Update: Regarding to Prof. Hamkins's guidance I restricted the questions to the "normal" measures to avoid trivial answers.

Definition: Let $\kappa$ be a measurable cardinal. Define:

$\mathbb{M}_{\kappa}:=\lbrace \mu:P(\kappa)\rightarrow \lbrace 0,1\rbrace~|~\mu~\text{is a non-trivial}~\kappa~\text{-additive measure on}~\kappa \rbrace$

$\mathbb{N}_{\kappa}:=\lbrace \mu:P(\kappa)\rightarrow \lbrace 0,1\rbrace~|~\mu~\text{is a non-trivial}~\kappa~\text{-additive normal measure on}~\kappa \rbrace$

A problematic problem on a measurable cardinal ‎$‎‎‎\kappa‎$ ‎is ‎the ‎number ‎of ‎its ‎two-valued non-trivial ‎‎$‎‎‎\kappa‎$-additive normal ‎measures. ‎There ‎are ‎many ‎classic ‎consistency results ‎by ‎Apter, ‎Ben ‎Neria, ‎Friedman, ‎Garti, ‎Hamkins, Kunen, Magidor and others on this topic. Here I want to ask this old question in a new way.‎ ‎

In some sense measures are tools for counting by weighing. If we find a measure on a set then we will have some useful information about the weight and size of its subsets and also about its own size which could be a two or real-valued measurable cardinal. Now consider the set ‎$‎‎‎\mathbb{N}_{‎\kappa‎}‎$‎ we want to determine its size. A possible way could be trying to find a two-valued non-trivial maximum-additive measure on ‎$‎‎‎\mathbb{N}_{‎\kappa‎}‎$. ‎In ‎the ‎other ‎words we are ‎asking ‎about possible ‎measurabilty ‎of ‎the ‎cardinal ‎‎$‎‎|‎\mathbb{N}_{‎\kappa‎}‎|$. ‎Now ‎define:‎ ‎

Definition: A ‎‎measurable ‎cardinal ‎‎$‎‎‎\kappa‎$ ‎called ‎"super measurable" ‎if ‎‎$‎‎|‎\mathbb{N}‎_{‎\kappa‎}|$ ‎be a‎ ‎measurable ‎cardinal too. ‎‎ ‎

Notation: ‎For ‎any ‎cardinal ‎‎$‎\kappa‎$ consider ‎$‎\kappa‎^{‎\oplus‎}‎‎$ ‎to be ‎the ‎least ‎measurable ‎cardinal ‎greater ‎than ‎‎$‎‎‎\kappa‎$‎. Also let ‎$‎‎m(x)$ ‎and ‎‎$‎‎sm(x)$ ‎to be ‎the ‎abbriviations ‎for ‎the ‎phrases ‎"‎$x‎‎$ ‎is a measurable cardinal‎" ‎and ‎"‎$‎‎x$ is a super measurable cardinal‎" respectively.‎ ‎ ‎

Question (1): ‎Can ‎"all" ‎measurable ‎cardinals ‎be ‎super ‎measurable? ‎In ‎the ‎other ‎words ‎is ‎the ‎following ‎statement ‎true?‎ ‎

$‎‎Con(ZFC+‎\exists~\text{some measurable cardinal}‎)\Longrightarrow ‎$‎‎‎ ‎ ‎ $‎Con(ZFC+‎\exists~\text{some measurable cardinal}~+‎$

$\forall ‎‎\lambda‎~~m(‎\lambda‎)‎\rightarrow ‎sm(‎‎\lambda‎‎)‎‎)$ ‎‎

‎‎Question (2): ‎Can ‎"all" ‎measurable ‎cardinals ‎be ‎super ‎measurable and moreover there are "many" such cardinals? ‎In ‎the ‎other ‎words ‎is ‎the ‎following ‎statement ‎true?‎ ‎

$‎‎Con(ZFC+‎\exists~\text{class many measurable cardinals}‎)\Longrightarrow ‎$‎‎‎ ‎ $‎Con(ZFC+‎\exists~‎‎‎\text{class many measurable cardinals}‎‎‎‎ +‎$

