# Trace Theorem

In the proof I would like to know why (1) is not trivial since $\Gamma \subset \partial U$ and $\partial U$ is a boundary which therefore has measure zero, does it not follow then that $\int_{\Gamma} |u|^{p}dx' \leq \int_{\partial U }|u|^{p}dx' = 0$, therefore $\int_{\Gamma} |u|^{p}dx' = 0\text{ } \leq\text{ } C\int_{U}|u|^{p} + |Du|^{p}dx$.
And then would the last line $||u||_{L^{p}(\Gamma_{i})} \text{ } \leq \text{ } C||u||_{W^{1,p}(U)} \text{ for i=1,...,N}$ not also follow easily since $0 = ||u||_{L^{p}(\Gamma_{i})} \text{ } \leq \text{ } C||u||_{W^{1,p}(U)} \text{ for i=1,...,N}$.