Problem about absolute continuity of a function

Question

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

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.