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Suppose that $f$ is a continuous function on $[a,b]$ . And suppose that there exists an integrable function $g$ on $[a,b]$ such that $\int_{[a,b]}fh'=-\int_{[a,b]}gh$ for any Lipschitz function $h$ on $[a,b]$ satisfying $h(a)=h(b)=0$. The conclusion is that $f$ is absolutely continuous.

But I just want to know is it possible to deduce that $f(x)=\int_{[a,x]}g$ for any $x\in(a,b)$ by finding some specific function $h$ and plugging in the equation in the problem? Since we have the Lebesgue integral here, we are not able to apply the integration by parts. So what confuses me is that how can I get rid of the integral sign?

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    I think the correct formulation of your second sentence should be: there exists an integrable function $g$, such that for any Lipschitz function $h$ ...2012-11-13
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    Yes, if $g$ depends on $h$, why would $f(x)=\int_{[a,x]} g(t)dt$? I mean, $h(x)=0$ is such and $h$, then $g(x)=0$ is a valid $g$.2012-11-13
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    Yes, that's what I mean. $g$ should not depend on $h$. I rewrite the question to avoid the confusion.2012-11-13

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Given $x\in(a,b)$, you may choose a sequence of Lipschitz functions $\{h_n\}$ in the following way. For every $n\ge \max(\frac{1}{x-a},\frac{1}{b-x})$, $$h_n(t)=\left\{\begin{array}{cc} n(t-a)& t\in[a,a+\frac{1}{n}]\\ 1&t\in[a+\frac{1}{n},x]\\ 1-n(t-x)&t\in[x,x+\frac{1}{n}]\\ 0&t\in[x+\frac{1}{n},b] \end{array}\right.\ .$$ Then you will find $f(x)=f(a)+\int_a^xg(t)dt$.

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    I was wondering how did you come up with this sequence. Is there any reason behind this? Thanks.2012-11-13
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    @user45955: Basically, since you want to get $\int_{[a,x]}g$ from $\int_{[a,b]}gh$, you have to choose $h$ as close to the indicator function of $[a,x]$ as possible.2012-11-13
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    @user45955: On the other hand, you want to get $f(x)-f(a)$ from $\int_{[a,b]}fh$, so you have to choose $h'$ behaving like $\delta_x-\delta_a$, where $\delta_x$ is the [Dirac measure](http://en.wikipedia.org/wiki/Dirac_measure) supported at $x$.2012-11-13
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    The choice of the sequence does make sense. But I still can't get $f(x)=f(a)-\int_{[a,x]}g$.2012-11-13
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    @user45955: $n\int_{[a,a+\frac{1}{n}]}f-n\int_{[x,x+\frac{1}{n}]}f=\int_{[a,b]}fh_n'=-\int_{[a,b]}gh_n$. Let $n\to\infty$, why cannot you get $f(x)-f(a)=\int_{[a,x]}g$?2012-11-14
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    I want to find the limit of $n\int_{[a,a+1/n]}f$ , so I apply L'hospital rule here. Then I have to differentiate $n\int_{[a,a+1/n]}f$. But can I do this just like in the Riemann integral?2012-11-14
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    @user45955: Why not? Note that $f$ is continuous, so you can apply Newton-Leibniz formula or mean value theorem for integration.2012-11-14