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

Solution 1:

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 ... .