Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a map $f: \mathbb{R}^n \to \mathbb{R}^m$ that is differentiable (usually even smooth). If $B \subset \mathbb{R}^m$ has measure zero (Lebesgue measure), then what types of maps $f$ satisfy $A = f^{-1}(B)$ also has measure zero?

To provide some context: I have a property $\mathcal{P}$ that holds almost everywhere in $\mathbb{R}^m$; now I want to characterize the class of maps $f$ such that $\mathcal{P}(f(\cdot))$ holds almost everywhere in $\mathbb{R}^n$

share|improve this question
1  
If the Jacobian $Df$ has rank $n$ (which necessarily implies $n \leq m$) at each point, then $f$ is locally a diffeomorphism, and hence, locally, has Lipschitz inverse. A Lipschitz function takes zero sets to zero sets, so $f^{-1}$ takes zero sets to zero sets (since we can break up the pre-image into countably many pieces contained in small open sets where $f$ is invertible). –  user15464 Apr 5 '12 at 2:34
    
I'm no expert, and certainly cannot provide an exhaustive list, but I know that absolutely continuous functions have this property –  Amin Saied Oct 15 '13 at 16:25
2  
Actually, they do not. Consider for instance the constant function. Preimage of a single point (value of the function) is the entire space and, hence, has positive (infinite) measure. You are confusing images and preimages: AC functions $R\to R$ map measure zero sets to measure zero sets. In higher dimensions this is called "Luzin N Property". –  studiosus Oct 15 '13 at 16:47
    
@Lord_Farin: This condition is clearly not sufficient, think about constant functions. Amin was confusing images and preimages. OP was asking about maps such that the preimage of a measure zero set again has measure zero. –  studiosus Oct 15 '13 at 16:56
    
@studiosus You are correct. I was in a review queue, so I had limited opportunity or motivation to verify the correctness of the assertion beyond that it didn't answer the question. –  Lord_Farin Oct 15 '13 at 17:04

1 Answer 1

I was searching for a solution for the problem that you are working on. I found an interesting paper about it:

Ponomarev, S. P. - Submersions and preimages of sets of measure zero.pdf

It says that, if a function is "sufficiently" smooth and the rank of the Jacobian matrix satisfy some properties, then the preimage has also measure zero.

Best,

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.