Let Z $\subseteq \mathbb{R}$ with with $m(Z) = 0$. Set $E=\left \{ x^{2}, x \in \mathbb{Z} \right \}$. Suppose Z is bounded, that is, $Z\subseteq \left [ -n,n \right ]$ for some integer n. Show that $E$ is Lebesgue measurable and $m(E)=0$.
My attempt:
$Z\subseteq \left [ -n,n \right ]$, then $E\subseteq \left [ 0,n \right ]$. Let $\epsilon > 0$.
Suppose there exist some sequence of the intervals that: $Z\subseteq \bigcup_{k=1}^{\infty}(a_{k}, b_{k})$ and that $\sum_{k=1}^{\infty}\left | b_{k}-a_{k} \right |< \epsilon $.
Then $m^{2}(a_{k}^{2}, b_{k}^{2})=\left |b_{k}^{2} - a_{k}^{2} \right |=\left | b_{k}-a_{k} \right |\left | b_{k}+a_{k} \right |$.
From this point I do not know how to continue and even if it is correct.
Can anyone help?
Note that $a_k <x<b_k$ does not imply $a_k^{2} <x^{2} <b_k^{2}$. So your approach needs a modification.
First note that by considering $(a_k,b_k) \cap (-n-1,n+1)$ (which is also an open integral we may suppose $|a_k|\leq n+1$ and $|b_k|\leq n+1$ for all $n$. Now $\{x^{2}: a_k <x<b_k\}$ is contained in
a) $(a_k^{2},b_k^{2})$ if $a_k \geq 0$
b) $(b_k^{2},a_k^{2})$ if $b_k \leq 0$ and
c)$(-\frac {\epsilon} {2^{k}},b_k^{2})$ if $a_k <0<b_k$.
Finally use the fact that $|a_k+b_k| |a_k-b_k| \leq 2(n+1) |a_k-b_k|$.
Now I believe you can put these together to complete the proof.