Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f:\mathbf{R} \to \mathbf{R}$ 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.

share|improve this question
1  
If $f$ 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 $f$. –  sos440 Aug 4 '11 at 18:08
4  
I do no know any integrable function $f$ satisfying your two limit conditions. –  GEdgar Aug 4 '11 at 18:49
    
I am sorry, it was a misprint that the function should be integrable, what we have is that the function is increasing. –  user14272 Aug 5 '11 at 2:07
    
I'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? –  Michael Hardy Aug 5 '11 at 2:26
    
And now it's appeared. –  Michael Hardy Aug 5 '11 at 2:57

1 Answer 1

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 $a<b$ (e.g., see Wheeden and Zygmund). If $f$ is not absolutely continuous on every bounded interval, then there exists $a<b$ such that the inequality is strict for this $a$ and $b$, i.e., $\int_a^b f'<f(b)-f(a)$ (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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.