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How to prove that an Absolutely continuous function admits weak derivative?

Absolutely continuous function: Let $(X, d)$ be a metric space and let I be an interval in the real line R. A function $f: I → X$ is absolutely continuous on I if for every positive number $\epsilon$, there is a positive number \delta such that whenever a finite sequence of pairwise disjoint sub-intervals $[x_k, y_k]$ for $k=1,2,...m$ of I satisfies

$$\sum_{k} \left| y_k - x_k \right| < \delta$$ then

$$\sum_{k} d \left( f(y_k), f(x_k) \right) < \epsilon.$$

If $f$ is an Absolutely continuous on $[a,b]$, then $f$ admits weak derivative $f'$ in $L^1([a,b])$ that is

$$ \int_a^b f(t) g'(t)dt=-\int_a^b f'(t)g(t)dt .......................(1)$$

for all absolutely continuous functions $g$.

I know that I have to use integration by parts and it is holds for absoltely continuous functions $f$ and $g$ (obvious),

$$ \int f(x) g'(x) \, dx = fg(b)-fg(a) - \int f'(x) g(x) \, dx ......(2) $$ My question is how to get (1) from (2).

Rahman
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    Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? – 5xum Jul 24 '14 at 10:42
  • There are a few ways to define absolute continuity. Please provide the definition you are working with. – Jonas Dahlbæk Jul 24 '14 at 13:03
  • Do you want to prove the integration formula of absolutely continuous functions? Notice that $fg$ is also absolutely continuous so it has derivative $f'g+fg'$ – Shine Jul 24 '14 at 14:22
  • I know that if f is absolutely continuous then integration by parts hold i am just using the integration by parts formula to prove that absolutely continuous function admits weak derivative, that is $∫baf(t)g′(t)dt=−∫baf′(t)g(t)dt$ – Rahman Jul 25 '14 at 07:18

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