# Measurable cardinal

A **measurable cardinal** $\kappa$ is an uncountable cardinal such that it is possible to "measure" the subsets of $\kappa$ using a 2-valued measure on the powerset of $\kappa$, $\mathcal{P}(\kappa)$. There exists several other equivalent definitions: For example, $\kappa$ can also be the critical point of a nontrivial elementary embedding $j:V\to M$.

Every measurable is a large cardinal, i.e. $V_\kappa$ satisfies $\text{ZFC}$, therefore $\text{ZFC}$ cannot prove the existence of a measurable cardinal. In fact $\kappa$ is inaccessible, the $\kappa$th inacessible, the $\kappa$th weakly compact cardinal, the $\kappa$th Ramsey, and similarly bears most of the large cardinal properties under Ramsey-ness. It is notable that every measurable has the mentioned properties in $\text{ZFC}$, but in $\text{ZF}$ they may not (but their existence remains consistency-wise *much* stronger than existence of cardinals with those properties), in fact under the axiom of determinacy, the first two uncountable cardinals, $\aleph_1$ and $\aleph_2$, are both measurable.

Measurable cardinals were introduced by Stanislaw Ulam in 1930.

## Contents

## Definitions

There are essentially two ways to "measure" a cardinal $\kappa$, that's to say we can require the measure to be $\sigma$-additive (a "classical" measure) or to be $\kappa$-additive (for every cardinal $\lambda$ such that $\lambda < \kappa$, the union of $\lambda$ null sets still has measure zero).

Let $\kappa$ be an uncountable cardinal.

Theorem 1 : The following are equivalent :

- There exists a 2-valued ($\sigma$-additive) measure on $\kappa$.
- There exists a $\sigma$-complete nonprincipal ultrafilter on $\kappa$.

The equivalence is due to the fact that if $\mu$ is a 2-valued measure on $\kappa$, then $U=\{X\subset\kappa|\mu(X)=1\}$ is a nonprincipal ultrafilter (since $\mu$ is 2-valued) and is also $\sigma$-complete because of $\mu$'s $\sigma$-additivity. Similarly, if $U$ is a $\sigma$-complete nonprincipal ultrafilter on $\kappa$, then $\mu:\mathcal{P}(\kappa)\to[0,1]$ defined by $\mu(X)=1$ whenever $X\in U$, $\mu(X)=0$ otherwise is a 2-valued measure on $\kappa$.

An uncountable cardinal which satisfies the equivalent conditions of theorem 1 is sometimes called a 2-measurable cardinal (because "2-valued"). This is not a traditional notation, but it was used in an article of Gustave Choquet : "Cardinaux 2-mesurables et cônes faiblement compacts", Annales de l'Institut Fourier, tome 17, n°2 (1967), P.383-393.

Note : It is clear that, il $\kappa$ is 2-measurable, then every cardinal $\lambda$ such that $\lambda > \kappa$ is also 2-measurable. Thus, the notion of 2-measurability separates the class $C$ of all cardinals in two subclasses : the "moderated" cardinals and the 2-measurable cardinals, the first one being an initial segment of $C$, and therefore this notion is of weak interest for the study of the hierarchy of large cardinals.

Theorem 2 : The following are equivalent :

- There exists a $\kappa$-complete nonprincipal ultrafilter on $\kappa$.
- There exists a nontrivial elementary embedding $j:V\to M$ with $M$ a transitive class and such that $\kappa$ is the least ordinal moved (the
*critical point*). - There exists a nonprincipal ultrafilter $U$ on $\kappa$ such that the ultrapower $(\text{Ult}_U(V),\in_U)$ of the universe is well-founded.

To see that the second condition implies the first one, one can show that if $j:V\to M$ is a nontrivial elementary embedding, then the set $\mathcal{U}=\{x\subset\kappa|\kappa\in j(x)\})$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, and in fact a normal fine measure. To show the converse, one needs to use ultrapower embeddings: if $U$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$, then the canonical ultrapower embedding $j:V\to\text{Ult}_U(V)$ is a nontrivial elementary embedding of the universe.

An uncountable cardinal $\kappa$ is called measurable if the equivalent conditions of theorem 2 are satisfied.

The two theorems are related by the fact (easy to prove) that the least cardinal $\kappa$ (if it exists) which carries a $\sigma$-complete nonprincipal ultrafilter is measurable, and in this case every $\sigma$-complete nonprincipal ultrafilter on $\kappa$ is $\kappa$-complete (see for example Patrick Dehornoy : "La théorie des ensembles", Calvage et Mounet, 2017).

In other words, the first 2-measurable cardinal is measurable.

