$f$ is a non-constant polynomial, $A $ is a set of measure zero, Is this true that $m(f^{-1}A)=0$, where $m$ stands for the Lebesgue measure. Let $f$ be a non-constant polynomial, and let $A \subset \mathbb{R}$ be a set of measure zero, Is this true that $m(f^{-1}A)=0$, where $m$ stands for the Lebesgue measure.
If $A$ is a countable set, it is easy to see that $Cardinality (f^{-1}A) \leq Cardinality (\mathbb{N} * \mathbb{N})$ and it means that $m(f^{-1}A)=0$.
My problem in proving that this is true, is sets like Cantor set, whose measure is zero but is uncountable.
Any hints or ideas is appreciated. Thanks !
 A: following @AlexBecker idea :
Suppose $f$ is a polynomial of degree $n$, so it has at most $n-1$ pieces, each of which is monotone. Partition $E$ such that $E=E_1 \cup E_2 \cup ... \cup E_{n-1}$ such that $f$ restricted to each $E_k$ is (strictly) monotone. We can say $$f=f|_{E_1}+f|_{E_2}+...+f|_{E_{n-1}}$$
Therefore we have : $$f^{-1}A=(f|_{E_1})^{-1}A \ \bigsqcup \ (f|_{E_2})^{-1}A \ \bigsqcup \ ... \bigsqcup \ (f|_{E_{n-1}})^{-1}A \ $$
If I show that for each $1 \leq k \leq n-1 $ the set $C_k=(f|_{E_{k}})^{-1}A$ has zero measure, we are done.
Suppose that's not the case, i.e. $mC_k=r_k>0$, and note that since $f|_{E_{k}}$ is strictly monotone on $E_k$, we can find $m_k>0$ (monotone increasing) or $m_k<0$ (monotone decreasing) such that $f'(x) \geq m_k$ for all $x \in E_k$. Therefore we have that $$|m_k| \leq \frac{m \left[f|_{E_{k}}C_k \right]}{m C_k} \Longrightarrow 0 < r_km_k \leq  m \left[f|_{E_{k}} C_k \right] \leq mA \ \ \ (by \ \ Monotonicity)$$
A contradiction. Therefore $$mf^{-1}A=m(f|_{E_1})^{-1}A \ + \ m(f|_{E_2})^{-1}A \ + \ ... + \ m(f|_{E_{n-1}})^{-1}A \ = 0 + 0 + ... + 0 = 0 $$
