TITLE: Lipschitz Implies $F(x)=F(0)+\int_0^x F'(t) dt$.
QUESTION [5 upvotes]: I've been working on the following question:
If $F : \mathbb{R} \rightarrow \mathbb{R}$ is a Lipschitz function, then $F(x)=F(0)+\int_0^x F'(t) dt$.
I've already proved that Lipschitz implies $F'$ is exists a.e., and $F'$ is essentially bounded, but for whatever reason I've been stumped on this one.  I looked at similar questions on here but couldn't seem to find too much that went into detail.

REPLY [4 votes]: If $F$ is Lipschitz it is absolutely continuous. From Rudin's "Real & Complex Analysis" Theorem 7.20, we have that $F$ is differentiable a.e. and $F(x)=F(0)+\int_0^x F'(t) dt$.