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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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    I'm aware of that, and did not understand his proof.2012-08-18
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    @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.2012-08-18
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    I know, I just included the relevant parts of the statement.2012-08-18
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    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 thing2012-08-18
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    @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?2012-08-18
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    @FrankWhite: Can you elaborate a bit regarding the problems you were having with his proof? The essential tool in his proof is the Radon-Nikodym theorem; and proving then the statement for monotonic functions first.2012-08-18
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    @copper.hat The problem is that the proof references two other theorems that each individually take quite a bit of effort to prove on their own. I was just wondering if there was a simpler, more direct way, like if this problem were on a qualifying exam.2012-08-18
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    @FrankWhite: Sorry, don't have a good answer for you. However, I imagine that you could take Radon-Nikodym as a given, and the other tricks are well worth knowing in general, so it might be a good investment to work through it?2012-08-18
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    @copper.hat that's a good point. I'll see what it yields.2012-08-18
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    Frankly, I find Rudin's proofs a bit opaque at times; I can follow, but it takes me a while to synthesize the idea behind the proof. Often the idea is sufficient for me to 'get it', but with Rudin (also in "Functional Analysis") I find it tough going at times. I am looking in my scattered library to see if I have an alternative proof.2012-08-18
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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