In section 5.5 of Evans PDE book he introduces the trace operator. He states the motivation behind this operator being that generally $u \in W^{1,p}(U)$ is not continuous or is only defined a.e. in $U$. Consequently, $\partial U$ may have Lebesgue measure zero and hence, as Evans puts it, there is no direct meaning we can give to the expression "$u$ restricted to $\partial U$".
Even before introducing trace class operators, how do the above two issues make the expression "$u$ restricted to $\partial U$" undefined?
The trace theorem states that if $U$ is bounded and $\partial U \in C^1$, then there exists a bounded linear operator $$T: W^{1,p}(U) \rightarrow L^p(\partial U)$$ such that $$Tu = u\vert_{\partial U}, \quad u \in W^{1,p}(U) \cap C(\overline{U})\tag{1}$$ and $$\|Tu\|_{L^p(\partial U)} \leq C\|u\|_{W^{1,p}(U)}. \tag{2}$$
Isn't (2) simply stating that $T$ is bounded? But this is already assumed as a hypothesis in the theorem. Secondly, how does the trace operator solve the "restriction to the boundary" problem exactly described by Evans, it seems (1) is trivially restricting $u$ to the boundary, so what's different?