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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).

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  • $\begingroup$ Hi W Rldt! Welcome to math.stackexchange. Before we can answer your question, could you please clarify what the square of a set is meant to mean here? I am uncertain and perhaps others will be as well. $\endgroup$ – Ragib Zaman Sep 15 '11 at 6:49
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    $\begingroup$ This is the definition: $E^2=\{x^2 : x \in E\}$ $\endgroup$ – wrldt Sep 15 '11 at 7:04
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    $\begingroup$ 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. $(5, 5.1)^2=(25,26.01)$. $\endgroup$ – Jonas Meyer Sep 15 '11 at 7:28
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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 ... .

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  • $\begingroup$ Perhaps this could be broken up into 3 steps, initially restricting to the case where $E$ is a bounded set of positive numbers, for the convenience that $[a,b]^2=[a^2,b^2]$ can be assumed for intervals in the covers. $\endgroup$ – Jonas Meyer Sep 15 '11 at 7:43
  • $\begingroup$ @Jonas: That’s what I get for posting too soon after I get up. Thanks again. $\endgroup$ – Brian M. Scott Sep 15 '11 at 7:51
  • $\begingroup$ @Brian I still have some sort of conceptual hurdle of trying to understand the connection between covering E and covering $E^2$. I get that we want to cover E with intervals $I_n$ such that $m^*(E)=\sum(E \cap I_n) \geq \sum I_n < \epsilon$ where $\epsilon$ is pretty small. I'm not sure how this translates into proving anything about $E^2$ however. $\endgroup$ – wrldt Sep 15 '11 at 14:57
  • $\begingroup$ @@ Rldt: Suppose that $E\subseteq [0,M]$ and you cover $E$ with intervals $[a_i,b_k]$ lying in $[0,M]$. Then the intervals $[a_k^2,b_k^2]$ will cover $M^2$. (Your inequality should be $m^*(E)\le\sum\vert I_n\vert<\epsilon$, where $\vert I_n\vert=m^*(I_n)$ is the length of $I_n$.) $\endgroup$ – Brian M. Scott Sep 15 '11 at 17:36
  • $\begingroup$ So does this mean that $m^*(E^2) \leq \sum I^2_n < M\epsilon$ $\endgroup$ – wrldt Sep 16 '11 at 0:13
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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.

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    $\begingroup$ $x\mapsto x^2$ is not Lipschitz ! $\endgroup$ – Mathronaut Jun 9 '16 at 17:02

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