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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $\\Omega \\subseteq \\R^n$ be an open set, and let $E \\subseteq \\R^n$ be a [[Lebesgue measurable|2Z2QVWZ262]] set.\nIts *perimeter in $\\Omega$* is the supremum below.\n$$P(E, \\Omega) = \\sup \\set{ \\int_E \\operatorname{div} \\varphi \\, dx : \\varphi \\in C^1_c(\\Omega; \\R^n), \\ \\abs{\\varphi(x)} \\le 1 \\text{ for all } x \\in \\Omega }.$$\nHere $\\operatorname{div}$ is the [[divergence|0MWV6ZTK6S]].\nThe set $C^1_c(\\Omega; \\R^n)$ is the space of continuously differentiable vector fields with compact support in $\\Omega$.\nThe value $P(E, \\Omega)$ lies in $[0, \\infty]$, and $P(E) = P(E, \\R^n)$ is the *perimeter* of $E$.\n\n## Remarks\n\nExample: let $E$ be a bounded open set with $C^1$ boundary and outer unit normal $\\nu$.\nBy the divergence theorem, $\\int_E \\operatorname{div} \\varphi \\, dx = \\int_{\\partial E} \\varphi \\cdot \\nu \\, d\\mathcal{H}^{n-1} \\le \\mathcal{H}^{n-1}(\\partial E)$.\nExtending $\\nu$ 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.\nSo the unit disc in $\\R^2$ has perimeter $2\\pi$.\n\nThe perimeter is not the size of the topological boundary.\nThe set $E = \\Q^2 \\cap [0, 1]^2$ has topological boundary $[0, 1]^2$, of positive area.\nBut $E$ has Lebesgue measure zero, so every integral above vanishes and $P(E) = 0$.\n\nEquivalently, $P(E, \\Omega)$ is the total variation $\\abs{D \\chi_E}(\\Omega)$ of the distributional gradient of the indicator function $\\chi_E$.\nIt is finite exactly when $\\chi_E$ has bounded variation in $\\Omega$.\nChanging $E$ on a set of measure zero does not change $P(E, \\Omega)$.\n",
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