I want to show that if $f$ is continuous on $[a,b]$ and differentiable a.e. on $(a,b)$. Then
$\int_{a}^{b}f'=f(b)-f(a)=>\int_{a}^{b}\lim_{n->\infty}Diff_{\frac{1}{n}}f=\lim_{n->\infty}\int_{a}^{b}Diff_{\frac{1}{n}}f$
Definition: $Diff_{h}=\frac{f(x+h)-f(x)}{h}$, the divided difference.
ATTEMPT:
By differentiability I know that $Diff_{\frac{1}{n}}f->f'$ pointwise a.e. on $(a,b)$. So if I can somehow show $f$ is absolutely continuous, then I will have the sequence of divided differences $(Diff_{\frac{1}{n}}f)_n$ uniformly integrable and by the Vitali Convergence Theorem I can pass the limit inside the integral. But from the information given, how can I show $f$ is absolutely continuous, or is that a faulty approach?
Any help is much appreciated.