Is every smooth function Lebesgue-Lebesgue measurable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:49:02Z http://mathoverflow.net/feeds/question/19459 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19459/is-every-smooth-function-lebesgue-lebesgue-measurable Is every smooth function Lebesgue-Lebesgue measurable? Sergei Ivanov 2010-03-26T20:44:18Z 2010-03-26T22:12:42Z <p>This is motivated by pure curiosity (triggered by <a href="http://mathoverflow.net/questions/19402" rel="nofollow">this question</a>). A map $f:\mathbb R^n\to\mathbb R^m$ is said to be <em>Lebesgue-Lebesgue measurable</em> 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.</p> <p><strong>Question</strong>: Is every $C^1$ function $f:\mathbb R^n\to\mathbb R$ Lebesgue-Lebesgue measurable? If not, what about $C^\infty$ functions?</p> <p>I could not figure out the answer even for $n=1$. However, there are some immediate observations (please correct me if I am wrong):</p> <ul> <li>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.</li> <li>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 <code>$df=0$</code>.</li> <li>If the answer is affirmative for $C^1$, it is also affirmative for Lipschitz functions (by an approximation theorem).</li> <li>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.</li> </ul> http://mathoverflow.net/questions/19459/is-every-smooth-function-lebesgue-lebesgue-measurable/19468#19468 Answer by Anton Petrunin for Is every smooth function Lebesgue-Lebesgue measurable? Anton Petrunin 2010-03-26T22:12:42Z 2010-03-26T22:12:42Z <p>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 $C^\infty$. </p> <p>Am I missing something?</p>