Therefore, the two notions are equiconsistent, but in the general case they differ : every measurable cardinal is 2-measurable, and the converse is false.

## Properties

If $\kappa$ is measurable, then it has a measure that take every value in $[0,1]$. Also there must be a normal fine measure on $\mathcal{P}_\kappa(\kappa)$.

Every measurable cardinal is regular, and (under AC) bears most large cardinal properties weaker than it. It is in particular $\Pi^2_1$-indescribable. However the least measurable cardinal is not $\Sigma^2_1$-indescribable. Independently of the truth of AC, the existence of a measurable cardinal implies the consistency of the existence of large cardinals with the said properties, even if that measurable is merely $\omega_1$.

If $\kappa$ is measurable and $\lambda<\kappa$ then it cannot be true that $\kappa<2^\lambda$. Under AC this means that $\kappa$ is a strong limit (and since it is regular, it must be strongly inaccessible, hence it cannot be $\omega_1$).

If there exists a measurable cardinal then $0^\#$ exists, and therefore $V\neq L$. In fact, the sharp of every real number exists, and therefore $\mathbf{\Pi}^1_1$-determinacy holds. Furthermore, assuming the axiom of determinacy, the cardinals $\omega_1$, $\omega_2$, $\omega_{\omega+1}$ and $\omega_{\omega+2}$ are measurable, also in $L(\mathbb{R})$ every regular cardinal smaller than $\Theta$ is measurable.

Every measurable has the following reflection property: let $j:V\to M$ be a nontrivial elementary embedding with critical point $\kappa$. If $x\in V_\kappa$ and $M\models\varphi(\kappa,x)$ for some first-order formula $\varphi$, then the set of all ordinals $\alpha<\kappa$ such that $V\models\varphi(\alpha,x)$ is stationary in $\kappa$ and has the same measure as $\kappa$ itself by any 2-valued measure on $\kappa$.

Measurability of $\kappa$ is equivalent with $\kappa$-strong compactness of $\kappa$, and also with $\kappa$-supercompactness of $\kappa$ (fragments of strong compactness and supercompactness respectively.) It is also consistent with $\text{ZFC}$ that the first measurable cardinal and the first strongly compact cardinal are equal.

If a measurable $\kappa$ is such that there is $\kappa$ strongly compact cardinals below it, then it is strongly compact. If it is a limit of strongly compact cardinals, then it is strongly compact yet not supercompact. If a measurable $\kappa$ has infinitely many Woodin cardinals below it, then the axiom of determinacy holds in $L(\mathbb{R})$, also the axiom of projective determinacy holds.

If $\kappa$ is measurable in a ground model, then it is measurable in any forcing extension of that ground model whose notion of forcing has cardinality strictly smaller than $\kappa$. Prikry showed however that every measurable can be collapsed to a cardinal of cofinality $\omega$ and no other cardinal is collapsed.

*See also: Ultrapower*

### Failure of $\text{GCH}$ at a measurable

Gitik proved that the following statements are equiconsistent:

- The generalized continuum hypothesis fails at a measurable cardinal $\kappa$, i.e. $2^\kappa > \kappa^+$
- The singular cardinal hypothesis fails, i.e. there is a strong limit singular $\kappa$ such that $2^\kappa > \kappa^+$
- There is a measurable cardinal of Mitchell order $\kappa^{++}$, i.e. $o(\kappa)=\kappa^{++}$

Thus violating $\text{GCH}$ at a measurable (or violating the SCH at any strong limit cardinal) is strictly stronger consistency-wise than the existence of a measurable cardinal.

However, if the generalized continuum hypothesis fails at a measurable, then it fails at $\kappa$ many cardinals below it.

## Real-valued measurable cardinal

A cardinal $\kappa$ is **real-valued** measurable if there exists a $\kappa$-additive measure on $\kappa$. The smallest cardinal $\kappa$ carrying a $\sigma$-additive 2-valued measure must also carry a $\kappa$-additive measure, and is therefore real-valued measurable, also it is strongly inaccessible under AC.

If a real-valued measurable cardinal is not measurable, then it must be smaller than (or equal to) $2^{\aleph_0}$. Martin's axiom implies that the continuum is not real-valued measurable.

Solovay showed that the existence of a measurable cardinal is equiconsistent with the existence of a real-valued measurable cardinal. More precisely, he showed that if there is a measurable then there is generic extension in which $\kappa=2^{\aleph_0}$ and $\kappa$ is real-valued measurable, and conversely if there exists a real-valued measurable then it is measurable in some model of $\text{ZFC}$.

## See also

## Read more

- Jech, Thomas -
*Set theory*

- Bering A., Edgar -
*A brief introduction to measurable cardinals*