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Recall that Cantor set can be defined as the set of numbers in $[0,1]$ that don't contain $1$ when written in ternary number system. Alternatively if we consider the map $\varphi: [0,1]\to [0,1]$, $x\to (3x\mod 1)$, then Cantor set consists of points whose orbits does not intersect the interval $(\frac{1}{3}, \frac{2}{3})$.

Suppose now we modify the map $\varphi$ in such a way that $x$ goes to $f(x)\mod 3$, where $f(x): [0,1]\to [0,3]$ is smooth and byjective, and $f'>1$. Call this map $\varphi_f$.

Question. Define a new Cantor set $C_f$ as the set of all points in $[0,1]$ whose orbit under $\varphi_f$ does not intersect $f^{-1}(1,2)$. Is it true that $C_f$ has measure zero? If not, what additional conditions could one impose on $f$ so that measure of $C_f$ is zero?

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This is called a cookie cutter. If by smooth, you mean $f$ is $C^{1+\epsilon}$ or smoother, then it's known that $f$ preserves a fully supported absolutely continuous invariant measure on $[0,1)$. In particular, almost every point enters the middle interval (there's no reason this should be $(\frac 13,\frac 23)$). So that the set of points that never enter the middle interval will be of Lebesgue measure 0 by the Birkhoff ergodic theorem.

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  • $\begingroup$ Thanks Anthony! Do you know some nice reference on cookie cutter? $\endgroup$
    – aglearner
    Commented Feb 22, 2017 at 23:28
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    $\begingroup$ A sufficient theorem to answer your question is Krzyzewski and Szlenk "On invariant measures for expanding differentiable mappings". It is not phrased in terms of cookie cutters. The more refined question for cookie cutters is: what is the dimension of the Cantor set? If you search for "cookie cutter", you will find many papers on this. $\endgroup$
    – Anthony Quas
    Commented Feb 23, 2017 at 1:16

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