Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be an  *[absolutely continuous](http://en.wikipedia.org/wiki/Absolute_continuity)*  map. Does $f$ map a nowhere dense subset of $[a, b]$ to a nowhere dense set?


Remarks:


1. The answer is "no" if $f$ is only assumed to be continuous and almost everywhere differentiable, e.g. take the  [Cantor function](http://en.wikipedia.org/wiki/Cantor_function) .
2. If $f$ is assumed to be $C^1$, then the answer is yes - a nice proof can be found at [this page.](http://sci.tech-archive.net/Archive/sci.math.research/2005-02/0171.html) Essentially the same proof works if it is assumed that $f$ is not differentiable at at most countably many points. 
 **Edit:**  I would like to retract the previous sentence. Now I don't see why it should be true.



Is the absolutely continuous image of a nowhere dense set is also nowhere dense?
I think this is a counter-example.


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 $[0,1]$. 


Let $f(x)=\int\_{0}^{x}\chi\_{C}(t)\mbox{ }dt,$ so $f$ is certainly absolutely continuous. 


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 $I$.)


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.


