm{\mid }\nu \mathrm{\mid }(\mathrm{\varnothing })=0$, step 2.1 gives countable additivity, and bounded variation in [L1] gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\le \mathrm{\mid }\nu \mathrm{\mid }(\mathrm{\Omega })<\mathrm{\infty }$. Thus $\mathrm{\mid }\nu \mathrm{\mid }$ is a finite positive measure, and step 1.2 supplies the asserted scalar-variation bound.

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 $\mathrm{\parallel }\mu (E\cap A)x\mathrm{\parallel }=\mu (E\cap A)\mathrm{\parallel }x\mathrm{\parallel }$.

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.

Convert nondentability into a uniformly separated convex bush. Choose $\eta >0$ such that no slice of $C$ has diameter below $\eta$, and put $r=\eta \mathrm{/}4$. For $x\in C$, if $x\mathrm{\notin }\stackrel{}{\mathrm{conv}}(C\setminus B(x,r))$, [L2] gives a slice lying inside $B(x,r)$ and hence of diameter at most $2r<\eta$, a contradiction. Thus every $x\in C$ belongs to that closed convex hull. Enlarge to $D=C+B(0,r\mathrm{/}2)$. Given $z=x+y\in D$, approximate $x$ by a finite convex combination ${\sum }_i{\alpha }_i{x}_i=x+e$ of points outside $B(x,r)$ with error $e$ satisfying $\mathrm{\parallel }e\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }<r\mathrm{/}2$, and put $z_i=x_i+y-e$. Then $z={\sum }_i{\alpha }_i{z}_i$, every $z_i\in D$, and $\mathrm{\parallel }{z}_i-z\mathrm{\parallel }\ge r-\mathrm{\parallel }e\mathrm{\parallel }>r\mathrm{/}2$. With $\delta =r\mathrm{/}2$, [A1] recursively chooses such finite successor families from an initial $z_0\in C$. The resulting node set is countable, bounded, and every child is at least $\delta$ from its parent.

Realize the bush as a separated interval martingale. Starting with $M_0\equiv z_0$, partition every atom at level $n-1$ into finitely many half-open subintervals in the successor proportions $({\alpha }_i)$, put the corresponding child value on each, and overlay the dyadic grid of mesh $2^{-n}$. Let ${\mathcal{P}}_n$ be the resulting refining finite interval partition. Parent averages equal parent values, so $(M_n)$ is a martingale on these finite algebras; it is uniformly bounded and $\mathrm{\parallel }{M}_n-M_{n-1}\mathrm{\parallel }\ge \delta$ away from the finitely many endpoints. The union algebra $\mathcal{R}={\bigcup }_n\sigma ({\mathcal{P}}_n)$ contains every dyadic interval algebra.

Define and extend the dominated vector measure. For $A\in \sigma ({\mathcal{P}}_n)$ set ${\nu }_0(A)={\int }_A{M}_{n}d\lambda$. The martingale identity makes this independent of $n$, and if $K={\sup }_n\mathrm{\parallel }{M}_n{\mathrm{\parallel }}_{\mathrm{\infty }}$, then $\mathrm{\parallel }{\nu }_0(A)-{\nu }_0(B)\mathrm{\parallel }\le K\lambda (A\mathrm{△}B)$. The class of Borel sets approximable in symmetric-difference measure by $\mathcal{R}$ is a sigma-algebra: complements preserve the distance, and countable unions reduce by [L5] to one large finite union. It contains the dyadic algebra and hence all Borel sets; the completion clause in [L5] adds Lebesgue sets. Choose such approximants. Completeness of $X$ gives a unique extension $\nu$ with $\mathrm{\parallel }\nu (E)\mathrm{\parallel }\le K\lambda (E)$; the same estimate proves norm countable additivity. Summing it over finite partitions gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\le K\lambda (E)$, so $\nu \ll \lambda$ and has bounded variation. Every algebra value is a finite linear combination of bush nodes, hence every extended value lies in the closed separable span $Y$ of the countable node set.

Assume a density and identify all its finite-partition averages. Suppose $f\in L^1([0,1];X)$ satisfies $\nu (E)={\int }_{E}f$ for all Lebesgue $E$. For every atom $A$ of ${\mathcal{P}}_n$, the construction gives ${\int }_{A}f=\nu (A)={\int }_A{M}_n=\lambda (A)M_n{\mathrm{\mid }}_A$. Thus the atomwise averaging operator $Q_n$ applied to $f$ equals $M_n$.

Prove that the atomwise averages of a Bochner density converge in $L^1$. Choose an integrable simple $s$ with $\int \mathrm{\parallel }f-s\mathrm{\parallel }<\epsilon$ by [L4]. By the approximation proved in step 3.1 and finiteness of its level family, approximate the level sets of $s$ by sets in $\mathcal{R}$, obtaining an $\mathcal{R}$-simple $t$ with $\int \mathrm{\parallel }f-t\mathrm{\parallel }<2\epsilon$. For all sufficiently large $n$, $Q_{n}t=t$. The norm inequality on each atom shows that $Q_n$ is an $L^1$ contraction, so $\mathrm{\parallel }{Q}_{n}f-f{\mathrm{\parallel }}_1\le \mathrm{\parallel }{Q}_n(f-t){\mathrm{\parallel }}_1+\mathrm{\parallel }t-f{\mathrm{\parallel }}_1<4\epsilon$. Hence $M_n=Q_{n}f\to f$ in $L^1$.