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 has bounded derivative, and thus is Lipschitz. Try then to take care of the case of finite intervals containing . Is this where you're stuck?
I would like to prove that f is absolute continuous over the closed interval [-1,1] .
Actually the derivative is unbounded there thus not Lipschitz.
I am trying to use the definition.