measure theory - Is every smooth function Lebesgue-Lebesgue measurable?

This is motivated by pure curiosity (triggered by this question). A map $f:mathbb R^ntomathbb R^m$ is said to be Lebesgue-Lebesgue measurable if the pre-image of any Lebesgue-measurable subset of $mathbb R^m$ is Lebesgue-measurable in $mathbb R^n$. This class of maps is terribly inconvenient to deal with but it might be useful sometimes. And maybe it is not that bad in the case $m=1$, especially if the answer to the following question is affirmative.

Question: Is every $C^1$ function $f:mathbb R^ntomathbb R$ Lebesgue-Lebesgue measurable? If not, what about $C^infty$ functions?

I could not figure out the answer even for $n=1$. However, there are some immediate observations (please correct me if I am wrong):

• Since the map is already Borel measurable, the desired condition is equivalent to the following: if $Asubsetmathbb R$ has zero measure, then $f^{-1}(A)$ is measurable.


• If $dfne 0$ almost everywhere, then $f$ is Lebesgue-Lebesgue measurable (because locally it is a coordinate projection, up to a $C^1$ diffeomorphism). So the question is essentially about how weird $f$ can be on the set where $df=0$.


• If the answer is affirmative for $C^1$, it is also affirmative for Lipschitz functions (by an approximation theorem).


• The answer is negative for $C^0$, already for $n=1$. An example is a continuous bijection $mathbb Rtomathbb R$ that sends a Cantor-like set $K$ of positive measure to the standard (zero-measure) Cantor set. There is a non-measurable subset of $K$ but its image is measurable since it is a subset of a zero-measure set.