I know the Gauss-Green theorem:
Let $U \subset \mathbb{R}^n$ be an open, bounded set with $∂U$ being $C^1$. Suppose $u ∈ C^1(\bar U)$, then $$∫_U u_{x_i} dx = \int_{∂U} u \nu^i dS,$$
where $\nu=(\nu^1,…\nu^n)$ denotes the outward-pointing unit normal vector field to the region $U$.
My question is: How to prove that this theorem is true with the weaker assumption that $u \in H^1(U)$?
Hint
$$\mathcal C^1(\bar U)\text{ is dense in }H^1(U).$$
Edit
$\mathcal C^1(\bar U)$ dense in $H^1(U)$ mean that if $u\in H^1(U)$, there exist a sequence $(u_n)\subset \mathcal C^1(\bar U)$ s.t. $$\|u_n-u\|_{H^1(U)}\underset{n\to \infty }{\longrightarrow } 0.$$ Therefore $$\int_{U}|u_n-u|^2+\sum_{i=1}^n\int_U|\partial _i u_n-\partial _i u|^2=0.$$
In paticular, since $U$ is bounded, $$\left|\int_{U}\partial _i u_n-\int_U \partial _iu\right|^2\leq C\int _U |\partial _iu_n-\partial _iu|^2,$$ by Jensen's inequality. Then you can get one limit. For $$\lim_{n\to \infty }\int_{\partial U}u_n\nu _i=\int_{\partial U}u\nu ^i,$$ it's a consequence of the continuity of the trace operator.