Is $f \in W^{1,1}[a,b]$ equivalent to $f$ absolutely continuous on $[a,b]$?

Question

$f$ is a function defined on $[a,b]$. Then $f \in W^{1,1}$ is equivalent to $f$ is absolutely continuous?

Answer 1

If $f\in W^{1,1}\big([0,1]\big)$, then $f$ is absolutely continuous, as $$ \sum_{k=1}^n |f(x_k)-f(y_k)|=\sum_{k=1}^n \Big|\int_{x_k}^{y_k}f'(x)\,dx\Big| \le\sum_{k=1}^n \int_{x_k}^{y_k}|f'(x)|\,dx, $$ and then use the fact that: If $g\in L^1$, then for every $\varepsilon>0$, there exists a $\delta>0$, such that $m(E)<\delta$ implies $\int_{E}|g|<\varepsilon$.

On the other hand, if $f$ is absolutely continuous, then there exists a $g\in L^1$, such that $$ f(x)=f(a)+\int_a^x g(t)\,dt, $$ in which case, one can show that $f$ possesses a weak derivative in $L^1$, which coincides with $g$ a.e., and hence $f\in W^{1,1}$.