Lipschitz continuity implies differentiability almost everywhere.

Question

I am running into some troubles with Lipschitz continuous functions.

Suppose I have some one-dimensional Lipschitz continuous function $f : \mathbb{R} \to \mathbb{R}$. How do I prove that its derivative exists almost everywhere, with respect to the Lebesgue measure?

I found on other places on the internet that any Lipschitz continuous function is absolutely continuous, and that this directly implies that the functions is differentiable almost everywhere. I don't quite see how this argument goes, though.

Any help with giving such a proof, or redirecting me to a source where I can find one, would be greatly appreciated.

Answer 1

An excellent reference for the properties of Lipschitz functions is Lectures on Lipschitz Analysis by Juha Heinonen. The differentiability a.e. is Theorem 3.2, page 19. I state the steps of the proof here:

1. Lipschitz continuity implies having bounded variation.
2. A function of bounded variation can be written as the difference of two increasing functions
3. An increasing function is differentiable almost everywhere: this is the main step of the proof, which uses the Vitali covering theorem. Concerning this step, see also $f$ continuous, monotone, what do we know about differentiability?