Does the pre image of a open interval is a open interval, if the function is absolutely continuous and non decreasing?

Assume that $u:[0,1]\to \mathbb{R}$ is a absolutely continuous (A.C. for short), non decreasing function. Suppose that $u(0)=\alpha$ and $u(1)=\beta$. Take any open interval $J\subset [\alpha,\beta]$. Is it possible to find a open interval $I\subset [0,1]$ such that $$u(I)=J.$$

The Cantor function shows that A.C is needed, however, I fail to see if it is sufficiently. Any idea is appreciated.

• What do you mean by "non decreasing"? – Stefan Mesken Jun 20 '14 at 15:29
• @user155124, $u(x)\le u(y)$ if $x\le y$. – Tomás Jun 20 '14 at 15:30
• Ok, so you actually mean "increasing". Then my counterexample doesn't work. – Stefan Mesken Jun 20 '14 at 15:32

Since $u$ is continuous and $J$ is open, it follows that $u^{-1}(J)$ is open.
Since $u$ is monotone and $J$ is an interval, it follows that $u^{-1}(J)$ is an interval.
Therefore, $u^{-1}(J)$ is an open interval. Since $J$ lies in the range of $u$, we have $u(u^{-1}(J))=J$.