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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.

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    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?) – 2017-01-17
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    The part that AC implies differentiability ae. :) – 2017-01-17

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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 a.e.: 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?
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    Wow, this turned out to be somewhat harder than I expected. Thanks for the references! – 2017-01-18
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    What if $f:R^n \to R$? – 2017-06-10
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    @Ashkan Theorem 3.1 in Lectures on Lipschitz Analysis that are linked in the post. – 2017-06-11