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If $m^*(E)=0$ show that $m^*(E^2)=0$ where $E^2=\{x^2:x\in E\};E\subseteq \Bbb R$.

EDITS(Please don't close this): I did not get a satisfactory answer.So I posted my approach. Here is my approach:

Since $m^*(E)=0\implies \exists $ a sequence of intervals $\{I_n\}$ such that $E\subset \cup I_n$ and $\sum l(I_n)<\epsilon $ for all $\epsilon>0$.

Now since $E\subset \cup I_n\implies E^2\subset \cup I_n^2$ because let $x^2\in E^2\implies x\in E\implies x\in I_n$ for at least one $n\in \Bbb N\implies x^2\in I_n^2\subset \cup I_n^2$.

Also $l(I_n)<\sum l(I_n)<\epsilon\forall n\in \Bbb N\implies l(I_n^2)<\epsilon^2\implies \sum l(I_n^2)<\epsilon^2\implies m^*(E^2)=0$.

Please help me to check whether it is correct or not .If not where are the loopholes in the proof.Please point them.

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The function $x\mapsto x^2$ is Lipschitz on compact sets. Note that a Lipschitz function maps null sets to null sets.

So consider $E_n^2=E^2\cap B_{n}(0)$ and $E_n$ defined similarly. By the second proof, $m(E_n^2)=0$. Note that $E^2=\cup_{n\in \Bbb{N}}E_n^2$ and the union of null sets is null.

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