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  "body": "\n## Definition\n\nLet $X$ be a topological space and let $\\mu$ be a [[Borel measure|DYR1QFNGNJ]] on $X$.\nA Borel set $A \\subseteq X$ is *inner regular* for $\\mu$ if\n$$\\mu(A) = \\sup \\set{\\mu(K) : K \\subseteq A, \\ K \\text{ compact and Borel}}.$$\nThe measure $\\mu$ is *inner regular* if every Borel set is inner regular for $\\mu$.\n\n## Remarks\n\nIn a Hausdorff space every [[compact set|0R6EWFKRCF]] is closed, hence Borel, so the condition \"Borel\" on $K$ is automatic.\n\nExample: counting measure on $\\R$ is inner regular.\nIf $A$ is finite, take $K = A$; if $A$ is infinite, its finite subsets are compact and have arbitrarily large counting measure, so the supremum is $\\infty = \\mu(A)$.\n\nNon-example: let $X$ be an uncountable set with the cocountable topology, whose open sets are $\\emptyset$ and the complements of countable sets.\nIts Borel sets are the countable sets and their complements; let $\\mu$ be $0$ on countable sets and $1$ on the others.\nCompact subsets of $X$ are finite, so they have measure $0$, while $\\mu(X) = 1$; thus $\\mu$ is not inner regular.\n\nVariants: some authors approximate by closed sets instead of compact sets, and some ask inner regularity only for open sets.\n",
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