# Preimage of zero measure sets

Denote by $|A|$ the measure of $A$ (Can be Lebesgue measure) under what conditions on a function $f:\mathbb{R}^m \to \mathbb{R}$ the preimage of a null set is zero. i.e.

$|A|=0 \Rightarrow |f^{-1}(A)| =0$

A special interest for conditions on not necessarily smooth functions

• What kind of conditions do you have in mind? In some sense constant functions are about as nice as you can have and they don't have this property. Apr 23, 2013 at 13:28
• There is a follow-up question: math.stackexchange.com/questions/3215996/… Apr 10, 2020 at 14:25
• It's Lusin's $N^{-1}$ condition. The answer is in $\S$ 4.4 "Lectures on Mappings of Finite Distortion" by S. Hencl, P. Koskela. Aug 25, 2023 at 17:29

For example, it is sufficient that $f\in C^1$ and the set $\lbrace x | \nabla f(x) = 0 \rbrace$ has measure zero.

To prove this, note that this is true locally, in a neighborhood of each point where $\nabla f \neq 0$, due to the implicit function theorem. Now the claim follows from the fact that $f^{-1}[A] \subset Z \cup \bigcup_n (U_n \cap f^{-1}[A])$, where $Z = \lbrace \nabla f = 0 \rbrace$ and $(U_n)$ is a countable covering of $\mathbb{R}^m \setminus Z$ by neighborhoods for which the implicit function theorem applies.

• What about $f \equiv 1$? Apr 23, 2013 at 13:39
• @Noah: I guess you didn't read carefully :) Apr 23, 2013 at 13:42
• What does $f'(x)=0$ mean if $m>1$? Apr 23, 2013 at 13:46
• @Emil: I mean that the gradient is zero. Apr 23, 2013 at 13:53
• @Noah,Emil: Sorry, I never knew that notation could be confusing. Apr 23, 2013 at 13:57

This isn't really an answer, but .... If you change "pre-image" to "image", then you are asking about Lusin's Condition (N). Despite being well-studied, a characterization of such functions would seem to be a fairly deep problem. Most results on condition (N), it would seem, are counterexamples, such as the construction of functions in certain Sobolev spaces which fail to obey condition (N). The positive results are somewhat scarcer. For instance, though the 1 dimensional case has been solved, as I understand it the 2 dimensional case is far from it, though there are some known sufficient conditions. Amongst these is that the Jacobian mapping $Jf$ of a $W^{1,n}$ function be strictly positive, or that a $W^{1,n}$ mapping be positive and open.

Some of these criteria could, obviously, imply something about what you are asking.

That said, this early in the morning, I am not certain whether your question is actually much easier than asking about condition (N), and could not, perhaps, receive a full resolution here on MO.