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Here by measureable cardinal I mean a cardinal $\kappa$ that admits a measure $\mu$ on $2^\kappa$ such that $\mu(\kappa)=1$ and $\forall x\in \kappa$ $\mu\{x\}=0$. My question is can such a cardinal be proved to exist in ZFC? When we additionally require that $\mu$ must be $\kappa$ additive then such a cardinal cannot be proved to exist as it must be weakly inacessible.

From theorem 1.4 on page 176 of Drake's book it is shown that the first measurable cardinal must have a measure that is $\kappa$ additive. It is then weakly inacessible and so cannot be shown to exist in ZFC. This is very close to showing the lack of existence of measurable cardinal but isn't quite there.

daRoyalCacti
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    Hi, welcome to MSE. I don't have access to Drake's book. Your question is a bit confusing because your definition of "measurable" is not the usual definition. I assume that your $\mu$ must be countably additive? It is true that ZFC proves that the least such cardinal has a $\kappa$-additive measure (ie is measurable in the usual sense). Isn't it quite straightforward to see from this that ZFC cannot prove the existence of such a cardinal? (If it did, ZFC would prove the existence of the least such cardinal...) – Izaak van Dongen Apr 23 '24 at 12:10

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The reason that $\sf ZFC$ cannot prove that there is an inaccessible cardinal, or a measurable cardinal, or any other of the many large cardinal axioms of this sort, is Gödel's second incompleteness theorem.

Namely, if $\kappa$ is an inaccessible cardinal, then $V_\kappa$ is a model of $\sf ZFC$, in which case, in that particular set theoretic universe, the statement "$\sf ZFC$ is consistent" is true.

If $\sf ZFC$ can prove the existence of an inaccessible cardinal, then in all of its models the statement "$\sf ZFC$ is consistent" holds, and therefore $\sf ZFC$ can actually prove "$\sf ZFC$ is consistent". But according to Gödel's second incompleteness theorem this cannot possibly happen with a theory that is recursively axiomatisable (which $\sf ZFC$ certainly is), interprets basic arithmetic (which $\sf ZFC$ most certainly does), and consistent.

So, this obstacle does not allow us to prove the existence of inaccessible cardinals. But to argue, philosophically, that this must mean that $\sf ZFC$ must be inconsistent, is just one small step from claiming that because $\sf ZFC$ can prove that $\sf PA$ is consistent, while $\sf PA$ itself cannot must mean that $\sf PA$ is inconsistent as well, and this argument goes well into "very reasonable theories" that hardly anyone in mathematics would think are inconsistent.

Instead, this is understood as giving a proper mathematical (and philosophical) strength to a mathematical theory. Namely, $\sf ZFC$ is a stronger theory than $\sf PA$ since it proves its consistency. $\sf ZFC$ + "There exists an inaccessible cardinal" is a stronger theory than $\sf ZFC$". And so on. This is what we mean when we say that large cardinal axioms provide us with a measuring stick for the strength of mathematical statements.

Now, back to the question at hand. If $\kappa$ is a measurable cardinal, not only that $\kappa$ is a strongly inaccessible cardinal, it is also Mahlo, greatly Mahlo, weakly compact, and much much more. So the theory $\sf ZFC$+"There exists a measurable cardinal" is significantly stronger than all of those, which themselves are already strong. But this is not close enough to showing that $\sf ZFC$ refutes their existence, and while this still might be the case, all of the arguments that were proposed this far tend to be based on some severe misunderstanding of how large cardinal axioms work, how the measuring stick work, and what does it mean for a theory to prove or refute a large cardinal axiom.

Asaf Karagila
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  • Hi Asaf. I think OP is a bit confused/is using confusing terminology. As best as I can understand it, they know that ZFC cannot prove the existence of a measurable cardinal in the standard sense: "When we additionally require that $\mu$ must be $\kappa$ additive then such a cardinal cannot be proved to exist as it must be weakly inacessible." It seems that by "measurable", they mean "cardinal that admits a countable additive measure". – Izaak van Dongen Apr 23 '24 at 12:12
  • Sorry if my question wasn't clear. I am using a non-standard definition of measurable. By measurable I mean a cardinal that admits a non-zero, countably additive, non-trivial set function. I don't fully understand of why (and this answer helped quite a bit) but I know that under the standard definition of measurable we cannot prove the existence of measurable cardinals. I wanted to know if we get the same conclusion if we use this different definition of measurable. – daRoyalCacti Apr 24 '24 at 00:07
  • @daRoyalCacti The standard modern set theory terminology for such a cardinal is "a cardinal greater than or equal to the least real-valued measurable cardinal". As you indicate, a real-valued measurable cardinal (i.e. with a $\kappa$-additive measure) is weakly inaccessible, so of course these things can't be proved to exist in ZFC. (Can't tell why you think this "isn't quite there"... it's there.) – spaceisdarkgreen Apr 24 '24 at 00:12
  • @daRoyalCacti (Or, "Ulam-measurable", per Noah's comment on your other question, though I don't know how common that usage is.) – spaceisdarkgreen Apr 24 '24 at 00:20
  • @spaceisdarkgreen Sorry I'm not very good a set theory yet. It is not obvious to me that if we can't prove the existence of the first Ulam-measurable cardinal then we can't prove the existence of any Ulam-measurable cardinal. In my head it might be possbile to prove the existence of infinitely many such cardinals but not possible to prove the existence of a smallest (and hence first). Something like we can prove the existence of the Ulam cardinals that aren't weakly inaccessible. – daRoyalCacti Apr 24 '24 at 00:42
  • @daRoyalCacti the cardinals are well-ordered. There is a least anything. – spaceisdarkgreen Apr 24 '24 at 00:50
  • @spaceisdarkgreen That makes sense. Thank you – daRoyalCacti Apr 24 '24 at 01:53