Q:

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? 

A:

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 $f(x)=f(a)+\int_a^xg(t)dt$. 

