Perimeter of a set

Remarks

Example: let $E$ be a bounded open set with $C^1$ boundary and outer unit normal $\nu$. By the divergence theorem, $\int_E \operatorname{div} \varphi \, dx = \int_{\partial E} \varphi \cdot \nu \, d\mathcal{H}^{n-1} \le \mathcal{H}^{n-1}(\partial E)$. Extending $\nu$ to a continuous field of norm at most $1$ and smoothing it shows that the supremum equals $\mathcal{H}^{n-1}(\partial E)$, the area of the boundary. So the unit disc in $\R^2$ has perimeter $2\pi$.

The perimeter is not the size of the topological boundary. The set $E = \Q^2 \cap [0, 1]^2$ has topological boundary $[0, 1]^2$, of positive area. But $E$ has Lebesgue measure zero, so every integral above vanishes and $P(E) = 0$.

Equivalently, $P(E, \Omega)$ is the total variation $\abs{D \chi_E}(\Omega)$ of the distributional gradient of the indicator function $\chi_E$. It is finite exactly when $\chi_E$ has bounded variation in $\Omega$. Changing $E$ on a set of measure zero does not change $P(E, \Omega)$.