Quick question. Let $I$ be an open bounded subset of $\mathbb{R}^{n}$. If I am given that $u_{m},u \in W^{1,\infty}(I)$ and I want to show that $u_{m} \rightharpoonup^{*} u$ in $L^{\infty}(I)$. Then I have to show that given any $\phi \in L^{1}(I)$ I can show $$\int_{I}u_{m}\phi dx \rightarrow \int_{I}u \phi dx$$
Would it be equivalent to instead show $$\int_{\bar{I}}u_{m}\phi dx \rightarrow \int_{\bar{I}}u \phi dx$$
where the integral is taken over $\bar{I}$ and I define $u_{m}$ and $u$ on $\partial I$ using the trace operator $T: W^{1,\infty}(I) \rightarrow L^{\infty}(\partial I)$?
If the answer to the above question is yes, would it be so because $L^{\infty}(I) = L^{\infty}(\bar{I})$?