Total variation of (weakly) differentiable functions

Let $\Omega\subset\mathbb{R}^n$ be an open set, $u\in W^{1,1}(\Omega)$, $\nabla u$ its weak gradient and $\int_\Omega |Du|$ its total variation as defined above. We want to show that $$ \int_\Omega |Du| = \int_\Omega |\nabla u|\;dx. $$

"$\leq$" clearly holds, since

$$ \int_\Omega u \; \text{div}g \; dx = - \int_\Omega \nabla u \cdot g \; dx \leq \int_\Omega |\nabla u|\; dx $$

due to $|g|\leq 1$. It remains to show that equality is achieved or, equivalently, that "$\geq$" holds. This can be done by constructing an appropriate sequence of functions $(g_\epsilon)$ living in $C_c^1$ that, in some sense, approximates $\frac{\nabla u}{|\nabla u|}$. Then, given we found a sequence that does the trick, we would be done, due to the following argument.

\begin{align} \int_\Omega |Du| &= \sup { \int_\Omega \nabla u \cdot g \; dx: g \in C_c^1(\Omega, \mathbb{R}^n), \; |g| \leq 1 } \newline &\geq \lim_{\epsilon \rightarrow 0} \int_\Omega \nabla u \cdot g_\epsilon \; dx \newline &=\int_\Omega|\nabla u|\;dx \end{align}

Thus, the crucial point is to find such $g_\epsilon$'s. Let $$ \tilde{g}:= \begin{cases} \frac{\nabla u}{|\nabla u|}, & \text{if } \nabla u \neq 0 \newline 0, & \text{otherwise}\end{cases} $$ and let $\tilde{g}_{\epsilon_1}$ be this function multiplied with the characteristic function of the set $\Omega_{\epsilon_1} \cap B_{\epsilon^{-1}_1}$, where $\Omega_{\epsilon_1} = \{x\in\Omega: \text{dist}(x,\partial\Omega) \geq \epsilon_1\}$ and $B_{\epsilon^{-1}_1}$ is the closed ball centered at the origin with radius $\frac{1}{\epsilon_1}$. For all positive $\epsilon_1$, $\tilde{g}_{\epsilon_1}$ is a compactly supported vector field satisfying $|\tilde{g}_{\epsilon_1}| \leq 1$ and $\tilde{g}_{\epsilon_1}\overset{\epsilon_1\rightarrow 0}{\rightarrow}\tilde{g}$. By convolution with a suitable mollifier $\eta_{\epsilon_2}$ (see e.g. Giusti 1984, pp. 10-11) the resulting functions $g_\epsilon := \eta_{\epsilon_2} \ast \tilde{g}_{\epsilon_1}$, $\epsilon := (\epsilon_1, \epsilon_2)$, are compactly supported, smooth and $g_\epsilon \overset{\epsilon_2\rightarrow 0}{\rightarrow}\tilde{g}_{\epsilon_1}$ in $L_1$. This sequence of functions (I believe) fulfills the above equality, which should conclude the proof.

Is this the right way? I have to emphasize that I am not very sure about the double limit, and might need some input to make it fully rigorous. Any ideas?

Let $\epsilon > 0$. Pick $f$ in $C^\infty(\Omega) \cap BV(\Omega)$ so $\|u-f\|_{L^1(\Omega)} < \epsilon$ and $\Big|\int_\Omega |Du|- \int_\Omega |Df|\Big| < \epsilon$. By integration by parts,

$$ \begin{aligned} \int_\Omega |Df|&= \sup\{ \int_\Omega f\text{ div}g \mid g\in C_0^1(\Omega, \mathbb{R}^n)|g| \leq 1\}\\ &=\sup\{ \int_\Omega -\nabla f \cdot g \mid g\in C_0^1(\Omega, \mathbb{R}^n),\ |g| \leq 1\}\\ &=\sup\{ \int_\Omega \nabla f \cdot g \mid g\in C_0^1(\Omega, \mathbb{R}^n)\, |g| \leq 1\} = \|\nabla f\|_{L^1({\Omega})}\\ \end{aligned} $$ % which can be confirmed by letting $g$ close to $\frac{1}{\lvert \nabla f \rvert} \nabla f$. Now let $\epsilon \to 0$