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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? –  Rudy the Reindeer Mar 22 '12 at 17:10
I thought that's what I did...no? –  Jack Mar 22 '12 at 17:12
See also math.stackexchange.com/q/85009/5363 –  t.b. Mar 22 '12 at 17:12
Thanks for the link. –  Jack Mar 22 '12 at 17:21
Note, you should argue why $f'$ is bounded. –  David Mitra Mar 22 '12 at 17:23

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