The question is can we prove the following: Consider $a,b\in {\bf R}$, $a<b$.
(1) If $u\in {\rm BV}([a,b])$, and
(2) if $u$ is an injection on $[a,b]$,
then there exists $u_0:[a,b]\rightarrow {\bf R}$ such that the following holds:
(3) $u(s)=u_0(s)$ (a.e. $s\in [a,b]$),
(4) $u_0$ is "piecewise strictly monotonic on $[a,b]$", meaning that there exists at most countably many disjoint sub-intervals $(a_k,b_k)$ such that $u_0$ is strictly monotonic on $(a_k,b_k)$ for every $k\in{\bf N}$ and such that $[a,b]=\cup_{k=1}^{+\infty}(a_k,b_k)\cup N$, where $\lambda(N)=0$.
Remark (I). Our question is a slight reformulation of the question
Bounded variation + injective implies strictly monotonic?
which was settled by counterexamples of Gerd and Kurt G., and it is inspired by the aforementioned counterexamples for the original version of the question.I note that examples of Gerd and Kurt G. do not provide a counterexample for the reformulated version of the question which we are considering here.
Remark (II). Here the set ${\rm BV}([a,b])$ is defined as the set of all functions on $[a,b]$ with finite total variation (not essential total variation!).Therefore, if two functions $u$ and $v$ agree almost everywhere on $[a,b]$, it can happen that $u$ belongs to ${\rm BV}([a,b])$, and that $v$ does not belong to ${\rm BV}([a,b])$.
Remark (III). We do not claim that $u_0$ belongs to ${\rm BV}([a,b])$, and we do not claim that $u_0$ is injective on $[a,b]$. Still, it would be good(but not essential) if we could obtain either of these additional properties(but this is a digression).
Remark (IV). If such a function $u_0$ exists, we could say that $u_0$ is "a piecewise strictly monotonic representative" of $u$, and we could say, in that sense, that "$u$ is piecewise strictly monotonic almost everywhere"(hence the title of this question).