$\forall ‎‎\lambda‎~~m(‎\lambda‎)‎\rightarrow ‎sm(‎‎\lambda‎‎)‎‎)$‎‎ ‎ ‎‎ ‎

Question (3): ‎Can ‎"all" ‎measurable ‎cardinals ‎be ‎super ‎measurable and moreover there are "many" such cardinals with some "control on size" of measure space? ‎In ‎the ‎other ‎words are ‎the ‎following ‎statements ‎true?‎ ‎

(a)‎ $‎Con(ZFC+‎\exists~\text{class many measurable cardinals}‎‎‎‎ +‎$

$\forall ‎‎\lambda‎~~m(‎\lambda‎)‎\rightarrow ‎sm(‎‎\lambda‎‎)‎) \Longrightarrow ‎Con(ZFC+‎\exists~‎‎‎\text{class many measurable cardinals}‎‎‎‎ +‎$

$\forall ‎‎\lambda‎~~m(‎\lambda‎)‎\rightarrow |‎\mathbb{N}_{‎‎\lambda‎‎}‎|=‎\lambda‎‎)‎$‎ ‎

(b) $‎Con(ZFC+‎\exists~\text{class many measurable cardinals}‎‎‎‎ +‎$

$\forall ‎‎\lambda‎~~m(‎\lambda‎)‎\rightarrow ‎sm(‎‎\lambda‎‎)‎) \Longrightarrow ‎Con(ZFC+‎\exists~‎‎‎\text{class many measurable cardinals}‎‎‎‎ +‎$

$\forall ‎‎\lambda‎~~m(‎\lambda‎)‎\rightarrow |‎\mathbb{N}_{‎‎\lambda‎‎}‎|=‎\lambda‎‎^{\oplus})‎$‎ ‎

Remark: ‎Obviously ‎it ‎is ‎impossible ‎to ‎find a‎ ‎direct (within ‎$‎‎ZFC$‎‎) ‎construction ‎of a‎ ‎non-trivial measure ‎on ‎the ‎measure ‎space ‎‎$‎‎‎\mathbb{N}_{‎\kappa‎}‎$ ‎because ‎it ‎is ‎like ‎proving ‎the ‎existence ‎of a‎ ‎real‎-valued measurable cardinal so we should think about "relative" measure producing methods. In order to do this we first need a "criterion" ‎‎‎‎‎‎‎‎to determine ‎that ‎which ‎one ‎of the ‎subsets ‎of ‎‎$‎\mathbb{N}_{‎\kappa‎}‎$‎ ‎are ‎"heavy weight" ‎or ‎"large in number". ‎In ‎this ‎direction ‎some ‎approaches ‎could ‎be ‎like ‎these (the arguments are about the space $\mathbb{M}_{\kappa}$, one can redefine a same argument for $\mathbb{N}_{\kappa}$):‎ ‎

(a) ‎We ‎know ‎that ‎every ‎measurable ‎cardinal ‎‎$‎\kappa‎‎‎$‎ ‎has ‎some ‎non-invariant ‎subsets ‎by ‎measures ‎in ‎‎$‎‎‎\mathbb{M}_{\kappa}‎$ ‎(for example the set of all successor ordinals in ‎$‎‎‎\kappa‎$‎). ‎So ‎we ‎can ‎say a‎ ‎subset ‎‎$‎‎S\subseteq ‎‎‎\mathbb{M}_{\kappa}‎$ ‎to ‎be ‎"large" ‎(measure ‎$‎1‎$‎) ‎if‎f for all non ‎$‎‎‎\mathbb{M}_{\kappa}‎‎‎$‎-invariant ‎set $X\subseteq ‎\kappa‎$‎ ‎there ‎are ‎at least two ‎measures ‎‎$‎‎‎\mu_{1},\mu_{2}\in S‎$ ‎such ‎that ‎‎the pair $(\mu_{1},\mu_{2})‎$‎ ‎"unfolds" ‎the ‎hidden ‎non-invariance ‎of ‎‎$‎‎X$ ‎(i.e. ‎$‎‎\mu_{1}(X)=0$ and ‎$‎‎\mu_{2}(X)=1$‎‎).

