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I'm attempting to find a solution to the following problem that doesn't involve splitting this into various cases. The question is:

"If $m^*(E) =0$, show that $m^*(E^2) = 0$, where $E^2 = \{x^2 \colon x \in E\}$ and $ E \subset \mathbb{R}$."

Here $m^*$ represents Lebesgue outer measure.

This problem can be reduced to the case where $E$ is bounded, however, when covering $E$ with a union of open intervals, say $(I_n)$, where $\sum_{n=1}^{\infty} \ell(I_n) \leq m^*(E) + \epsilon$ for any given $\epsilon >0$, I find that I reach a problem if any elements of $E$ belong to $(-1,1)$. This can be remedied by breaking the real line up into parts, however, if there's a more efficient way to approach this problem, please let me know.

mathgenesis22813
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2 Answers2

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The quickest way I know how to solve this problem makes use of Lipschitz continuity, but you'll still need to break things down into intervals.

There is a theorem in basic measure theory that says Lipschitz continuous functions preserve measure $0$ sets (you can read a proof on a previous stack exchange thread). The function $f$ given by $x\mapsto x^2$ is Lipschitz continuous on compact intervals $[-L, L]$: $$|x^2-y^2| = |x+y|\cdot|x-y|\leq 2L|x-y|.$$ Therefore $f(E\cap [-k,k])$ is measure $0$ for every $k$. Now we have $$E^2 = f(E) = f\left(\bigcup_{k=1}^\infty E\cap [-k, k]\right)\subset \bigcup_{k=1}^\infty f(E\cap [-k,k]).$$ Unions of measure $0$ sets are measure $0$, so we may conclude.

Even if using the fact about Lipschitz functions isn't quite kosher in your situation (this sounds like a homework question), you could probably adapt the estimate and the proof in the thread I linked to write a new self-contained one.

Community
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Glare
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There is no need to worry about if $E$ is bounded. $m^*(E) = 0$ means that for any $\varepsilon > 0$ there is a colllection $\{I_n\}$ of intervals with $E \subseteq \bigcup I_n$ and $\sum_{n=1}^{\infty} \ell(I_n) \leq \epsilon$. Now, look at $\{I_n \times I_m\}_{n,m}$, and recall that $\Bbb N \times \Bbb N \sim \Bbb N$.

Paul Sinclair
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    Is $I_n \times I_m$ the Cartesian product? It is about $E^2 = { x^2 : x\in E }\subseteq\mathbb R$, not about $E\times E \subseteq \mathbb R \times \mathbb R$. – user251257 Aug 19 '15 at 01:50
  • @user251257 - Good grief. I've got to start reading these questions more carefully! – Paul Sinclair Aug 19 '15 at 02:07