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This comes as a complement to: Relation between total variation and absolute continuity; I was wondering if the following holds:

Let $F$ be a function of bounded variation on $[a,b]$, then $\int_{a}^{b}{|F'(x)|dx} = T_{F}(a,b)$ implies $F$ is absolutely continuous (same notations).

Any help is welcomed.

I guess that we actually have that if $G$ is an increasing continuous function for which $G'(x) < \infty$ a.e, then $G$ is absolutely continuous. (?)

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NVM, got it :). –  Anna Apr 4 '12 at 5:20

1 Answer 1

Let $F$ be a function of bounded variation on $[a,b]$, then $\int_{a}^{b}{|F'(x)|dx} = T_{F}(a,b)$ implies $F$ is absolutely continuous (same notations).

Yes. $T_F(a,b)$ is the norm of the measure $|dF|$. In general we have the decomposition $F(x) = F_s(x) + \int_a^x F'(t)\ dt$ where $F_s$ is singular with respect to Lebesgue measure, and $T_F(a,b) = \||dF_s|\| + \int_a^b |F'(x)|\ dx$. $T_F(a,b) = \int_a^b |F'(x)|\ dx$ iff $dF_s = 0$ iff $F(x) = \int_a^x F'(t)\ dt$ iff $F$ is absolutely continuous.

I guess that we actually have that if $G$ is an increasing continuous function for which $G'(x) < \infty$ a.e, then $G$ is absolutely continuous. (?)

No, that's wrong. $G'(x) < \infty$ a.e. for any increasing function.

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