Continuous function which satisfies the Luzin N property, but which does not satisfy the Banach S property

Question

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

Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?

), 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

Banach Indicatrix Function

). 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?

Answer 1

For an $\epsilon>0$ and finite unions of half-open intervals (FOHOI for short) $J=\cup_{n=1}^N [a_n, b_n)$ and $J'=\cup_{k=1}^K [c_k, d_k)$, where $a_1<b_1<a_2<\cdots <a_N<b_N$ and $c_1<d_1<c_2<\cdots <c_K<d_K$, let us say that $J$ $\epsilon$-evenly covers $J'$ if

• $J\subseteq J'$;
• $\{c_k \mid 1\leq k \leq K\}\subseteq \{a_n \mid 1\leq n\leq N\}$;
• $\{b_n \mid 1\leq n\leq N\}\cap \{d_k \mid 1\leq k \leq K\}=\emptyset$;
• For all $n$ such that $[a_{n}, a_{n}+\epsilon]\subseteq J'$ we have $n<N$ and $a_{n+1}-a_n\leq \epsilon$.

Fix a sequence of strictly positive numbers $\alpha_n$ such that $\sum_n \alpha_n <1/2$. Inductively we construct a sequence of disjoint FOHOIs $\{J_n\}_n$ contained in $[0, 1)$, such that $\lambda(J_n)<\alpha_n$ and $J_n$ is a $1/n$-even covering of $[0, 1)\backslash \cup_{m=1}^{n-1} J_m$ (observe that the latter set is again a FOHOI. Interpret it as $[0,1)$ for $n=1$).

Define a function $h: \bigcup_n J_n \rightarrow \mathbb{R}$ as follows. For each $n$ write $J_n=\bigcup\limits_{k=1}^{K_n} [a_k^n, b_k^n)$, $a_1^n<b_1^n<a_2^n<\cdots<a_{K_n}^n<b_{K_n}^n$, and on each $[a_k^n, b_k^n)$ interval let $h$ be the tent function which is $1/n$ at the midpoint of that interval and $0$ at the endpoints. Extend $h$ by $0$ to a function $[0,1] \rightarrow \mathbb{R}$.

Consider the function $$f(x)=x+h(x), \quad x\in[0,1].$$ By construction $f(J_n)\supseteq [0, 1]\backslash \cup_{m=1}^{n-1}J_{m}$ for each $n$ (*), therefore $f$ does not have the Banach (S) property. However, since $f$ is Lipschitz on each $J_n$ and is the identity on $[0,1]\backslash \cup_{n=1}^\infty J_n$, it has the Luzin (N) property. The continuity of $f$ follows from that of $h$; $h$ is evidently continuous on $\bigcup_n J_n=\bigcup_n \overline{J_n}$. If $x\notin\bigcup_n \overline{J_n}$ then for any $N$ we can find a neighbourhood of $x$ avoiding $J_1, J_2, \dots$ and $J_N$ and on that neighbourhood we have $h\leq \frac{1}{N+1}$, hence $h$ is continuous at $x$.

(*) The point is that, by the $1/n$-evenly covering property, as $x$ races over the interval $[a_k^n, b_{k}^n)$, $f(x)$ either races over (at least) the interval $[a_k^n, a_{k+1}^n)$ or (at least) everything from $a_k^n$ to next left endpoint in $\bigcup_{m=1}^{n-1} J_m$ or everything from $a_k^n$ to $1$.