(b) Another criterion for the "largness" of a subset ‎$‎‎S\subseteq ‎\mathbb{M}_{‎\kappa‎}‎$ ‎could ‎be ‎defining a‎ ‎notion ‎of ‎"addition" ‎on ‎‎$‎‎‎\mathbb{M}_{‎\kappa‎}‎$ ‎and ‎considering ‎the "convergence" of the ‎series ‎‎$‎‎\sum_{i\in S}\mu_{i}$ ‎i‎n the space ‎$‎\mathbb{M}_{‎\kappa‎}‎$‎ ‎and ‎in ‎the ‎last ‎step ‎defining a‎ ‎natural two-valued ‎measure ‎on ‎‎$‎\mathbb{M}_{‎\kappa‎}‎$ ‎like ‎‎$‎‎\Omega :P(‎\mathbb{M}_{‎\kappa‎}‎)‎\longrightarrow ‎‎‎‎‎‎\lbrace ‎‎0,1\rbrace‎‎‎$ ‎such ‎that ‎‎$‎‎‎\forall ‎S\subseteq ‎\mathbb{M}_{‎\kappa‎}‎‎~~~~~\Omega(S)=1‎\Longleftrightarrow‎ ‎‎\sum_{i\in S}\mu_{i} \notin ‎\mathbb{M}_{‎\kappa‎}‎‎‎$‎ ‎‎ ‎ ‎‎

(c) Unfortunately both of above ‎approaches are "direct" measure constructions and don't work perfectly. Perhaps ‎one ‎can‎ ‎refine them ‎by ‎considering ‎some ‎"special" ‎two-valued ‎$‎‎‎\kappa‎$-additive‎ measure ‎‎$‎‎\mu_{0}$‎ on ‎‎$‎‎‎\kappa‎$ ‎as a‎ ‎"shepherd measure" ‎which controls the ‎non-triviality ‎and ‎maximum-additivity ‎of ‎measures ‎on ‎‎‎‎$‎\mathbb{M}_{‎\kappa‎}‎$ ‎like ‎‎$‎‎\Omega$ ‎when ‎this ‎measures ‎are ‎"defined" ‎from ‎‎$‎‎\mu_{0}$ ‎in ‎some ‎way.‎ ‎‎

Question (4): ‎Is ‎there a ‎‎known‎ ‎"addition" (or "convolution") operator ‎on ‎‎$‎‎‎\mathbb{M}_{‎\kappa‎}‎$ (or ‎‎$‎‎‎\mathbb{N}_{‎\kappa‎}‎$) ‎to "‎amalgamate" ‎two ‎given ‎two-valued ‎non-trivial ‎‎$‎‎‎\kappa‎$-additive (normal) ‎measure (in a non-trivial way) ‎to ‎produce a new ‎measure on ‎‎‎‎$‎‎‎\kappa‎$‎‎?‎


1 Answer 1


If you mean only to consider the measures on $\kappa$, in the sense of a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, then of course there are at most $2^{2^\kappa}$ many of them, and when $\kappa$ is measurable one can prove that every measure has at least $2^\kappa$ many isomorphic copies. So the number of measures on $\kappa$ will never be a measurable cardinal itself. With this interpretation of your question, there are no supermeasurable cardinals.

However, in the large cardinal context we do have ways of measuring the largeness of a given measure. Suppose that $\kappa$ is measurable, witnessed as the critical point of an elementary embedding $j:V\to M$, and that $\kappa$ is measurable in $M$. This is what it means to say that $\kappa$ has nontrivial Mitchell rank, since the measure on $\kappa$ induced by $j$ will concentrate on measurable cardinals. More generally, one defines the Mitchell order $\mu\triangleleft\nu$ if $\mu\in M_\nu$, which is well-founded, and the rank of this order is the Mitchell rank of $\kappa$. Measures with high Mitchell rank concentrate on cardinals with high-but-not-quite-as-high Mitchell rank.

The Mitchell rank is connected with some of the ideas you mention at the end of your post. This Mitchell rank concept is quite general, and applies to many different large cardinal concepts, such as strongness extenders and supercompactness measures.

  • $\begingroup$ Dear Joel, Thanks. It seems that if I replace $\mathbb{M}_{\kappa}$ by $\mathbb{N}_{\kappa}$ everywhere the question will be more interesting because it is consistent to have a measurable cardinal $\kappa$ with exactly $\kappa$ many normal measure which says that the existence of a super measurable cardinal (in the sense of measurability of $|\mathbb{N}_{\kappa}|$) is consistent. $\endgroup$
    – user36136
    Oct 17, 2013 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.