On the total variation of a differentiable function

Question

According to Wikipedia the total variation of a differentiable function defined on a bounded open set $\Omega \subset \mathbb{R}^n$ can be expressed as $$V(f, \Omega) = \int_\Omega \left| \, \nabla \, f \, \right|  \, \mathrm{d}x. $$ Does this, under some conditions, hold if $\Omega$ is unbounded? Is there some good reference work I should consult?

Answer 1

Note that, by the very definition of $V(f, \Omega)$, we have that $$ V(f, \Omega) = \sup\{V(f|_B, B) : B \subseteq \Omega \text{ bounded, open} \}$$ To see this, let $\epsilon > 0$, then there is a $\phi \in C^1_c(\Omega, \mathbf R^n)$, with $\def\abs#1{\left|#1\right|}\abs \phi \le 1$, such that $$ \int f\operatorname{div} \phi \, dx \ge V(f,\Omega) - \epsilon $$ As $\phi$ has compact support, there is an open and bounded $B$ with $\operatorname{supp}\phi \subseteq B \subseteq \Omega$ (to see this, note that the compact set $\operatorname{supp} \phi$ and the closed set $\mathbf R^n - \Omega$ have postive distance $\delta$. Now let, for example $B = \bigcup_{x \in \operatorname{supp}\phi} U_{\delta/2}(x)$). Then $\phi|_B \in C^1_c(\Omega, \mathbf R^n)$, hence $$ V(f|_B,B) \ge \int_B f\operatorname{div}\phi \, dx = \int_\Omega f\operatorname{div}\phi \, dx \ge V(f,\Omega) - \epsilon $$ Hence, $$ V(f,\Omega) = \sup_{B\subseteq \Omega\text{ open, bounded}} V(f|_B,B) = \sup_{B \subseteq \Omega} \int_B \abs{\nabla f}\, dx = \int_\Omega \abs{\nabla f}\, dx $$ for unbounded $\Omega$.