Let $E\subseteq\mathbb{R}$ be a borel measurable set with $m(E)=0$ and $f(x)=x^{2}$. Is $m(f(E))=0$?

Question

Let $E\subseteq\mathbb{R}$ be a Borel measurable set with $m(E)=0$ and $f(x)=x^{2}$. Is $m(f(E))=0$?

I think it is true, but I do not know how to prove it. The only think I have got is that, if $m(f(E))>0$, then there exists $A\subseteq E$ such that $A$ is not Lebesgue measurable. I don't know how to follow from that.

Answer 1

Yes. The function $f(x) = x^2$ is absolutely continuous, because it can be expressed as an integral function: $$f(x) = \int_0^x 2t\ {\rm d}{\frak m}(t).$$ And if $f$ is absolutely continuous and ${\frak m}E = 0$, then ${\frak m}(f(E)) = 0$.

Answer 2

It's enough to show that $f(E)$ has measure $0$ if $E$ is a bounded set of measure $0$. Assume $E$ bounded, $E \subset [-r, r]$. On the interval $[-r,r]$ $f$ has a bounded derivative so it's Lipschitz with constant $k$. But then it's easy to see that for every subset $A$ of $[-r,r]$ we have $m^*(f(A)) \le k\cdot m^*(A)$.

For Lipschitz maps $f\colon \mathbb{R}^m$ to $\mathbb{R}^m$ the method does not work if $m>1$ since the diameter and the measure of sets do not coincide for images of rectangles. However, the result about image of sets of measure $0$ still holds if $f$ is smooth, basically an application of the change of variables (in)equality: $m(f(E)) \le \int_E | \det J_f|$