Preimage of null sets under a monotone increasing function

Question

Let $I\subseteq \mathbb{R}$ be a closed bounded interval and $f:I \to I$ a monotonic increasing function and $S$ the countable set of points $s$ such that $|f^{-1}(s)| > 1$. Is the following conjecture true?

For every Lebesgue null set $N \subseteq I\setminus S$, the preimage $f^{-1}(N)$ is again a Lebesgue null set.

Edit: As pointed out by Christian Remling in the comments, it seems that this is false in general. Are there assumptions on $f$ that guarantee that the conjecture holds? I am particularly interested in whether absolute continuity of $f$ is sufficient.

Answer 1

I will answer in the negative the question "If $f:I\to I$ is a strictly increasing absolutely continuous function, does $f^{-1}$ send Lebesgue null sets to Lebesgue null sets?"

Indeed, suppose that $f$ is an absolutely continuous strictly increasing bijection on $I$, such that all preimages of nullsets are null. Write $g:=f^{-1}$. Note that, under the hypotheses, $f$ and $g$ are homeomorphisms. Additionally, $m(f^{-1}(A))=m(g(A))$ is null for all $A\subseteq I$ null, so $g$ maps null sets to null sets, i.e. $g$ satisfies the Luzin property.

It is a classical fact that a function is a.c. if and only if it is continuous, BV, and has the Luzin property. Since $g$ is continuous and has the Luzin property, and is a bounded increasing function, it follows that $g$ is a.c.

Thus, $f$ has to also have the property that its inverse is a.c. On the other hand, strictly increasing a.c. functions need not have a.c. inverses; indeed, if $g=c(x)+x$ with $c:[0,1]\to[0,1]$ the Cantor function, and $f=g^{-1}$, then $g$ is clearly not a.c., but $f$ is Lipschitz (hence a.c.) and strictly increasing.

As suggested in the preceding, the correct condition has something to do with $g=f^{-1}$ being absolutely continuous, which might be thought of as $f'$ being not too small (i.e. $\frac{1}{f'}$ being integrable).

Answer 2

My colleague and I came up with a proof for the following related Theorem:

Let $I \subseteq \mathbb{R}$ be an interval, $f:I\to I$ an absolutely continuous monotone increasing function. Define $I_{f'>0}:=\{t \in I\mid f'(t) \text{ exists and } f'(t) >0\}$. Then $f$ fulfills Lusin's $N^{-1}$ property on $f(I_{f'>0})$, i.e. for every null set $\mathcal{N}\subseteq f(I_{f'>0})$, we have that $f^{-1}(\mathcal{N})$ is a null set as well.

Proof sketch: One can show that $f$ is injective on $I_{f'>0}$ and hence admits an inverse $g: f(I_{f'>0}) \to I$. Extending this $g$ to the whole set in a monotone manner to $\tilde{g}$ yields a function that is differentiable on every (not almost every) point in $f(I_{f'>0})$. This property allows for every null set $N \subseteq f(I_{f'>0})$ the estimate $\sigma(\tilde{g}(N)) \leq \int_N \tilde{g}'d\sigma = 0$ where $\sigma$ is the Lebesgue measure (see (1), Chapter 7). Observing that $f^{-1}(N) = \tilde{g}(N)$ due to $N \subseteq f(I_{f'>0})$ finishes the proof.

Remark: It seems that the absolute continuity of $f$ may be relaxed.

(1) Bruckner, Andrew M.; Bruckner, Judith B.; Thomson, Brian S., Real analysis, International Edition. Upper Saddle River, NJ: Prentice-Hall International. xiv, 713 p. (1997). ZBL0872.26001.