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.

No. If $f$ is strictly increasing, then you are asking if $g=f^{-1}$ maps null sets to null sets, which is false for singular continuous functions. — Christian Remling
– Christian Remling, Commented Jan 18, 2024 at 22:01
@ChristianRemling : Do you know the answer if we additionally require that $f$ be continuous? — Iosif Pinelis
– Iosif Pinelis, Commented Jan 19, 2024 at 1:51
@IosifPinelis: This is the situation I had in mind: $g=f^{-1}$ a strictly increasing, singular continuous (as in singular + continuous) function, so that $f$ will also be strictly increasing and continuous. — Christian Remling
– Christian Remling, Commented Jan 19, 2024 at 20:25
A simple and standard counterexample: let $C'$ be a fat Cantor set obtained by removing the middle $1/2^n$ at each stage, and $C$ be the standard (null) Cantor set obtained by removing the middle third at each stage, both in $I=[0,1]$. Construct $f\colon I\to I$ by taking the increasing bijective affine map from each open interval component of the complement of $C'$ to the corresponding one of $C$, and extend it by continuity to $I$. Then $f\colon I\to I$ is a continuous increasing bijection and $f^{-1}(C) = C'$, yet $C$ is null while $C'$ is not. — Gro-Tsen
– Gro-Tsen, Commented Jan 19, 2024 at 23:01

Ben Johnsrude · Accepted Answer · 2024-01-19 21:09:58Z

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).

Julian · Accepted Answer · 2024-01-24 13:31:31Z

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.

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