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  "body": "\n## Definition\n\nLet $X$ be a topological space and let $\\mu$ be a [[Borel measure|DYR1QFNGNJ]] on $X$.\nThe measure $\\mu$ is *regular* if both of the following hold.\n\n1. $\\mu$ is [[inner regular|28E5Q5BM69]]: $\\mu(A) = \\sup \\set{\\mu(K) : K \\subseteq A, \\ K \\text{ compact and Borel}}$ for every Borel set $A$.\n2. $\\mu$ is [[outer regular|33C1KEK3SC]]: $\\mu(A) = \\inf \\set{\\mu(U) : U \\supseteq A, \\ U \\text{ open}}$ for every Borel set $A$.\n\n## Remarks\n\nExample: Lebesgue measure on the Borel sets of $\\R^n$ is regular.\nOuter regularity follows from its construction by covers with open boxes.\nFor inner regularity, a Borel set $A$ contains a closed set $F$ with $\\lambda(A \\setminus F)$ arbitrarily small when $\\lambda(A) \u003c \\infty$.\nThe sets $F \\cap [-k, k]^n$ are compact, and their measures tend to $\\lambda(F)$.\nThe case $\\lambda(A) = \\infty$ follows by cutting $A$ into bounded pieces.\n\nNon-example: counting measure on $\\R$ satisfies clause 1 but not clause 2.\nFinite sets are compact, which gives clause 1; the set $\\{0\\}$ has measure $1$, but every open set around it has measure $\\infty$.\n\nConventions vary.\nMany texts use closed sets instead of compact sets in clause 1.\nOn a metric space, every finite Borel measure is regular in that closed-set sense.\n",
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