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I know the statement " If $E$ is a subset of real numbers with measure zero,then so is $E^2$={$x^2$: $x\in E$} " is true. How about " If $E$ is a subset of real numbers with positive measure, then so is $E^2$={$x^2$: $x\in E$} " ? Is true or false ?

C-HJ
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  • I think it's true, prove the contrapositive, if the measure of $E^2$ is $\le 0$, i.e. it is 0, then the measure of $E$ is 0. – simonzack Jun 03 '15 at 13:18

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If $E$ has positive measure, then at least one of $E\cap\mathbb{R}_+$ and $E\cap\mathbb{R}_-^\ast$ has positive measure. Assume without loss of generality that $E\cap\mathbb{R}_+$ has positive measure. Then $(E\cap\mathbb{R}_+)^2$ must have positive measure as $f\colon x\in\mathbb{R}_+\mapsto x^2$ is a differentiable bijection.

But $(E\cap\mathbb{R}_+)^2\subseteq E^2$, so $E^2$ also has positive measure.

Clement C.
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    In general, bijections do not preserve positive measure. But differentiable bijections do. – GEdgar Jun 03 '15 at 13:35
  • My bad -- I edited the answer. – Clement C. Jun 03 '15 at 13:35
  • Expanding on @GEdgar's comment, a great example of this is the Cantor set, which can have a bijective correspondence to the interval $[0,1]$. The Lebesgue measure of the cantor set is zero, while the Lebesgue measure of the interval is 1. – Joel Jun 03 '15 at 14:23