My question is: can we find the function $g\in{\rm C}([a,b])$ which satisfies the Luzin N property on $[a,b]$, but which does not satisfy the Banach S property on [a,b]? Here $[a,b]$ is a compact non-empty interval in ${\bf R}$.
Reminder:
Def. 1. A function $g:[a,b]\rightarrow {\bf R}$ is said to satisfy the Luzin N property on $[a,b]$ if the following holds: For arbitrary measurable set $E\subseteq [a,b]$ it follows that $\lambda(E)=0$ implies $\lambda(g(E))=0$.
Def. 2. A function $g:[a,b]\rightarrow {\bf R}$ is said to satisfy the Banach S property on $[a,b]$ if the following holds: For every $\varepsilon>0$ there exists $\delta>0$ such that for arbitrary measurable set $E\subseteq [a,b]$ it follows that $\lambda(E)\leq\delta$ implies $\lambda(g(E))\leq\varepsilon$.
Remark.0. By assuming the opposite, it is easy to see that the Banach S property implies the Luzin N property. The converse in general should not be true. So I am looking for a counterexample(if possible, the counterexample should be constructed within the class of continuous functions on compact interval $[a,b]$.
Remark.1. Note that the counterexample can not be a function which belongs to $BV([a,b])$. This is beacuse Banach-Zaretsky Theorem says that a function is continuous, BV and satisfies N property on $[a,b]$, if and only if it is absolutely continuous on $[a,b]$. On the other hand, absolutely continuous functions ($AC$-functions) satisfy the Banach S property (see statement 5 in the question
Extension of Luzin N-property of absolutely continuous functions and measure preserving mappings
and the answer therein);
Remark.2. Counterexample can not be a function which satisfies $g'(s)=0$ almost everywhere on $[a,b]$. This is because Theorem 4.6.4. ("Measure-theoretic characterization of 0-derivative") in textbook Benedetto, Czaja: "Integration and Modern Analysis"(Birkhauser, 2009) says that we have: $g'(s)=0$ (a.e. $s\in[a,b]$) if and only if $\lambda(g([a,b]))=0$. As a consequence, such a function $g$ satisfies the Banach S property, so it can not be a counterexample.
Remark.3. Here is somewhat related question:
Example of a function that has the Luzin $n$-property and is not absolutely continuous.
It is mentioned therein that $f_0(x)=x{\rm sin}({{1}\over{x}})$, $x\neq 0$, $f_0(0):=0$, is an example of the continuous function which satisfies Luzin N property (this follows from the fact that $f_0$ is locally Lipschitz in ${\bf R}$∖$\{0\}$, compare
), while $f_0$ is not $BV$, so it is not $AC$.
My comment (which is open to re-checking) here is that this particular $f_0$ satisfies the Banach S property (this also follows from the fact that $f_0$ is locally Lipschitz in ${\bf R}$∖$\{0\}$).
Remark.4. In particular, this example shows that we can find a function $g\in {\rm C}[a,b]$ which satisfies the Banach S property on $[a,b]$, but which does not belong to ${\rm AC}([a,b])$.
Remark.5. My motivation in considering this issue comes from the statement in
https://encyclopediaofmath.org/wiki/Luzin-N-property
where it is claimed that (and I quote):
assertion $(*)$: "S. Banach proved that a function $f$has the S property if and only if the inverse image $f^{\leftarrow}({x})$ is finite for almost-all $x$ in $f([a,b])$." (My remark. Here it is assumed that $f\in {\rm C}([a,b])$).
I suspect that such an equivalence is not true after all. I was not able to find this equivalence in literature (for instance, in Banach's original paper from 1927., it is only established that$g\in {\rm C}([a,b])$ belongs to ${\rm BV}([a,b])$ if and only if $\xi\mapsto {\rm card }g^{\leftarrow}({\xi})$ is Lebesgue integrale on ${\bf R}$ (compare also
). So, in particular, if $g\in {\rm C}([a,b])$ and ${\rm BV}([a,b])$, then ${\rm card }\{g^{\leftarrow}({\xi})\}<+\infty$ (a.e. $\xi$). Now, if the assertion $(*)$ is true, it follows that $g$ satisfies Banach S property, so $g$ satisfies Luzin N property, and so, by the Banach-Zaretsky Theorem, $g$ is AC-function, which is obviously not a true conclusion.
On the other hand, in the paper G. Mokobodzki: Ensembles à coupes dénombrables et capacités do minées par une mesure (Sém Proba XII, L.N. n°649, p 491-508, Springer 1978)},I would think that it is proved that the following implication holds: If a function $f$ satisfies the Banach S property, then it holds that $f^{\leftarrow}({\xi})$ is finite for almost-all $\xi$ in $f([a,b])$. But I do not claim to understand the proof therein, since it is given in much more general setting then the one which we discuss here.
My conclusion: I think that there is definitely an issue to be settled here. Am I on the right track here, or not?