Nondentability produces a vector measure without density

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 $L^1$.

Contradict the fixed separation and conclude. [discharge-contradiction: step 5.1, L3, step 2.1, step 4.1, step 5.1] Step 5.1 would imply $\mathrm{\parallel }{M}_n-M_{n-1}{\mathrm{\parallel }}_1\to 0$, whereas step 2.1 gives $\mathrm{\parallel }{M}_n-M_{n-1}{\mathrm{\parallel }}_1\ge \delta$ for every $n$. Thus no density exists. The measure in step 3.1 is the witness required by [L3], with separable range. The exact uses of [A1] are strict separation through [L2], recursive bush and partition choices, and the countable approximation choices in extending ${\nu }_0$. The empty interval endpoints form null sets; the one-child case cannot occur because every child is $\delta$-separated.