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 $(0,1)$. Show that $f$ must be absolutely continuous on $[0,1]$ and construct a counter example for the case without the "monotone" condition.
My attempt:
$f'(x)$ exists for all $x$ in $(0,1)$ anld continuous on (0,1), then $f'(x)$ is integrable on (0,1) and
$$f(x) = \int_{0}^{x}f'(x) + f(0),\forall x \in [0,1]$$
Since $f$ is an indefinite integral on [0,1] hence $f$ is absolutely continuous.
I want a different approach with the direct proof using the definition and not using so many theorems.
for the counterexample I have:
$f(x) = x\sin(1/x)$ and we must show that $f$ is not of bounded variation.
Your argument for the first part is not correct. You did not use monotonicity. Assume that $f$ is monotonically increasing. Note that $f' \geq 0$ on $(0,1)$. We can write $f(x)=f(\frac 1 2)+\int_{1/2} ^{x} f'(t)\, dt$ for $x<\frac 1 2 <1$. Since LHS tends to the finite limit $f(1)$ as $ x \to 1$ it follows that $\int_{1/2}^{1} |f'(t)|\, dt=\int_{1/2}^{1} f'(t)\, dt <\infty$. Similarly, $\int_0^{1/2} |f'(t)|\, dt <\infty$. This proves that $f'$ is integrable. Now it is easy to see that $f(x)=f(0)+\int_0^{x} f'(t)\, dt$ for all $x \in [0,1]$ from which absolute continuity follows.
For the second part consider the intervals $(\frac 1 {\frac {(2n+1)\pi} 2},\frac 1 {n\pi})$ and apply definition of absolute continuity.