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

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I'm aware of that, and did not understand his proof. –  Frank White Aug 18 '12 at 19:40
@copper.hat: Just a nitpick, but F being absolutely continuous implies not only that F' exists a.e., but also that it's Lebesgue integrable. –  Bey Aug 18 '12 at 19:47
I know, I just included the relevant parts of the statement. –  copper.hat Aug 18 '12 at 19:48
But it is relevant to say F' is integrable, otherwise the expression involving the integral of F' doesn't necessarily make sense. In any case, like I said, it's a nitpicky little thing –  Bey Aug 18 '12 at 19:49
@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