$\mathbf{Theorem:}$ Let $f$ be defined on $I=[a,b]$ and absolutely continuous on $I$. Then $f$ is differentiable a.e in $(a,b)$, and its derivative is integrable over $[a,b]$ with $$\int_{[a,b]}f'=f(b)-f(a).$$ I am ok with this theorem. But I need an example to show the converse is not true, $i.e$, we are looking for a function that is differentiable a.e on $(a,b)$ with $$\int_{[a,b]}f'=f(b)-f(a),$$ but it is not absolutely continuous on its domain, which we want it to be compact. I have an example which i am not sure about it:
Define $f(x)=x^{\frac{1}{3}}$ for $x \in [-1,1]$ and $g(x)=\begin{cases} x^2 \cos(\pi/2x) & x \in [-1,1]\sim \{0\}\\ 0 & x =0\end{cases}$.
Clearly, $f$ is increaing on $[-1,1]$, and we have that for any $1>\epsilon>0$, $|f'(x)|=|\frac{1}{3 \sqrt[3]{x^2}}|\leq |\frac{1}{3\sqrt[3]{\epsilon^2}}|$ for all $x \in [\epsilon,1]\cup [-1,-\epsilon]$, thus $f$ is Lipschitz on $[\epsilon,1]\cup [-1,-\epsilon]$, hence $f$ is absolutely continuos on $[\epsilon,1]\cup [-1,-\epsilon]$. Thus, $f$ is absolutely continuous on $[0,1]\cup [-1,0]=[-1,1]$ since it is increaing.
Now, we see obviously, $g$ is differentiable on $[-1,1]$ with $g'(x)=2x \cos(\pi/2x)+\frac{\pi}{2}\sin(\pi/2x)$, and we have $|g'(x)|\leq 2+1=3$. Therefore, $g$ is Lipschitz on $[-1,1]$, thus it is absolutely continuous on $[-1,1].$
Now, consider the partition $\mathcal{P}_n=\{-1,0,\frac{1}{2n},\frac{1}{2n-1},...,\frac{1}{2},1\}$ of $[-1,1]$. We have $(f \circ g)(x)=x^{^{\frac{2}{3}}}\cos^{^{\frac{1}{3}}}(\frac{\pi}{2x})$, so $$\sum_{i=1}^{2n-1}|f(x_{i})-f(x_{i-1})|=|0-0|+|\frac{\cos^{1/3}(n \pi)}{2n^{2/3}}-0|+|\frac{\cos^{1/3}(\frac{(2n-1)\pi}{2})}{(2n-1)^{2/3}}-\frac{\cos^{1/3}(n \pi)}{2n^{2/3}}|+...$$ $$+ |\frac{\cos^{1/3}( \pi)}{2^{2/3}}-\frac{\cos^{1/3}(\pi/2)}{1^{2/3}}|=\sum_{i=0}^n2^{^{1/3}}\frac{1}{i^{2/3}}=2^{^{1/3}}\sum_{i=0}^n\frac{1}{i^{2/3}}.$$ since $2/3<1$, then the sum diverges as $n \rightarrow \infty$, thus $f\circ g$ fails to be of a bounded variation on $[-1,1]$, and hence it is not absolutely continuous on $[-1,1].$
But I can not see that $$\int_{[0,1]}(f \circ g)'=f(g(b))-f(g(a))$$
Thanks for any help.