# Sufficient condition for a function to be absolutely continuous

A function $F:\mathbb{R} \rightarrow \mathbb{R}$ is absolutely continuous if there exists a function $f:\mathbb{R} \rightarrow [0, \infty)$ such that $\forall x \in \mathbb{R}$, we have:

$$F(x) = \int \limits_{-\infty}^{x} f(t) dt$$

We know that such a function must be everywhere continuous and almost everywhere differentiable. However, the converse is not true; the Devil's Staircase is continuous and almost everywhere differentiable but not absolutely continuous.

I would like to know if the following condition is sufficient for a function $F$ to be absolutely continous: it is continous everywhere and differentiable everywhere except on a finite set.

We could ask the same question for $F$ continous everywhere and differentiable everywhere except on a countable set, but I suspect the answer in this case is no.

• Is the derivative continuous in between the points it doesn't exist? If so then you have a locally $C^1$ function hence it is locally Lipschitz hence locally absolutely continuous. (Making these global requires some integrability of the derivative.) – Ian Oct 19 '17 at 15:22
• I was asking the question pretty generally, so I didn't have a particular function in mind. You're saying if $F'$ is continuous and integrable, $F$ is absolutely continuous? – Sambo Oct 19 '17 at 15:27
• If $f$ is integrable then $F$ is absolutely continuous. – copper.hat Oct 19 '17 at 15:28
• My goal is to have a general way to justify stating that a function is absolutely continuous for my Probability class; we'll be dealing with functions that are most likely not too exotic. – Sambo Oct 19 '17 at 15:28
• Sure, that is just the regular FTC. But I am adding that if F' exists except on a discrete set and is continuous where it exists then F is absolutely continuous. I am wrong about it being Lipschitz though because of examples like $|x|^{1/2}$. – Ian Oct 19 '17 at 15:29

## 1 Answer

In the usual definition of absolute continuity, $f$ may take negative values. I think it should be stressed that the function $f$ is required to be Lebesgue integrable, i.e. its positive and negative parts each have finite integral. Even if Lebesgue is the default notion of integrability, it is still an important feature here.

The condition you suggest - $F$ continuous and differentiable except on a finite set - is not sufficient. A simple counterexample for $\mathbb R$ is $F(x)=x.$ Even being differentiable everywhere does not imply absolute continuity on a compact interval.

It is well known that Lipschitz compactly supported functions are absolutely continuous.

Here's a different sufficient condition. If $F$ is continuous, has bounded variation (e.g. bounded and "unimodal": increasing then decreasing), and is differentiable except on a countable set, then it is absolutely continuous. Bounded variation lets us take the Lebesgue decomposition of the distributional derivative; the continuity of $F$ implies there is no discrete part; the differentiability implies there is no singular continuous part.

• Can you give a reference for the last paragraph in your answer please? – user661541 May 12 at 22:54