$f:\mathbf{R} \to \mathbf{R}$ 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.
Problem about absolute continuity of a function
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measure-theory
problem-solving
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1If $f$ is merely integrable, there is no guarantee that $f'$ exists almost everywhere. And even both existence and integrability of $f'$ given, we are still unable to deduce the absolute continuity of $f$. – 2011-08-04
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4I do no know any integrable function $f$ satisfying your two limit conditions. – 2011-08-04
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0I'd have edited this to say $\lim_{x\to-\infty}$ instead of $lim_{x\to-\infty}$, and $[a,b]$ instead of [a,b$, but the "edit" button isn't there. Why not? – 2011-08-05
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0And now it's appeared. – 2011-08-05
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0I am sorry, it was a misprint that the function should be integrable, what we have is that the function is increasing. – 2011-08-05
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0@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
1 Answers
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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 $aWheeden and Zygmund). If $f$ is not absolutely continuous on every bounded interval, then there exists $aon 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.