How to prove that x^1/3 is absolutely continuous without using
the fact that it is the integral of a derivative.
I remember this being annoying to do. You know everything's fine on any union of intervals bounded away from zero, since $\displaystyle x^{\frac{1}{3}}$ has bounded derivative, and thus is Lipschitz. Try then to take care of the case of finite intervals containing $\displaystyle 0$. Is this where you're stuck?