Let $\Omega\subseteq \mathbb{R}^N$ be a regular domain (a bounded open set with $C^1$ boundary) and let $X$ be a $W^{1,1}(\Omega,\mathbb{R}^N)$ be a Sobolev vector field. I want to prove the following version of the divergence theorem:
$$\int_{\Omega} \text{div}(X)=\int_{\partial\Omega} \text{Tr}(X)\cdot \nu$$
where $\text{Tr}$ is the trace operator. Clearly the theorem is true if $X\in C^{\infty}(\overline{\Omega},\mathbb{R}^N)$ because of the classical divergence theorem. Otherwise we can find a sequence of $\{X_n\}\subseteq C^{\infty}(\overline{\Omega},\mathbb{R}^N)$ converging to $X$. And clearly
$$\lim_{n\to \infty} \int_{\Omega} \text{div}(X_n)=\lim_{n\to \infty} \int_{\partial\Omega} \text{Tr}(X_n)\cdot \nu$$
If we can exchange limits and integral we've done because $\text{div}$ and $\text{Tr}$ are continous operators. But why can we do it?
I'm trying to apply dominated convergence but without any success.