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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 $F:\mathbb R\to \mathbb R$?2012-08-18

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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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    @FrankWhite: I found a less opaque proof (my opinion, of course). There are still some preliminary results needed. Kolmogorov & Fomin, "Introductory Real Analysis", Ch.9, Section 33.2, Theorem 6. In general, I like the style of authors like Kolmogorov, Fomin, Kantorovich, Akilov, etc. Their emphasis seems to be distributed nicely between theorem statement and proof.2012-08-18