"Suppose f is a continuous, but not absolutely continuous, function from [a,b] to $\mathbb{R}$, then find a set $E$ with Lebesgue measure $m(E)=0$ such that $m(f(E))>0$" I'm having trouble answering this question, especially with the last step of my tentative proof. My reasoning went like this:
I've started by negating the AC hypothesis so: $\exists\epsilon>0$ such that $\forall\delta>0$ there exists disjoint intervals $\ (a_k,b_k)$ such that $\sum^n_{k=1}|b_k -a_k|< \delta$ and $\sum^n_{k=1}|f(b_k) -f(a_k)|> \epsilon$
Supposing this first step is correct, then I considered a sequence $\delta_n=1/n$ and a corresponding set of disjoint intervals $I_n$ such that$\sum^n_{k=1}|b_k,_n -a_k,_n|< \delta_n$ for every $(a_k,_n , b_k,_n)\in I_n$ with the added condition that $I_{n+1} \subset I_n$, and I do this by taking smaller intervals contained in the intervals of $I_n$ that satisfy $\sum^n_{k=1}|b_k,_{n+1} -a_k,_{n+1}|< \delta_{n+1}$ for every $(a_k,_{n+1} , b_k,_{n+1})\in I_{n+1}$. Now I take $\lim_{n\to\infty} I_n = E$ and I have that $m(E)=0$
(Edited after comment) Since $m(f(I_n))>\sum^n_{k=1}|f(b_k,_n) -f(a_k,_n)|> \epsilon$ for $(a_k,_n , b_k,_n)\in I_n$, and since f is continuous I can use the intermediate value theorem, so $m(f(E))$ is at least $\epsilon$.
(PreEdit)Am I on the right track or is my reasoning completely wrong? I also didn't use the continuity hypothesis, and it feels like I had to but I don't know how.