8
$\begingroup$

Currently I'm just working through measure theory just to see if its something I would like to take.

Unfortunately I am stuck on this problem from Carothers.

If $m^*(E)=0$, then $m^*(E^2)=0$.

Where $m^*$ denotes outer measure and $E^2=\{x^2:x\in E\}.$

I toyed with the idea that $I_k < 1 \Rightarrow I^2_k < I_k$. However I am at a loss as to how to set up a chain of inequalities (which is what I am assuming I need).

  • 1
    What do you mean by I_k<1? Is $I_k$ an interval? If so, do you really mean that the length of $I_k$ is less than $1$? If so, note that this would not imply that $I_k^2$ has length less than $1$. E.g. $(5, 5.1)^2=(25,26.01)$.2011-09-15

2 Answers 2

9

Here’s one approach.

(1) First show that if $E\subseteq [0,M]$ for some real number $M>0$, and $m^*(E)=0$, then $m^*(E^2)=0$. HINT: You can cover $E$ with arbitrarily short intervals $[a,b]$ in such a way that for each of these intervals $m^*([a^2,b^2])=b^2-a^2<(M+1)(b-a)$.

(2) Use (1) to take care of the case $E\subseteq [M,0]$ for some $M<0$.

(3) Use (1) and (2) to show that for any $\epsilon > 0$ and any positive integer $n$, $m^*(E\cap [-n,n])<2^{-n}\epsilon$. Then $E = \bigcup_{n=1}^\infty(E\cap [-n,n])$ and $\epsilon = \sum_{n=1}^\infty 2^{-n}\epsilon$, so ... .

  • 0
    @W Rldt: *If* $E\subseteq [0,M]$ and all of your intervals are contained in $[0,M]$, so that their endpoints are non-negative, then $E^2\subseteq\bigcup_n I_n^2$, so $m^*(E^2)\le\sum_n \vert I_n^2\vert$.2011-09-16
-2

See that

  • any set of Outer Measure zero is measurable
  • any Lipshcitz continuous function takes measure zero sets to measure zero set.
  • $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $x\mapsto x^2$ is Lipschitz
  • Image of $E$ under $f$ is $E^2$.

Done!

This is a bit more general question than you have asked.

  • 2
    $x\mapsto x^2$ is not Lipschitz !2016-06-09