# Is every smooth function Lebesgue-Lebesgue measurable?

This is motivated by pure curiosity (triggered by this question). A map $f:\mathbb R^n\to\mathbb R^m$ 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 $f:\mathbb R^n\to\mathbb R$ 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 $A\subset\mathbb R$ has zero measure, then $f^{-1}(A)$ is measurable.
• If $df\ne 0$ 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 $\mathbb R\to\mathbb R$ 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.
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It seems that your example of bijection that sends one Cantor set with positive measure to an other Cantor set with zero measure can be made $C^\infty$.
Yup, got it. The function is an antiderivative of a nonnegative $C^\infty$ function whose set of zeroes is a Cantor set of positive measure. –  Sergei Ivanov Mar 26 '10 at 23:18