Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be an absolutely continuous map. Does $f$ map a nowhere dense subset of $[a, b]$ to a nowhere dense set?
Remarks:
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The answer is "no" if $f$ is only assumed to be continuous and almost everywhere differentiable, e.g. take the Cantor function .
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If $f$ is assumed to be $C^1$, then the answer is yes – a nice proof can be found at this page. 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.
Best Answer
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.