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?



A modified Cantor and its measure
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.


