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$f:\mathbf{R} \to \mathbf{R}$ is an increasing function with $\lim_{x\to -\infty}f=0$ ,$\lim_{x\to \infty}f=1$, and \int_{R}f'=1. Prove that $f$ is absolutely continuous on every interval $[a,b]$. Any help is appreciated.

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    @Michael: There was an edit suggested at 2:03 UTC waiting for review (it was approved at 2:39) see [here](http://math.stackexchange.com/suggested-edits/1927) (you can get there by clicking on "edited xx time ago" and then clicking on "suggested xx time ago" - the precise times can be seen by hovering your mouse over xx time ago and waiting until a tool-tip containing the exact time appears). There can only be one suggested edit per question and your first comment is from 02:26 UTC, so you couldn't suggest a second edit until after 02:39 when the first suggestion was approved by Qiaochu.2011-08-05

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Because $f$ is increasing, it is differentiable a.e., its derivative is measurable, and \int_a^b f'\leq f(b)-f(a) for all $a (e.g., see Wheeden and Zygmund). If $f$ is not absolutely continuous on every bounded interval, then there exists $a such that the inequality is strict for this $a$ and $b$, i.e., \int_a^b f' (this follows from the characterization of AC functions as integrals of their derivatives, as seen e.g. on Wikipedia). Let c=(f(b)-f(a)) - \int_a^b f' >0.

Now you can show that for all $M>\max\{|a|,|b|\}$, \int\limits_{-M}^Mf'\leq 1-c, by breaking it up into 3 parts and applying the results of the previous paragraph along with the fact that $0\leq f\leq 1$. Once you have this, the proof by contraposition is almost complete.