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Let E be a subset of $\mathbb{R}$ with lebesgue measure zero. How can I prove that $\lbrace x^2 : x\in E\rbrace$ also has lebesgue measure zero?

Let $\epsilon>0$, I should find a cover of $\lbrace x^2 : x\in E\rbrace$ with total length at most $\epsilon$. since $m(E)=0$ so there exists a countable collection of intervals $\{ (a_k, b_k)\}$ such that $\sum b_k-a_k<\epsilon$. Then what is the next step? How can I find a cover for $\lbrace x^2 : x\in E\rbrace$? If without loss of generality we think that $E\subset \mathbb{R}^+$ then $\bigcup (a_k^2, b_k^2)$ is a cover for $\lbrace x^2 : x\in E\rbrace$. But how to prove $\sum b_k^2-a_k^2<\epsilon$?

Jonas Meyer
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maysor
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  • Use that a set has Lebesgue measure zero if it can be covered by countably many intervals of total length at most $\epsilon$ for each $\epsilon>0$. – Michael Greinecker Feb 13 '14 at 07:25
  • @MichaelGreinecker: Thank you.Let $\epsilon>0$, I should find a cover of $\lbrace x^2 : x\in E\rbrace$ with total lenght at most $\epsilon$. since $m(E)=0$ so there exists a countable collection of intervals $\lbrace (a_k, b_k)\rbrace$ such that $\sum b_k-a_k<\epsilon$. Then what is the next step? How can I find a cover for $\lbrace x^2 : x\in E\rbrace$? – maysor Feb 13 '14 at 07:37
  • If without loss of generality we think that $E\subset \mathbb{R}^+$ then $\bigcup (a_k^2, b_k^2)$ is a cover for $\lbrace x^2 : x\in E\rbrace$. But How to prove $\sum b_k^2-a_k^2<\epsilon$? – maysor Feb 13 '14 at 07:46
  • You could try using this. – goblin GONE Jul 02 '15 at 17:47

1 Answers1

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Let $f(x)=x^2$. If $I$ is an interval, then $f(I)$ is also an interval, and if $I=(a,b)$, then $f(I)$ is one of $(a^2,b^2)$, $(b^2,a^2)$, $[0,a^2)$ and $[0,b^2)$, the latter two cases occurring only if $a$ and $b$ are of opposite sign. Suppose now that $b-a<1$. If $a<0<b$, then $-1<a$ and $b<1$, so $b^2<b$, $a^2<-a$, and the latter two intervals have length strictly smaller than $b-a$. If $I\subseteq[-M,M]$ and $a$ and $b$ have the same sign, then $|b^2-a^2|=|b+a||b-a|\le 2M|b-a|$.

Let now $E$ be a set of measure $0$. Since $E=\bigcup_n E\cap (-n,n)$, we may assume that $E$ is bounded: $\{x^2\mid x\in E\}$ will not be null iff $\{x^2\mid x\in E\cap(-n,n)\}$ is not null for some $n$. Let $M\ge 1$ be a bound for $E$, that is, $E\subseteq [-M,M]$, and let's further assume that $-M,M\notin E$. Since $E$ has measure $0$, for any $\epsilon>0$ we can cover $E$ with intervals whose lengths add up at most to $\min\{1/2,\epsilon/2M\}$. We may also assume that all these intervals are also contained in $[M,M]$, and since the sum of their lengths is at most $1/2$, they all have length less than $1$. It follows from the previous paragraph that if $I$ is any of these intervals, then $f(I)$ has length at most $2M$ times the length of $I$. Note that the $f(I)$ cover $f(E)$, and their lengths add up to at most $2m(\epsilon/2M)=\epsilon$. It follows that $f(E)$ also has measure zero.

(There is a small technical detail, depending on whether one insists that the intervals that cover are open. In that case, we can remove from each $f(I)$ its endpoints if necessary. This leaves at most countably many points of $f(E)$ uncovered, but this is irrelevant, since countable sets are null.)

Note that the same argument shows that if $f$ is any differentiable function whose derivative is bounded on bounded sets, then $f(E)$ is null for any null set $E$: All we need to do to adapt the argument is to use the mean value theorem to note that there is a constant $K$ (expressible in terms of $M$ and $B$, where $B$ is a bound for $f$ in $[-M,M]$) such that for any interval $I\subseteq [-M,M]$, we have that the length of $f(I)$ is at most $K$ times the length of $I$. A bit more work shows that the assumption on boundedness of the derivative can be eliminated. These are special cases of results that go back to Luzin, who investigated the question of when a function $f$ maps null sets to null sets, and said that $f$ had property (N) if that was the case.

Andrés E. Caicedo
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  • Andres: Thank you so much. I just don't understand why we can assume that $E$ is bounded? – maysor Feb 13 '14 at 13:27
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    Write $E=E_1\cup E_2\cup E_3\cup\dots$ where $E_i=E\cap(-i,i)$, so each $E_i$ is bounded. Then $f(E)=f(E_1)\cup f(E_2)\cup\dots$ We argued that when $E$ is bounded, then $f(E)$ has measure $0$. It follows that $f(E_1)$ has measure $0$, since $E_1$ is bounded. Similarly, $f(E_2)$ has measure zero, since $E_2$ is bounded, etc. Since $f(E)$ is the union of countably many null sets, it is null as well. – Andrés E. Caicedo Feb 13 '14 at 14:32