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  "body": "\n## Definition\n\nLet $A \\subseteq \\R^n$, and let $\\mathcal{L}^n$ be [[Lebesgue measure|2Z2QVWZ262]] on $\\R^n$.\nFor $r \u003e 0$, let $A_r = \\set{x \\in \\R^n : \\abs{x - a} \u003c r \\text{ for some } a \\in A}$.\nLet $0 \\le m \\le n$ be an integer, and let $\\alpha(k)$ be the volume of the unit ball of $\\R^k$, with $\\alpha(0) = 1$.\nThe *upper* and *lower $m$-dimensional Minkowski contents* of $A$ are\n$$\\mathcal{M}^{*m}(A) = \\limsup_{r \\to 0^+} \\frac{\\mathcal{L}^n(A_r)}{\\alpha(n - m) \\, r^{n - m}}, \\qquad \\mathcal{M}^m_*(A) = \\liminf_{r \\to 0^+} \\frac{\\mathcal{L}^n(A_r)}{\\alpha(n - m) \\, r^{n - m}}.$$\nIf they are equal, their common value $\\mathcal{M}^m(A)$ is the *$m$-dimensional Minkowski content* of $A$.\n\n## Remarks\n\nThe [[limit superior|0T3RZQXDYD]] and limit inferior always exist in $[0, \\infty]$.\nThe same formulas with a real exponent $s$ in place of $m$ are also used, with a suitable constant.\n\nExample: let $A \\subseteq \\R^2$ be a segment of length $L$ and $m = 1$.\nThen $A_r$ is a rectangle of area $2 r L$ with two half-discs of radius $r$, so $\\mathcal{L}^2(A_r) = 2 r L + \\pi r^2$.\nSince $\\alpha(1) = 2$, the ratio is $L + \\pi r / 2 \\to L$, so $\\mathcal{M}^1(A) = L$.\n\nMinkowski content does not ignore countable sets.\nFor $A = \\Q \\cap [0, 1]$ in $\\R$ with $m = 1$, one has $A_r = (-r, 1 + r)$, so $\\mathcal{M}^1(A) = 1$, although $A$ has Lebesgue measure $0$.\nThus Minkowski content is not countably additive.\n\nFor a closed set $A \\subseteq \\R^n$ that is the image of a bounded subset of $\\R^m$ under a Lipschitz map, $\\mathcal{M}^m(A)$ exists and equals its normalized $m$-dimensional [[Hausdorff measure|Q1TGZK8RN6]].\n\n## Explanations\n\n### L1 human\n\nThicken the set a little, measure the volume of the thickening, and divide by the volume one would expect from an $m$-dimensional set.\nThe limit of this ratio is the $m$-dimensional size of the set.\n",
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