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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 $\mathbb{R}^n$

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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 – AS_Butler Oct 15 '13 at 16:25
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

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.


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