# Difference between revisions of "Measurable"

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 $j:V\to M$.

Measurable cardinals were introduced by Stanislaw Ulam in 1930.

## 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 :

1. There exists a 2-valued ($\sigma$-additive) measure on $\kappa$.
2. 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, if $\kappa$ 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.

### Embedding Characterization

Theorem 2 : The following are equivalent :

1. There exists a $\kappa$-complete nonprincipal ultrafilter on $\kappa$.
2. 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).
3. 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.

### Hayut Property

There is also another quite interesting model-theoretic characterization of measurability. Let a theory $T$ be $\kappa$-unboundedly satisfiable iff for every $\lambda<\kappa$, there is a model $\mathcal{M}\models T$ with $\lambda\leq|M|<\kappa$. In other words, the sizes of models of $T$ are unbounded in $\kappa$.

A class of formulae $Q$ is $\kappa$-Hayut iff for any $\kappa$-unboundedly satisfiable theory $T\subseteq Q$, there is a model of $T$ of size at least $\kappa$. More intuitively, $\kappa$-many small models of size less than $\kappa$ can combine to make one big $\kappa$-sized model.

An abstract logic $\mathcal{L}$ is called almost $\kappa$-favorable iff there is some way to represent every sentence of $\mathcal{L}$ with vocabulary $\tau$ as a sequence of length below $\kappa$ of symbols of $\tau$ and ordinals in $\kappa$ in such a way that the satisfaction relation is upward absolute for inner models $M$ of ZFC elementarily equivalent to $V$ with $M^{<\kappa}\subset M$. If $\kappa$ is an uncountable regular cardinal, the following are almost $\kappa$-favorable:

1. $\mathcal{L}_{\lambda,\mu}$ for any $\lambda,\mu\leq\kappa$
2. $\mathcal{L}_{\kappa,\omega}(q_{<\kappa})$, which is $\mathcal{L}_{\kappa,\omega}$ with universal cardinality quantifiers $q_\lambda$ for every $\lambda<\kappa$ (where $M\models q_\lambda$ iff $|M|\geq q_\lambda$)
3. $\mathcal{L}_{\kappa,\kappa}$ with the addition of a single existential 2nd-order quantifier, where negation on the resulting sentences is not allowed

Assuming $V=L$, every $\mathcal{L}$ where sentences are represented as sequences of length below $\kappa$ of symbols of $\tau$ and ordinals in $\kappa$ ($\kappa$-sequential logic) that has an extension with an $\mathcal{L}_{\omega,\omega}$-definable satisfaction relation is almost $\kappa$-favorable. For example: if $V=L$, then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is almost $\kappa$-favorable, but if a measurable exists then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is not almost $\kappa$-favorable, and in fact if $\kappa$ is the least measurable then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is not $\kappa$-Hayut; however, if $\kappa$ is extendible, then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is $\kappa$-Hayut, though it still isn't almost $\kappa$-favorable.

An uncountable regular cardinal $\kappa$ is measurable if and only if $\mathcal{L}_{\kappa,\kappa}$ is $\kappa$-Hayut, if and only if $\mathcal{L}_{\kappa,\omega}(q_{<\kappa})$ is $\kappa$-Hayut up to $2^\kappa$. Furthermore, an uncountable regular cardinal $\kappa$ is measurable if and only if every almost $\kappa$-favorable logic is $\kappa$-Hayut.

### Other Embedding Characterizations

There are also other embedding characterizations of measurable cardinals. Namely (under NBG or ZFC + $j$) the following are equivalent for any cardinal $\kappa$:

1. $\kappa$ is measurable.
2. $\kappa$ is the critical point of some $j:V\prec_{\Delta_0}V$.
3. $\kappa$ is the critical point of some $j:V\prec_{\Delta_1}V$.
4. $\kappa$ is the critical point of some $j:V\rightarrow V$ such that for any $\Sigma_1$-formula $\varphi$, $\varphi[x,y,z...]\rightarrow\varphi[j(x),j(y),j(z)...]$.

Proof:

(1 implies 4). If $\kappa$ is measurable, then $\kappa$ is the critical point of a $j:V\prec M$ for some inner model $M$. Therefore $\kappa$ is the critical point of a $j:V\prec_{\Sigma_1}M$ and so for any $\Sigma_1$-formula $\varphi$, $\varphi[x,y,z...]\rightarrow M\models\varphi[j(x),j(y),j(z)...]$. Then, let $\varphi$ be a $\Sigma_1$-formula. If $\varphi[x,y,z...]$ then $M\models\varphi[j(x),j(y),j(z)...]$ and because $\Sigma_1$-formulae are upward absolute for inner models, $\varphi[j(x),j(y),j(z)...]$. Therefore: $$\varphi[x,y,z...]\rightarrow\varphi[j(x),j(y),j(z)...]$$

(4 implies 3). Let $j:V\rightarrow V$ be as described. Then any $\Delta_1$ formula $\varphi$ is $\Sigma_1$ and $\neg\varphi$ is also $\Sigma_1$. So: $$\varphi[x,y,z...]\rightarrow\varphi[j(x),j(y),j(z)...]$$ $$\neg\varphi[x,y,z...]\rightarrow\neg\varphi[j(x),j(y),j(z)...]$$ $$\therefore\varphi[x,y,z...]\Leftrightarrow\varphi[j(x),j(y),j(z)...]$$

(3 implies 2). If $j:V\prec_{\Delta_1}V$ then $j:V\prec_{\Delta_0}V$ by definition.

