The total variation of a function $u\in L^1_{loc} (\Omega)$ is defined as $$\text{TV}(u,\Omega)= \sup \, \bigg\{ -\int_{\Omega} u\, div \phi \, dx : \phi \in C_c^{\infty} (\Omega,\mathbb{R}^N), \, \lvert \phi (x) \rvert \leq 1\, \forall x\in \Omega \bigg \}. $$
Now if $E \subseteq \Omega$, the perimeter of $E$ is defined as the total variation of its characteristic function $\chi_E$: $$ \text{Per}(E)= \text{TV}(\chi_E,\Omega)= \sup \, \bigg\{ -\int_{E}\, div \phi \, dx : \phi \in C_c^{\infty} (\Omega,\mathbb{R}^N), \, \lvert \phi (x) \rvert \leq 1\, \forall x\in \Omega \bigg \}.$$ My question is, how do you practically calculate the perimeter of a given set, using this formula? For example, for a tyical square $\square$ in $\mathbb{R}^2$ with each of its sides having length 1, we'd expect the Per to equal 4 but I get this: $$ \text{Per}(\square)= \text{TV}(\chi_{\square},\Omega)= \sup \, \bigg\{ -\int_{\square}\, div \phi \, dx : \phi \in C_c^{\infty} (\Omega,\mathbb{R}^2), \, \lvert \phi (x) \rvert \leq 1\, \forall x\in \Omega \bigg \} = \int 1 \ dx dy = 1 $$ which is more like the area of the square, not its perimeter. Or am I wrong?
Thanks bunches.