Outer Lebesgue measure Let $E\in\mathbb{R}$ and let $\lambda^*(E)$ denotes the outer Lebesgue measure of $E$. Let $F:=\{x^2:x\in E\}$. If $\lambda^*(E)=0$, how can we show that $\lambda^*(F)=0$?
 A: Assume $E$ is bounded by $[0,M]$ or $[-M , 0]$, with $M$ positive, since if it is not, we may split $E$ into $E \cap [k , k+1 ]$ for $k \in \mathbb Z$.  Since nullsets form a $\sigma$-algebra, if each $E \cap [k,k+1]$ has measure $0$, then $E \subseteq \bigcup_{k \in \mathbb Z} E \cap [k,k+1]$ does.
Without loss of generality, $E \subseteq [0,M]$, since if $E \subseteq [-M,0]$, then $\{ -x : x \in E \} \subseteq [0,M]$ is a nullset and $\{ (-x)^2 : x \in E \} = F$.
Suppose we have some closed intervals $I_n = [a_n,b_n]$ with $b_n \ge a_n \ge 0$ which cover $E$ such that $\sum |I_n| < \epsilon$.
Then, we have intervals $J_n = [a_n^2 , b_n^2]$ that cover $F$, since if $0 \le a_n \le x \le b_n$, then $a_n^2 \le x^2 \le b_n^2$.
So $\sum |J_n| = \sum | b_n^2 - a_n^2 | = \sum | b_n + a_n | | b_n - a_n |$.
But $|b_n + a_n|$ is bounded above by $2M+1$, by the bounds on $E$.  So $\sum |J_n| < (2M+1)\epsilon$.
Since $\epsilon$ was selected arbitrarily, as $\epsilon \to 0$, $\sum |J_n| \to 0$ and $F$ is a nullset.
A: Let $\epsilon>0$. Then $\lambda^*(E \cap [-\frac{\epsilon}{4},\frac{\epsilon}{4}]) \le \frac{\epsilon}{2}$.
Now consider the intervals $I_n=[n+\frac{\epsilon}{4},n+1+\frac{\epsilon}{4}] \subset [\frac{\epsilon}{4}, \infty)$. Then $f(x) = x^2$ is uniformly Lipschitz on $I_n$ with rank $2(n+1+\frac{\epsilon}{4})$. Furthermore, since $E$ has outer measure zero, it follows that $E_n' = E \cap I_n$ has outer measure zero. Hence there is a sequence of open intervals $J_k$ such that $E_n' \subset \cup_k J_k$ and $\sum_k l(J_k) < \frac{\epsilon}{2(n+1+\frac{\epsilon}{4})2^{n+2}}$. If we let $J_k' = ((\inf J_k)^2, (\sup J_k)^2)$, we see that (i) $f(E') \subset  \cup_k J_k'$, and (ii) $l(J_k') \leq 2(n+1+\frac{\epsilon}{4}) \,l(J_k) < \frac{\epsilon}{2^{n+2}}$. It follows that $\lambda^*f(E_n') < \frac{\epsilon}{2^{n+2}}$.
It should be clear that the same reasoning applies to $(-\infty, -\frac{\epsilon}{4}]$.
Hence we obtain the bound $\lambda^* f(E) < \frac{\epsilon}{2} + 2 \sum_n \frac{\epsilon}{2^{n+2}}= \epsilon$. Since $\epsilon>0$ was arbitrary, the result follows.
It should be clear from the proof that this is true for any function that is Lipschitz continuous on compact intervals.