(2 implies 1). If $j:V\prec_{\Delta_0}V$ has critical point $\kappa$, then $\{X\subseteq\kappa:\kappa\in j(X)\}$ is a $\kappa$-complete measure on $\kappa$.

This gives further characterizations (which are weakenings of the $j:V\prec M$ characterizations). Namely, $\kappa$ is measurable iff it is the critical point of some $j:V\prec_{\Delta_0} M$ for some inner model $M$.

Proof:

(1 implies 2). Let $\kappa$ be measurable. Then $\kappa$ is the critical point of $j:V\prec M$ for some inner model $M$, and so $j:V\prec_{\Delta_0} M$.

(2 implies 1). Let $\kappa$ be the critical point of a $j:V\prec_{\Delta_0}M$. Then, $j:V\rightarrow V$ and for any $\Delta_0$ formula $\varphi$, $\varphi[x,y,z...]\Leftrightarrow M\models\varphi[j(x),j(y),j(z)...]$. Of course, since $\varphi$ is $\Delta_0$, $M\models\varphi[j(x),j(y),j(z)...]$ iff $\varphi[j(x),j(y),j(z)...]$. So: $$\varphi[x,y,z...]\Leftrightarrow\varphi[j(x),j(y),j(z)...]$$ $$j:V\prec_{\Delta_0}V$$ $$\therefore\kappa\text{ is measurable.}$$

Finally, this gives much more general (and astonishing) characterizations of measurability in terms of nontrivial elementary embeddings of classes into themselves. The following are equivalent:

1. $\kappa$ is measurable.
2. There is some transitive class (or set) $M\models\text{ZFC}$ with $\mathcal{P}^2(\kappa)\subseteq M$ and some $j:M\prec_{\Delta_0}M$ with critical point $\kappa$.
3. For every transitive class (or set) $M\models\text{ZFC}$ with $\kappa\subseteq M$ there is a $j:M\prec_{\Delta_1}M$ with critical point $\kappa$.

Proof:

(1 implies 3). Let $\kappa$ be measurable and let $j:V\prec_{\Delta_1}V$. Then, let $M\models\text{ZFC}$ be a transitive class with $\kappa\subseteq M$. Consider $j\upharpoonright M:M\rightarrow M$. Let $\varphi$ be a $\Delta_1$ formula. Then, $\varphi[x,y,z...]\Leftrightarrow M\models\varphi[x,y,z...]$ (because $\Delta_1$ formulae are upward and downward absolute for transitive classes). Since $j$ is $\Delta_1$-elementary, for any $x,y,z...\in M$:$$\varphi[x,y,z...]\Leftrightarrow\varphi[j(x),j(y),j(z)...]\Leftrightarrow\varphi[j\upharpoonright M(x),j\upharpoonright M(y),j\upharpoonright M(z)...]$$ $$\Leftrightarrow M\models\varphi[x,y,z...]\Leftrightarrow M\models\varphi[j\upharpoonright M(x),j\upharpoonright M(y),j\upharpoonright M(z)...]$$ Therefore $j\upharpoonright M:M\prec_{\Delta_1}$. Now note that $j$ has critical point $\kappa$ and so $j\upharpoonright M$ does too.

(3 implies 2). Simply choose $M=V$ and note that any $j:V\prec_{\Delta_1}V$ with critical point $\kappa$ is already a $j:M\prec_{\Delta_0}M$ with critical point $\kappa$.

(2 implies 1). If $j:M\prec_{\Delta_0}M$ is as described, then simply choose $U=\{X\subseteq\kappa:\kappa\in j(X)\}$ as usual. The proof that $U$ is a $\kappa$-complete nonprincipal ultrafilter is a little more difficult than usual:

