# Are measures of a measurable cardinal measurable? (Edited and Updated Version)

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 ‎‎‎‎$‎‎‎\kappa‎$‎‎?‎

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.
• 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. – user36136 Oct 17 '13 at 14:51