Editing Measurable cardinal (section)

== Properties ==
It is trivial to note that if ''κ'' admits a non-trivial ''κ''-additive measure, then ''κ'' must be [[regular cardinal|regular]]. (By non-triviality and ''κ''-additivity, any subset of cardinality less than ''κ'' must have measure 0, and then by ''κ''-additivity again, this means that the entire set must not be a union of fewer than ''κ'' sets of cardinality less than ''κ.'')  Finally, if ''λ''&nbsp;<&nbsp;''κ,'' then it can't be the case that ''κ''&nbsp;≤&nbsp;2<sup>''λ''</sup>. If this were the case, we could identify ''κ'' with some collection of 0-1 sequences of length ''λ.'' For each position in the sequence, either the subset of sequences with 1 in that position or the subset with 0 in that position would have to have measure&nbsp;1. The intersection of these ''λ''-many measure 1 subsets would thus also have to have measure 1, but it would contain exactly one sequence, which would contradict the non-triviality of the measure. Thus, assuming the Axiom of Choice, we can infer that ''κ'' is a [[strong limit cardinal]], which completes the proof of its [[inaccessible cardinal|inaccessibility]].

Although it follows from [[ZFC]] that every measurable cardinal is [[inaccessible cardinal|inaccessible]] (and is [[Ineffable cardinal|ineffable]], [[Ramsey cardinal|Ramsey]], etc.), it is consistent with [[Zermelo–Fraenkel set theory|ZF]] that a measurable cardinal can be a [[successor cardinal]]. It follows from ZF&nbsp;+&nbsp;[[axiom of determinacy|AD]] that ω<sub>1</sub> is measurable,<ref name="JechDeterminacy81">T. Jech, "[https://projecteuclid.org/journalArticle/Download?urlid=bams%2F1183548432 The Brave New World of Determinacy]" (PDF download). Bulletin of the American Mathematical Society, vol. 5, number 3, November 1981 (pp.339--349).</ref> and that every subset of ω<sub>1</sub> contains or is disjoint from a [[club set|closed and unbounded]] subset.

Ulam showed that the smallest cardinal ''κ'' that admits a non-trivial countably-additive two-valued measure must in fact admit a ''κ''-additive measure. (If there were some collection of fewer than ''κ'' measure-0 subsets whose union was ''κ,'' then the induced measure on this collection would be a counterexample to the minimality of ''κ.'') From there, one can prove (with the Axiom of Choice) that the least such cardinal must be inaccessible.

If ''κ'' is measurable and ''p''&nbsp;∈&nbsp;''V''<sub>''κ''</sub> and ''M'' (the ultrapower of ''V'') satisfies ''ψ''(''κ,&nbsp;p''), then the set of ''α''&nbsp;<&nbsp;''κ'' such that ''V'' satisfies ''ψ''(''α,&nbsp;p'') is stationary in ''κ'' (actually a set of measure 1). In particular if ''ψ'' is a Π<sub>1</sub> formula and ''V'' satisfies ''ψ''(''κ,&nbsp;p''), then ''M'' satisfies it and thus ''V'' satisfies ''ψ''(''α,&nbsp;p'') for a stationary set of ''α''&nbsp;<&nbsp;''κ.'' This property can be used to show that ''κ'' is a limit of most types of large cardinals that are weaker than measurable. Notice that the ultrafilter or measure witnessing that ''κ'' is measurable cannot be in ''M'' since the smallest such measurable cardinal would have to have another such below it, which is impossible.

If one starts with an elementary embedding ''j''<sub>1</sub> of ''V'' into ''M''<sub>1</sub> with [[critical point (set theory)|critical point]] ''κ,'' then one can define an ultrafilter ''U'' on ''κ'' as {&nbsp;''S''&nbsp;⊆&nbsp;''κ''&nbsp;| ''κ''&nbsp;∈&nbsp;''j''<sub>1</sub>(''S'')&nbsp;}. Then taking an ultrapower of ''V'' over ''U'' we can get another elementary embedding ''j''<sub>2</sub> of ''V'' into ''M''<sub>2</sub>. However, it is important to remember that ''j''<sub>2</sub>&nbsp;≠&nbsp;''j''<sub>1</sub>. Thus other types of large cardinals such as [[strong cardinal]]s may also be measurable, but not using the same embedding. It can be shown that a strong cardinal ''κ'' is measurable and also has ''κ''-many measurable cardinals below it.

Every measurable cardinal ''κ'' is a 0-[[huge cardinal]] because <sup>''κ''</sup>''M''&nbsp;⊆&nbsp;''M'', that is, every function from ''κ'' to ''M'' is in ''M''. Consequently, ''V''<sub>''κ''+1</sub>&nbsp;⊆&nbsp;''M''.