# Non-decreasing and everywhere differentiable on $[0,1]$ implies absolutelly continuous?

I know that if a function is everywhere differentiable on $[0,1]$ it need not be absolutely continuous. Also a function that is almost everywhere differentiable, continuous and non-decreasing may fail to be absolutely continuous. But what if $f$ is differentiability everywhere and non-decreasing? Are these enough to guarantee I can use the Fundamental Theorem of Calculus? I'm grateful for any references you might have.

## 1 Answer

If $f$ is nondecreasing on $[0,1]$ then $f$ has bounded variation and $f' \in L^1([0,1])$. If such a function is differentiable everywhere in $[0,1]$ then it satisfies $$f(x) = f(0) + \int_0^x f'(t) \, dt$$ for all $x \in [0,1]$ and is consequently absolutely continuous.

I don't have a short proof handy, but this is Theorem 7.21 in Real and Complex Analysis, 3rd ed. by Rudin.

• Quick clarification: assuming $f(0)\ge 0$ for simplicity, the reason $f^\prime \in L^1([0,1])$ is because of the inequality $+\infty>f(1)-f(0)\ge \int_0^1 f^\prime (x) dx=\int_0^1 | f^\prime(x)| dx$ ? – Sergio Parreiras Apr 29 '15 at 17:22
• Yes, and what you wrote doesn't require $f(0) \ge 0$. – Umberto P. Apr 29 '15 at 17:46