f is super absolutely continuous on [a,b] if for each $\epsilon > 0$ there exists $\delta > 0$ such that for every finite collection of intervals $\{(a_i,b_i)\}_{i=1}^n$ of open intervals in $(a,b)$, if $\sum_{i=1}^n [b_i-a_i] < \delta$ then $\sum_{i=1}^n |f(b_i)-f(a_i)|< \epsilon$.
Edit: Does this definition absolutely continuous imply super absolutely continuous. I believe it doesn't as it drops the assumption of disjointness from the collection. However, I am having a hard time coming up with a good counterexample.