• If $X\in U$ and $X\subseteq Y\subseteq\kappa$, then $M\models (j(X)\subseteq j(Y)\subseteq j(\kappa))$ so $j(X)\subseteq j(Y)\subseteq j(\kappa)$ and therefore $\kappa\in j(Y)$ and $Y\in U$ ($U$ is closed upwards under subset).
• If $X,Y\in U$ then $\kappa\in j(X)\cap j(Y)$ and by $\Delta_0$-ness of $\cap$, $j(X\cap Y)=j(X)\cap j(Y)$ so $\kappa\in j(X\cap Y)$ and therefore $X\cap Y\in U$ ($U$ is a filter).
• Assume $X\in U$ is a finite set. Then it is easily verified that $j(X)=j"X$ and so $\kappa\in j"X$. This is a contradiction as $\kappa\neq j(x)$ for any set $x$; if it were then it would be an ordinal below $\kappa$ and so $\kappa$ would not be the critical point. ($U$ is nonprincipal).
• If $X\notin U$ then $\kappa\notin j(X)$ so $\kappa\in (j(\kappa)\setminus j(X))$ and by $\Delta_0$-ness of $\setminus$, $\kappa\in j(\kappa\setminus X)$ meaning $\kappa\setminus X\in U$. ($U$ is an ultrafilter).
• Finally, if $F\subseteq U$ is a family of size $\lambda<\kappa$, then $j(F)=j"F$ and since $\kappa\in j(X)$ for every $X\in F$, $\kappa\in X$ for every $X\in j"F=j(F)$ and so $\kappa\in\cap j(F)$. By $\Delta_0$-ness of $\cap$, $\kappa\in j(\cap F)$ and so $\cap F\in U$. ($U$ is $\kappa$-complete).

### Category Theoretic Characterization

Interestingly, measurable cardinals have certain category theoretic properties about them. This connection is likely at heart due to the fact that certain embeddings $j:V\rightarrow V$ share connections with functors $F:\text{Set}\rightarrow\text{Set}$, and measurable cardinals can be characterized in terms of those embeddings.

Specifically a measurable cardinal exists if and only if there is a nonidentity exact functor $F:\text{Set}\rightarrow\text{Set}$ .

## Properties

Every measurable is a large cardinal, i.e. $V_\kappa$ satisfies $\text{ZFC}$, therefore $\text{ZFC}$ cannot prove the existence of a measurable cardinal.

$\kappa$ 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.

Measurable cardinals are ineffable and stationary limits of ineffable cardinals.

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.

Any measurable cardinal $κ$ has super Ramsey M-rank $κ^+$, is super completely Ramsey and a stationary limit of super completely Ramsey cardinals and is a limit of regular cardinals $α$ which are $α$-Ramsey.

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.

A cardinal $\kappa$ is measurable if and only if it is $\kappa^+$-hypermeasurable.

Measurable cardinals are not necessarily tall, but it is relatively consistent that the least measurable cardinal is tall.

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.

If $κ$ is uncountable, $κ = κ^{<κ}$ and $2^κ = κ^+$, then the following are equivalent:

• $κ$ is measurable.
• $κ$ satisfies the $κ^+$-filter property.
• $κ$ satisfies the strategic $κ^+$-filter property.

On the other hand, starting from a $κ^{++}$-tall cardinal $κ$, it is consistent that there is a cardinal $κ$ with the strategic $κ+$-filter property, however $κ$ is not measurable.

If $κ$ is a measurable cardinal, then $κ$ is $κ$-very Ramsey. If a cardinal is $ω_1$-very Ramsey (=strategic $ω_1$-Ramsey cardinal), then it is measurable in the core model unless $0^\P$ exists and an inner model with a Woodin cardinal exists.[6, 7]

The existence of a strategic $ω+1$-Ramsey cardinal (and of a strategic fully Ramsey cardinal) is equiconsistent with the existence of a measurable cardinal.

A cardinal $κ$ is $C^{(n)}$-measurable iff there is a transitive class $M$ and an elementary embedding $j : V → M$ with critical point $crit(j) = κ$ and with $j(κ) ∈ C^{(n)}$. Every measurable cardinal is $C^{(n)}$-measurable for all $n$.

If $\mathrm{I}_4^0(\kappa)$, then $\kappa$ is measurable and $\{\alpha\lt\kappa|\alpha\text{ is measurable}\}$ has measure 1. If $i$ witnesses $\mathrm{I}_4^n(\kappa)$, then $i^n(\kappa)$ is measurable.

Measurable cardinals are equiconsistent with weakly measurable cardinals. If GCH holds, then all weakly measurable cardinals are measurable.

Without the axiom of choice $\omega_1$ can be measurable. ZFC+“There is a measurable cardinal” is equiconsistent to ZF+DC+“$\omega_1$ is measurable”. (Jech, 1968, Takeuti, 1970, after )

### 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 $\text{ZFC}$.

## Virtually measurable cardinal

(all information from )

A cardinal $κ$ is virtually measurable iff for every regular $ν > κ$ there exists a transitive M and a forcing $\mathbb{P}$ such that, in $V^\mathbb{P}$, there is an elementary embedding $j : H_ν^V → M$ with $\mathrm{crit}(j) = κ$.

Every virtually measurable cardinal is strategic $ω$-Ramsey, and every strategic $ω$-Ramsey cardinal is virtually measurable in $L$.

If κ is virtually measurable, then either $κ$ is remarkable in $L$ or $L_κ \models \text{“there is a proper class of virtually measurables”}$.