I know there is a theorem saying if $f$ defined on $[a,b]$ is of bounded variation, then it is differentiable on $(a,b)$ a.e and $f'$ is integrable over $[a,b]$.
I wonder whether the converse is true, say, if $f$ defined on $[a,b]$ is differentiable on $(a,b)$ a.e and $f'$ is integrable, must $f$ have bounded variation?
In fact, I run into such concrete question: $ f=x^\alpha \text{sin}\left(\frac{1}{x^\beta}\right) \text{ for $x\in(0,1]$} $ and $f(0)=0$. The hint says that we can prove that when $\alpha > \beta$, $f$ is of bounded variation by showing $f'$ is integrable.
So here comes my question above, is such argument true?