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  "body": "\n## Definition\n\nLet $\\kappa$ be a [[cardinal|0EBACQFAZW]].\nAn [[ultrafilter|J64Y3JC0RE]] $U$ on $\\kappa$ is *$\\kappa$-complete* if $\\bigcap \\mathcal{F} \\in U$ for every family $\\mathcal{F} \\subseteq U$ with $0 \u003c \\abs{\\mathcal{F}} \u003c \\kappa$.\nIt is *nonprincipal* if it contains no set $\\set{\\alpha}$ with $\\alpha \u003c \\kappa$.\nThe cardinal $\\kappa$ is *measurable* if:\n\n1. $\\kappa$ is [[uncountable|0W7FBYS1F1]].\n2. There is a $\\kappa$-complete nonprincipal ultrafilter on $\\kappa$.\n\n## Remarks\n\nThe cardinal $\\aleph_0$ satisfies clause 2: every ultrafilter is closed under finite intersections, and nonprincipal ultrafilters on $\\omega$ exist by the axiom of choice; it fails only clause 1.\nThe cardinal $\\aleph_1$ fails clause 2, because every measurable cardinal is inaccessible, and $\\aleph_1$ is not.\nEquivalent form: $\\kappa$ is measurable if and only if there is an elementary embedding $j \\colon V \\to M$, with $M$ a transitive class, whose critical point is $\\kappa$.\nThe critical point condition means that $j(\\alpha) = \\alpha$ for $\\alpha \u003c \\kappa$ and $j(\\kappa) \u003e \\kappa$.\nThe embedding comes from the ultrapower of $V$ by $U$, and conversely $U = \\set{X \\subseteq \\kappa : \\kappa \\in j(X)}$.\nScott proved that if a measurable cardinal exists, then $V \\neq L$.\n",
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