# $f$ 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 !

• This is a nice one. How did you find this question? – Patrick Da Silva Dec 23 '13 at 17:45
• This is a part of my solution to another problem, so I "need" this to be true, otherwise my attempt would not be the best. Original Question is a very general statement : If $E$ is a measurable set, and $f$ is a polynomial, Prove that $f^{-1}E$ is measurable. – the8thone Dec 23 '13 at 17:49
• You can probably take advantage of the fact that a polynomial is invertible almost everywhere with continuous inverse. – Alex Becker Dec 23 '13 at 17:51
• @PatrickDaSilva That's what I meant. I think you can piece together the local inverses to get only finitely many neighborhoods total, but I haven't worked it out. – Alex Becker Dec 23 '13 at 17:55
• Thanks you guys ! I'm gonna try that idea – the8thone Dec 23 '13 at 17:58

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$$
• non-constant ${}{}$ – mike Dec 23 '13 at 19:40