# Lipschitz Implies $F(x)=F(0)+\int_0^x F'(t) dt$.

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.

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You mean $F:\mathbb R\to \mathbb R$? –  Pedro Tamaroff Aug 18 '12 at 19:19

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 $F(x)=F(0)+\int_0^x F'(t) dt$.
@Bey: Since $F$ is Lipschitz, we have $|F'(x)| \leq L$ where $L$ is the Lipschitz rank. Using the DCT shows that it is integrable, but again, that wasn't the focus, I think? –  copper.hat Aug 18 '12 at 19:53