Show a property for open intervals of absolute continuous functions

Question

I would like to show that for an absolutely continuous function $u \in AC(I)$, with $I=(a,b)\subset \mathbb{R}$ an open interval, the inequality $V(I)<\infty$ holds. We define $$V(I)=\sup\{\sum_{i=1}^N{|u(x_i)-u(x_{i-1})|}:a<x_0<x_1<\dots<x_N<b\}$$ We cannot use the property of bounded variation, because $(a,b)$ is an open interval. I tried to show it by contradiction and suppose that $\forall\ a<x_0<x_1<\dots<x_N<b $ there exists $a<y_0<y_1<\dots<y_N<b$ such that $$\sum_{i=1}^N{|u(x_i)-u(x_{i-1})|}<\sum_{i=1}^N{|u(y_i)-u(y_{i-1})|}$$ but I don't know where to search for the contradiciton.

Answer 1

Since $u$ is absolutely continuous, it is uniformly continuous in $(a,b)$. But this implies that there exists $\lim_{x\to a^+}u(x)=\ell$ and $\lim_{x\to b^-}u(x)=L$. So now extend $u$ to $[a,b]$. The extended function is absolutely continuous in $[a,b]$ (not difficult to check) and so you can apply the standard proof that an absolutely continuous function in $[a,b]$ has finite variation in $[a,b]$ and in particular also in $(a,b)$.