Let $f: [0,1] \to \mathbb{R}$ be a continuous monotone function such that $f$ is differentiable everywhere on $(0,1)$ and $f'(x)$ is continuous on $(0,1)$.
(a) Prove that $f$ is absolutely continuous on [0,1].
(b) Construct a counter example where this does not hold if we drop the monotonic assumption.
My attempt:
Let $\epsilon > 0$, and $\{(a_n,b_n)\}_{n=1}^{k}$ be a disjoint sequence of open sets in $[0,1]$, then by the continuity of $f$ in $[0,1]$ we might pick $\delta_n > 0$ such that $|b_n - a_n| < \delta_n \Rightarrow |f(b_n) - f(a_n)| < \frac{\epsilon}{k}$ $\forall$ $n = 1,...,k$. Then we can choose $\delta = \sum_{n=1}^{k}\delta_n$, then
$\sum_{n=1}^{k}|b_n - a_n| < \delta \rightarrow \sum_{n=1}^{k}|f(b_n) - f(a_n)| < \sum_{n=1}^{k}\frac{\epsilon}{k} = \epsilon$,
Hence, $f$ is absolutely continuous. I know this is not correct because there are continuous functions which are not absolutely continuous if we drop the monotone condition as part $2$ suggest but I never use any of the other conditions. What am I doing wrong? What am I missing?
Part $(2)$ The function is $x\sin(\frac{1}{x})$ if $x \neq 0$, $0$ if $x = 0$.
The point which is wrong is the implication
$$\sum_{n=1}^k |b_n-a_n| <\delta \rightarrow \sum_{n=1}^k |f(b_n)-f(a_n)|$$
here you "implicitely" assume the wrong claim that if $$\sum_{n=1}^k |b_n-a_n| <\sum_{n=1}^k \delta_n$$ then each $|b_n-a_n|<\delta_n$.