Let $A \subseteq \mathbb{R}$ be measurable and $f: A \to \mathbb{R}$ be a Lipschitz function with constant K, i.e., for any x, y in A, $|f(x) - f(y)| \leq K|x - y|$. Prove that for any $E \subseteq A,$ $|f(E)|_e \leq K|E|_e$.
I can directly use the fact that $|f(x) - f(y)| \leq K|x - y|$ to show the inequality holds when E is an interval, and use the interval case to show that this is true for open sets as well, but it's not clear where to go from there. I know that for any $\epsilon > 0,$ any set E can be covered by countably many closed intervals $Q_j$ so that $|E|_e \leq \sum_{j=1}^\infty |Q_j| \leq |E|_e + \epsilon$, and that an open set within $\epsilon$ of E's exterior measure exists as well, but the problem is that these approximate from outside of E and possibly outside of A, and I can't work out the right way to "extend" f out to the intervals or open sets and be sure that the result is still a Lipschitz function. I think it would be best if I could show the result holds for closed sets and work upward instead, but I'm not quite sure how to do that, either.
Nice statment of what you've tried. It looks to me like $f$ can be extended to all of $\mathbb {R}$ with the same Lipschitz constant. First, extend $f$ to $\bar {A};$ the extension is the same as for any uniformly continuous function on $A.$ Then $\mathbb {R}\setminus \bar {A}$ is open, hence is the pairwise disjoint union of open intervals $I_1, I_2, \dots.$ On any $I_n = (a_n,b_n),$ just connect $(a_n,f(a_n)),(b_n,f(b_n))$ with a line segment. There's the desired extension, and it would seem you're good to go. (I'm not sure why $A$ needs to be measurable though. Where does that come in?)