# Absolute continuity

• December 3rd 2011, 04:30 PM
hgd7833
Absolute continuity
How to prove that x^1/3 is absolutely continuous without using
the fact that it is the integral of a derivative.
• December 3rd 2011, 05:36 PM
Drexel28
Re: Absolute continuity
Quote:

Originally Posted by hgd7833
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 $x^{\frac{1}{3}}$ has bounded derivative, and thus is Lipschitz. Try then to take care of the case of finite intervals containing $0$. Is this where you're stuck?
• December 4th 2011, 07:58 AM
hgd7833
Re: Absolute continuity
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.