Absolute continuity of a nondecreasing function Can anyone give me a hint on how to approach this problem? It's another problem from an old qualifying exam.
Suppose that $f\colon \mathbb R \to \mathbb R$ is nondecreasing, $\int_{\mathbb R} f' = 1$, $f(-\infty) = 0$, and $f(\infty) = 1$. Prove that $f$ is absolutely continuous on any interval $[a,b]$.

One approach is as follows:
Since $f$ is non-decreasing, it is differentiable ae. [$m$] and $f(x)-f(y) \ge \int_y^x f'(t) dt$ for all $x>y$.
Use the fact that $1 = \lim_{x \to + \infty} f(x) - \lim_{x \to - \infty} f(x) = \int_{\infty}^\infty f'(t) dt$ to show that,
$f(x)-f(y) = \int_y^x f'(t) dt$ for all $x>y$.
Now use the fact that $f'$ is integrable and positive to conclude absolute continuity (note that $\lim_{M \to \infty} \int_{\{x | f'(x) \ge M\}} f' = 0$, and if $f'(t) \le M$, the result is straigtforward).
