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.

1 Answer 1


It seems that your example of bijection that sends one Cantor set with positive measure to another Cantor set with zero measure can be made $C^\infty$.

Am I missing something?

  • $\begingroup$ Probably you are right, this possibility did not occur to me. My first impression is that it will stretch the complement intervals too much for that. Should that bigger Cantor set be specially crafted somehow? $\endgroup$ Mar 26, 2010 at 22:36
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    $\begingroup$ 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. $\endgroup$ Mar 26, 2010 at 23:18

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