Sobolev Space $W_0^{1,p}(I)$ and the boundary of $I$ Given $I \subset\mathbb{R}$ an open interval, the Sobolev Space $W_0^{1,p}(I)$ is defined as $W_0^{1,p}(I)=\overline{C^1_c(I)}^{W^{1,p}(I)}$ (The closure of $C^1_c(I)$ on the space $W^{1,p}(I)$) .
There is a basic result that tells us:

Given $u \in W^{1,p}(I)$, then $u \in W_0^{1,p}(I)$ if and only if $u=0$ on $\partial(I)$.

Considering the fact that the boundary of $I$ is at most consisting of two points and that two points is a set of meassure zero, Is the boundary the usual topological boundary? Does this result mean there is some representative $\tilde{u}$ of the equivalence class of $u$ which actually banishes at $\partial(I)$?
 A: Yes, the boundary is the usual topological boundary. To answer your second question, as stated, yes, it means that there is a representative that attains the value $0$ on the boundary of the interval. As you probably recall, every Sobolev function in one variable has a representative that is absolutely continuous, hence the statement you quoted is correct, but it is tacitly implied that you choose this representative. (Also recall that an absolutely continuous function defined in $I$ has a unique continuous extension to $\bar{I}$, this is useful to keep in mind!)
It is worth mentioning that this doesn't quite work in more dimensions. For a generic open set $\Omega \subset \mathbb{R}^N$, $N \ge 2$, with Lipschitz boundary, the best you can say is that $u = 0$ on $\partial \Omega$ for $u \in C(\bar{\Omega}) \cap W_0^{1,p}(\Omega)$.
I hope this answers your questions! :)
A: The points do not have measure zero. On the (topological) boundary you typically have to consider the (so called Hausdorff-) measure of one dimension less (for, e.g., defining equivalence classes). And the zero dimensional measure is the "counting measure". Every point has measure one.
