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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.

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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 ${\frak m}(f(E)) = 0$.

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  • $\begingroup$ Ok, thank you. Do you know any book where this is proved? $\endgroup$ – Srinivasa Granujan Jan 14 '15 at 15:14
  • $\begingroup$ If I recall correctly, in Royden's Real Analysis... $\endgroup$ – Ivo Terek Jan 14 '15 at 15:16
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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: $m(f(E)) \le \int_E | \det J_f|$

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  • $\begingroup$ Can you use the change of variable inequality if $f$ is not a diffeomorphism? $\endgroup$ – Srinivasa Granujan Jan 14 '15 at 17:07
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    $\begingroup$ You can, that's why you get an inequality. The change of variable is mostly stated, rarely proved; so this is not readily noticed, although known by specialists. $\endgroup$ – orangeskid Jan 15 '15 at 1:12
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    $\begingroup$ Btw, I corrected a typo, should be only about maps $\mathbb{R}^m \to \mathbb{R}^m$. $\endgroup$ – orangeskid Jan 15 '15 at 1:15

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