If $E$ has measure zero, then does $E^2$ have measure zero? 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 $I_k &lt; 1 \Rightarrow I^2_k &lt; I_k$.  However I am at a loss as to how to set up a chain of inequalities (which is what I am assuming I need).
Here’s one approach. 

(1) First show that if $E\subseteq [0,M]$ for some real number $M&gt;0$, 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 $m^*([a^2,b^2])=b^2-a^2&lt;(M+1)(b-a)$.

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

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