AC zero-derivative constancy and integrable-derivative reconstruction under Countable Choice

If $a=b$, both conclusions hold directly. Otherwise fix $a\le s<t\le b$ and $\epsilon >0$ for claim 1. Let $E$ be a null exceptional set for $F^{\mathrm{\prime }}=0$ on $[s,t]$, enlarged by $s,t$; it is null by [F4]. Restricting [F1] to $[s,t]$, choose $d>0$ so that total interval length below $d$ implies total absolute increment below $\epsilon \mathrm{/}2$. Put $\eta =\epsilon \mathrm{/}(2(t-s+1))$. By [F3], choose one open $U\supseteq E$ with $\lambda (U)<d$.

For $x\mathrm{\notin }E$, define $k(x)$ as the least $k\in {\mathbb{N}}_0$ such that $\mathrm{\mid }F(y)-F(x)\mathrm{\mid }\le \eta \mathrm{\mid }y-x\mathrm{\mid }$ whenever $y\in [s,t]$ and $\mathrm{\mid }y-x\mathrm{\mid }<2^{-k}$. This set is nonempty by $F^{\mathrm{\prime }}(x)=0$, [F7] and [F5], including $y=x$ by equality. For $x\in E$, instead take the least $k$ with $(x-2^{-k},x+2^{-k})\subseteq U$; openness and [F5] give existence. Set $\gamma (x)=2^{-k(x)}$. These uniquely specified least integers define one positive gauge; no arbitrary pointwise radius selections are needed. Obtain one finite $\gamma$-fine partition $\{([x_{i-1},x_i],{\xi }_i){\}}_{i=1}^m$ from [F2].

Cells tagged in $E$ lie in $U$. Their interiors are disjoint, and their total length is the measure of their finite union, since shared endpoints form a finite null set. By [F4] this sum is at most $\lambda (U)<d$, so [F1] bounds their total absolute $F$-increment by $\epsilon \mathrm{/}2$; for an empty family the sum is zero.

For every other cell, strict fineness and step 2.1 give $\mathrm{\mid }F(x_i)-F(x_{i-1})\mathrm{\mid }\le \mathrm{\mid }F(x_i)-F({\xi }_i)\mathrm{\mid }+\mathrm{\mid }F({\xi }_i)-F(x_{i-1})\mathrm{\mid }\le \eta (x_i-x_{i-1})$, because the tag lies in the cell. The sum over these cells is at most $\eta (t-s)<\epsilon \mathrm{/}2$. Telescoping all cells and applying step 3.1 yields $\mathrm{\mid }F(t)-F(s)\mathrm{\mid }<\epsilon$. Arbitrary positive $\epsilon$ gives equality, and arbitrary $s,t$ prove claim 1.

For claim 2 with $a<b$, fix a coordinate $j$ and put $I_j(t)={\int }_a^t{h}_j(s)ds$. By [F6], $I_j$ is absolutely continuous and $I_j^{\mathrm{\prime }}=h_j$ almost everywhere. The function $H_j=y_j-y_j(a)-I_j$ is absolutely continuous: for each positive tolerance use the [F1] moduli for $y_j,I_j$ at half that tolerance and take their minimum. Outside the union of the two exceptional null sets, [F7] gives $H_j^{\mathrm{\prime }}=0$. Claim 1 makes $H_j$ constant, and $H_j(a)=0$, so $y_j(t)=y_j(a)+I_j(t)$ for all $t$. Applying this argument to the finitely many coordinates proves the displayed vector identity.