# absolutely continuous

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 $h$ ... –  23rd Nov 13 '12 at 16:56
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 $g$. –  Thomas Andrews Nov 13 '12 at 17:23
Yes, that's what I mean. $g$ should not depend on $h$. I rewrite the question to avoid the confusion. –  user45955 Nov 13 '12 at 17:37

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 $f(x)=f(a)+\int_a^xg(t)dt$.
@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. –  23rd Nov 13 '12 at 19:07
@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 supported at $x$. –  23rd Nov 13 '12 at 19:12
The choice of the sequence does make sense. But I still can't get $f(x)=f(a)-\int_{[a,x]}g$. –  user45955 Nov 13 '12 at 21:37
@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 $f(x)-f(a)=\int_{[a,x]}g$? –  23rd Nov 14 '12 at 5:15