User ben hayes - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:39:57Z http://mathoverflow.net/feeds/user/5994 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24059/is-the-absolutely-continuous-image-of-a-nowhere-dense-set-is-also-nowhere-dense/24063#24063 Answer by Ben Hayes for Is the absolutely continuous image of a nowhere dense set is also nowhere dense? Ben Hayes 2010-05-10T05:28:49Z 2010-05-10T05:28:49Z <p>I think this is a counter-example.</p> <p>Let $C$ be a cantor set of positive measure, so $C$ is nowhere dense, perfect and is the countable decreasing intersection of sets $C_{n}$ each of which are a finite union of closed disjoint intervals in $[0,1]$. </p> <p>Let $f(x)=\int_{0}^{x}\chi_{C}(t)\mbox{ }dt,$ so $f$ is certainly absolutely continuous. </p> <p>Assuming I did this right $f(C)$ contains $[0,m(c)).$ Consider each $C_{n},$ since $f$ is increasing and continuous $f([0,1])$ contains $[0,m(C)].$ Also $[0,1]\setminus C_{n}$ is a union of open intervals each of which are in the compliment of $C$ so it follows that $f$ is constant on each such interval. Since $f([0,1])=[0,m(C)]$ it follows that for any $x\in [0,m(C))$ we can find $t\in C_{n}$ so that $f(t)=x.$ Indeed we already know we can do this with $t\in [0,1]$ but since $f$ is constant on the intervals in the complement of $C_{n}$ we can force $t\in C_{n}$ (for instance if $x$ is in some interval $I$ in $[0,1]\setminus C_{n}$ then its left endpoint is in $C_{n}$ and since $f$ is constant on $I$ we have that $f$ has the same value at the left-endpoint of $I$ as on $I$.)</p> <p>Now fix $y\in [0,m(C)).$ Since $\lbrace x\in C_{n}:f(x)=y\rbrace$ is non-empty and these sets are decreasing, (since the $C_{n}$ are decreasing) by compactness we can find $x\in C$ so that $f(x)=y.$ Thus $f(C)=[0,m(C))$ and we have found a nowhere dense set which is mapped to a set which is not nowhere dense.</p>