# Lipschitz continuity implies differentiability almost everywhere.

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.

• Which part of the reasoning using absolute continuity do you not get? The fact that Lipshitz implies AC, or the fact that AC implies differentiability ae? (Or both?) Commented Jan 17, 2017 at 19:33
• The part that AC implies differentiability ae. :) Commented Jan 17, 2017 at 20:53
• This is known as Rademacher's theorem. Commented Jun 30, 2020 at 12:27

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?
• What if $f:R^n \to R$? Commented Jun 10, 2017 at 22:59
• Hi, concerning the part where the vitali covering is used. The part $q|E_q| \leq |f(E_q)|$ is not so clear to me. Do you maybe understand how the author came to this conclusion? @user357151 Commented Jan 4, 2022 at 20:36