Assume that $ f: R \to R $ is a non-decreasing function with $ \int_R f' dm =1, $ $ \lim_{x \to-\infty} f(x) =0 $ , $ \lim_{x \to\infty}f(x)=1 $. Then Prove that $f$ is absolutely continuous on any interval $[a,b]$.
If we can show $ f(x) - f(a) =\int_{a}^{x} g(t) dt $ for some Lebesgue integrable function $g$ on $[a,b]$ then we are done. I could not figure out this.
Define $F(x) = \int_{-\infty}^x f'(t) dt$ for all $x \in \mathbb{R}$. By the condition $\int_{-\infty}^\infty f'(t) dt = 1$, we conclude that $\lim\limits_{x \to -\infty} F(x) = 0$ and $\lim\limits_{x \to \infty} F(x) = 1$. On the other hand, since $f$ is nondecreasing, we know that $f'$ is nonnegative almost everywhere, and $$f(b) - f(a) \geq \int_a^b f'(t) dt = F(b) - F(a).$$ It thus can be seen that the function $g(x) = f(x) - F(x)$ is also nondecreasing on $\mathbb{R}$. Meanwhile, we have $$\lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} f(x) - \lim_{x \to -\infty} F(x) = 0 - 0 = 0$$ and $$\lim_{x \to \infty} g(x) = \lim_{x \to \infty} f(x) - \lim_{x \to \infty} F(x) = 1 - 1 = 0.$$ Hence by monotonicity of $g$, $g(x) \equiv 0$ on $\mathbb{R}$. That is $f(x) = F(x) = \int_{-\infty}^x f'(t) dt$ for every $x \in \mathbb{R}$, i.e., $f$ is absolutely continuous. A fortiori, the property in your statement holds.