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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.

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In the second paragraph, you describe uniform continuity, not absolute continuity. The counterexample provided by Arthur is indeed not absolutely continuous, but it is uniformly continuous (even Lipschitz) – Martijn Dec 12 '12 at 10:46
In the second paragraph he describes absolute continuity, not uniform continuity. Any continuous function on a compact domain is uniformly continuous, since given an $\epsilon$, there is a smallest necessary $\delta$, by compactness. – Arthur Dec 12 '12 at 11:10
To recap, uniform continuity is the same as regular continuity, only you have to pick a $\delta$ which works for any $x$. Absolute continuity is stronger, since you can divide the (in total $\delta$ wide) interval into smaller pieces. – Arthur Dec 12 '12 at 11:13
up vote 1 down vote accepted

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.

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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.

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