RNP and almost-everywhere differentiability of Lipschitz curves

Make the inherited choice assumptions explicit. By [L1], [A1] supplies both Countable Choice and Dependent Choice. Countable Choice is used in [L2], [L5], and [L11]; both principles are hypotheses of the scalar FTC in [L9].

Associate a dominated vector measure to a Lipschitz curve. Suppose first that $X$ has RNP, let $F:[0,1]\to X$ be $L$-Lipschitz, and put $G(t)=F(t)-F(0)$. Then $G(0)=0$ and [L2] gives a vector measure ${\nu }_G$ with $\mathrm{\mid }{\nu }_G\mathrm{\mid }\le L\lambda$ and ${\nu }_G((a,b])=G(b)-G(a)$.

Reduce an arbitrary interval vector measure to bounded-density levels. For the converse direction, let $\nu$ be a bounded-variation vector measure on $[0,1]$ with $\nu \ll \lambda$. If $\lambda (E)=0$, every member of every finite partition of $E$ is null and has $\nu$-value zero, so $\mathrm{\mid }\nu \mathrm{\mid }(E)=0$. Thus $\mathrm{\mid }\nu \mathrm{\mid }\ll \lambda$. By [L8] and AC there is an integrable scalar density $g$ with $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_{E}gd\lambda$. Positivity of $\mathrm{\mid }\nu \mathrm{\mid }$ makes $g\ge 0$ almost everywhere: applying the representation to $\{g\le -1\mathrm{/}m\}$ for each $m\ge 1$ makes each such set null. Replace $g$ by zero on their null union. Put $A_n=\{n-1\le g<n\}$ for $n\ge 1$ and $Z=\{g=+\mathrm{\infty }\}$. The $A_n$ are disjoint, $Z$ is null, and they cover $[0,1]\setminus Z$. Define ${\nu }_n(E)=\nu (E\cap A_n)$. Directly from finite partitions, $\mathrm{\mid }{\nu }_n\mathrm{\mid }(E)=\mathrm{\mid }\nu \mathrm{\mid }(E\cap A_n)\le n\lambda (E)$.

Obtain a Bochner density in the RNP-to-differentiability direction. The interval test [L3] applied to ${\nu }_G$ supplies a Bochner-integrable $f:[0,1]\to X$ with ${\nu }_G(E)={\int }_{E}fd\lambda$. Hence $G(b)-G(a)={\int }_{(a,b]}fd\lambda$ for every $a<b$.

Turn each bounded level measure into a Lipschitz curve. For each $n$, set $F_n(t)={\nu }_n((0,t])$. The converse part of [L2] and the bound in step 1.3 show that $F_n(0)=0$ and that $F_n$ is $n$-Lipschitz.

Prepare a common set of vector Lebesgue points. Choose integrable simple $s_m$ with $\int \mathrm{\parallel }f-s_m\mathrm{\parallel }\to 0$ as in [L4]. Passing to a subsequence if necessary, the scalar errors $e_m=\mathrm{\parallel }f-s_m\mathrm{\parallel }$ converge to zero almost everywhere: choose least indices with $L^1$ errors below $2^{-2m}$, and the sets where the corresponding pointwise error exceeds $2^{-m}$ have summable measures, so their tail unions decrease to a null set. Extend $e_m$ and the finitely many indicator functions of the level sets of $s_m$ by zero outside $[0,1]$. Apply [L5] to every one of this countable family and remove the union of their exceptional null sets. At each remaining interior point $t$, every $e_m$ differentiates by interval averages, $e_m(t)\to 0$, and

Construct measurable derivative fields for the bounded level curves. By the assumed differentiability property, for each $n$ there is a measurable null set $N_n$ off which $F_n^{\mathrm{\prime }}$ exists in norm. For $k\ge 2$ put $q_{n,k}(t)=k(F_n(t+1\mathrm{/}k)-F_n(t))$ when $t\le 1-1\mathrm{/}k$, and put it equal to zero on the remaining interval. On the first piece $q_{n,k}$ is $2nk$-Lipschitz; a finite interval partition of sufficiently small mesh, together with the constant-zero last piece, therefore gives a measurable simple function within $1\mathrm{/}k$ uniformly of $q_{n,k}$. These simple functions converge to $F_n^{\mathrm{\prime }}$ off $N_n$. Define $f_n=F_n^{\mathrm{\prime }}$ there and $f_n=0$ on $N_n$. This proves strong measurability in the sense of [L4]. Difference quotients give $\mathrm{\parallel }{f}_n\mathrm{\parallel }\le n$ off $N_n$, so [L4] makes $f_n$ Bochner integrable.

Differentiate the indefinite Bochner integral in norm. At a point $t$ retained in step 3.1, for fixed $m$ the triangle inequality gives

Show that each derivative field represents its level measure. Fix $n$ and $x^{\ast }\in X^{\ast }$. The real-valued function $x^{\ast }{F}_n$ in the real case, and its real and imaginary parts in the complex case, are Lipschitz and hence absolutely continuous by [L9]. Their derivatives agree almost everywhere with the corresponding scalar parts of $x^{\ast }{f}_n$. The scalar FTC, whose choice hypotheses were supplied in step 1.1, and commutation in [L7] give

Complete the forward implication, including its boundary cases. Step 4.1 proves almost-everywhere norm differentiability of every Lipschitz curve when $X$ has RNP. Adding the constant $F(0)$ does not affect difference quotients. If $L=0$, the curve is constant and has derivative zero everywhere; the endpoints are excluded from the derivative assertion and have measure zero. The zero Banach space and the one-point interval cause no exception.

Paste the bounded derivative fields into one density. Define $h(t)=h_n(t)$ on $A_n$ and $h=0$ on $Z$. The explicit simple approximants from step 3.2, multiplied by ${\mathbf{1}}_{A_n}$ and summed for $n\le k$, form a simple sequence converging to $h$ off the countable union of the $N_n$ and $Z$; [L5] makes that union null. Hence $h$ is strongly measurable. Moreover $\mathrm{\parallel }h\mathrm{\parallel }\le {\sum }_{n\ge 1}n{\mathbf{1}}_{A_n}\le g+1$, so [L4] makes $h$ Bochner integrable. Let $H_N={\sum }_{n=1}^N{h}_n$. Then $H_N\to h$ pointwise and $\mathrm{\parallel }{H}_N\mathrm{\parallel }\le g+1$. Applying [L11] to ${\mathbf{1}}_E{H}_N$ for any measurable $E$ gives ${\int }_E{H}_N\to {\int }_{E}h$. Finite linearity follows by combining the simple approximations in [L4], so step 4.2 gives ${\int }_E{H}_N={\sum }_{n=1}^N\nu (E\cap A_n)$. Norm countable additivity of $\nu$ and $\nu (E\cap Z)=0$ make the latter sums converge to $\nu (E)$. Thus $h$ is a Bochner density of $\nu$.

Conclude the equivalence and record the exact AC use. [A1, L3, step 5.1, step 5.2] Step 5.1 proves RNP implies almost-everywhere differentiability. Conversely, step 5.2 gives a density for every vector measure in the interval test [L3], so $X$ has RNP. AC is used by the scalar Radon--Nikodym theorem, the interval RNP test, and through step 1.1 for countable null-set, dominated-convergence, and scalar-FTC suppliers. Empty and zero measures give the zero density, and both directions of the equivalence have been proved.