# Measuring More

It’s quite standard nowadays to characterise the measurable cardinals as the cardinals $\kappa$ such that there exists a normal $\kappa$-complete non-principal measure on $\kappa$. As we continue climbing the large cardinal hierarchy we get to the strong cardinals, Woodin cardinals and superstrong cardinals, all of which are characterised by extenders, which can be viewed as particular sequences of normal measures on $\kappa$. This trend then stops, and there’s a shift from measures on $\kappa$ to measures on $\mathcal P_\kappa(\lambda)$, being the set of subsets of $\lambda$ of cardinality less than $\kappa$. Now, how does one work with such measures? Where are the differences between our usual measures and these kinds? And how can we view this shift as expanding the amount of things that we can measure?

Before we start, let’s note that the step from measures on $\kappa$ to measures on $\mathcal P_\kappa(\lambda)$ is a simple matter of generalisation. Because if $\mu$ is a non-principal measure on $\kappa$,then we can associate to $\mu$ a filter $\nu$ on $\mathcal P_\kappa(\kappa)$ by setting

$X\in\nu$   iff   $\{\sup(\sigma)\mid\sigma\in X\}\in\mu$.

Note that non-principality implies that $\nu$ is a measure as well and it furthermore satisfies that whenever $\xi\in\kappa$, the set of all $\sigma\in\mathcal P_\kappa(\kappa)$ satisfying that $\xi\in\sigma$ is in $\nu$; we say that $\nu$ is fine. Now, if we started off with a fine measure $\nu$ on $\mathcal P_\kappa(\kappa)$ then we could’ve defined $\mu\subseteq\mathcal P(\kappa)$ as

$X\in\mu$   iff   $\{\sigma\mid\sup(\sigma)\in X\}\in\nu$.

Note here fineness of $\nu$ ensures that $\mu$ is a non-principal measure on $\kappa$, showing the equivalence between the existence of these two different types of measures. We furthermore see that $\kappa$-completeness for $\mu$ is equivalent to $\kappa$-completeness for $\nu$. This leads us to the definition of strong compactness.

Definition. Let $\kappa\leq\lambda$ be cardinals. Then $\kappa$ is $\lambda$-strongly compact if there exists a $\kappa$-complete fine measure on $\mathcal P_\kappa(\lambda)$. We also simply say that $\kappa$ is strongly compact if it’s $\lambda$-strongly compact for all $\lambda\geq\kappa$.

The above argument then shows that $\kappa$ being measurable is equivalent to it being $\kappa$-strongly compact, and considering larger $\lambda$ then gives us a natural way to improve the strength of measurability. This is somehow orthogonal to the extender approach, as that approach ensures that we can measure subsets of $\kappa$ in many different (coherent) ways, and this other approach is instead about measuring subsets of cardinals greater than $\kappa$.

Recall that a measure $\mu$ on $\kappa$ is normal if for every $\kappa$-sequence $\langle X_\alpha\mid\alpha<\kappa\rangle\in{^\kappa}\mu$ of measure one sets, the diagonal intersection $\triangle\vec X$ has measure one as well, where

$\triangle\vec X:=\{\xi<\kappa\mid\xi\in\bigcap_{\alpha<\xi}X_\alpha\}$.

Analogously, normality of a measure $\nu$ on $\mathcal P_\kappa(\lambda)$ is precisely the same, except that we define the diagonal intersection slightly differently. If we let $\langle Y_\alpha\mid\alpha<\kappa\rangle\in{^\lambda}\nu$ be a $\lambda$-sequence of measure one sets, then we set

$\triangle\vec Y:=\{\sigma\in\mathcal P_\kappa(\lambda)\mid\sigma\in\bigcap_{\alpha\in\sigma}Y_\alpha\}$,

and again $\nu$ is normal if $\triangle\vec Y\in\nu$ for every $\vec Y\in{^\lambda}\nu$.

Now note that, in the $\lambda=\kappa$ case, the normality of $\mu$ is equivalent to normality of $\nu$, in the sense of the above. The argument uses $\kappa$-completeness, which we may assume since normality of a measure on $\mathcal P_\kappa(\lambda)$ does imply $\kappa$-completeness (here it’s important that we’re looking at ${<}\kappa$-sized subsets). We arrive at supercompacts.

Definition. Let $\kappa\leq\lambda$ be cardinals. Then $\kappa$ is $\lambda$-supercompact if there exists a normal fine measure on $\mathcal P_\kappa(\lambda)$. We also simply say that $\kappa$ is supercompact if it’s $\lambda$-supercompact for all $\lambda\geq\kappa$.

As the existence of a normal measure on $\kappa$ is equivalent to the existence of a $\kappa$-complete one, the above argument then also shows that measurability is also equivalent to being $\kappa$-supercompact! As we increase $\lambda$ however, $\lambda$-supercompactness diverges from $\lambda$-strongly compactness.

Taking a step back, we can also start talking about club and stationary subsets of $\mathcal P_\kappa(\lambda)$, in complete analogy with the usual terminology. We say that a subset $C\subseteq\mathcal P_\kappa(\lambda)$ is closed if it’s closed under unions, and unbounded if for every $\sigma\in\mathcal P_\kappa(\lambda)$ there’s a $\tau\in C$ such that $\sigma\subseteq\tau$. $C$ is then club if it’s both closed and unbounded, and a subset $S\subseteq\mathcal P_\kappa(\lambda)$ is stationary if it meets all clubs. If we then look at the collection of stationary sets of $\mathcal P_\kappa(\lambda)$ this suddenly makes an appearance in the realm of Woodin cardinals, as the stationary tower forcing.