If $f$ is continuous over $[a,b]$ and $\vert f\vert$ has bounded variation, is $f$ absolutely continuous?
Given $\varepsilon >0$. I need to find a $\delta$ such that $\sum\vert f(b_i)-f(a_i)\vert<\varepsilon$ when $\sum(b_i-a_i)<\delta,$ but this doesn't use the fact that $\vert f\vert $ has bounded variation.
On
here is a related theorem
Theorem: $f: I → R$ is absolutely continuous if and only if it is continuous, is of bounded variation and has the Luzin N property.
The Cantor function is a counter-example. It is continuous over $[0, 1]$, and has bounded variation (in fact, the arc length is $2$, according to wikipedia), but it fails to be absolutely continuous.