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I want help in showing that $f$ is Lipschitz on $[0,1]$ $\implies$ that $f$ can written in the form $$f(x) = f(0) + \int_0^x h(x)~dt$$ for some bounded Lebesgue measurable function $h$ on $[0,1]$.

$f$ being Lipschitz on $[0,1]$ implies there is some constant $K$ such that $|f(x)-f(y)|\leq K |x-y|$ for every $x,y \in [0,1]$.

Can I argue as follows:

I know that if $f$ is Lipschitz on $[0,1]$, then $f$ is abosolutely continuous on $[0,1]$ and so $f$ is a definite integral. i.e. $$f(x) = f(0) +\int_0^x f'(t)~dt,$$ So I can take $f' = h$. Can I do this or there is a better way of approaching it.

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    Can't you use that if $f$ is absolutely continuous then it's differentiable almost everywhere and then apply the fundamental theorem of calculus?2012-03-22
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    I thought that's what I did...no?2012-03-22
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    See also http://math.stackexchange.com/q/85009/53632012-03-22
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    Note, you should argue why $f'$ is bounded.2012-03-22
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    @DavidMitra: right. can I get help with that? thanks.2012-03-22
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    @Jack It follows directly from the Lipschitz condition.2012-03-22
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    @DavidMitra: I guess I'd have to use $$f'(x) = \lim_{y\to x} \frac{f(y)-f(x)}{y-x}$$ and the fact that $f$ is Lipschitz.2012-03-22
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    @Jack Yes, the limit expression is at most $K$ in absolute value.2012-03-22
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    @DavidMitra: Thanks very much.2012-03-22

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