while doing some exercises about measurable cardinals, I got stuck on this one:

If $κ$ is the minimal cardinal that carries a non-trivial two-valued measure, then how can one prove that $κ$ is measurable?

I do not really have an idea on how to approach this, and am grateful for help.

  • $\begingroup$ How about $A\in \mathcal{U}\iff \mu (A) = 1$ ? $\endgroup$ – Max Jan 10 at 22:58
  • $\begingroup$ @Max I think the question is about how to prove the measure is $\kappa$-additive. The difference between encoding the measure as a subset of $\mathcal P(\kappa)$ or as a function $\mu:\mathcal P(\kappa)\to\{0,1\}$ is insignificant. $\endgroup$ – bof Jan 11 at 0:48
  • $\begingroup$ What properties do you have to prove to show that $\kappa$ is measurable? Which ones have you done, and which are you stuck on? $\endgroup$ – bof Jan 11 at 0:55
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    $\begingroup$ Suppose that, for some cardinal $\lambda\lt\kappa$, there are $\lambda$ null sets whose union has measure $1$. Can you use that to show that $\lambda$ carries a non-trivial two-valued measure? $\endgroup$ – bof Jan 11 at 0:56
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    $\begingroup$ Let $\mu$ be a non-trivial two-valued measure on $\kappa$ and assume for a contradiction that $\mu$ is not $\kappa$-additive, so there is a cardinal $\lambda\lt\kappa$ and a family $(A_\xi,\ \xi\in\lambda)$ of subsets of $\kappa$ such that $\mu(A_\xi)=0$ for each $\xi\in\lambda$ while $\mu(\bigcup_{\xi\in\lambda}A_\xi)=1$. Now let's try something wild and crazy. Let's define a function $\nu:\mathcal P(\lambda)\to\{0,1\}$ by $$\nu(X)=\mu\left(\bigcup_{\xi\in X}A_\xi\right).$$ I wonder what properties $\nu$ has. $\endgroup$ – bof Jan 13 at 3:35

Let $m:P(k)\to \{0,1\}$ be a non-trivial measure. Let $I=m^{-1}\{0\}.$

Suppose there exists cardinal $j$ with $\omega <j<k$ and $A=\{A_x:x<j\}\subset m^{-1}\{0\}$ such that $m(\cup A)=1.$

Let $i$ be the least such $j.$

Let $B=\{A_x\setminus \cup_{y<x}A_y: y\in i\}\setminus \{\emptyset\}.$ Then $m(\cup B)=1$ and $B\subset m^{-1}\{0\}.$ Now $B$ is a pair-wise disjoint family and $\emptyset \not \in B,$ and by the minimality of $j$ we have $|B|=j.$ So let $B=\{B_y:y<j\}.$

For $S\in P(j)$ let $m^*(S)=m(\cup_{y\in S}B_y).$ We may confirm that $m^*:P(j)\to \{0,1\}$ is a non-trivial measure. But $j<k$ so this contradicts the minimality of $k.$

So there is no such $j$.

Therefore $I=m^{-1}\{0\}$ is a free (non-principal) maximal ideal on $k,$ and $\cup A\in I$ whenever $A\subset P(k)$ and $|A|<k,$ so $F= \{k\setminus i:i\in I\}$ is a free maximal filter on $k,$ with $\cap C\in F$ whenever $C\subset F$ and $0<|C|<k.$

Remark: Regarding $m^*,$ if $S=\{S(n):n\in \omega\}\subset (m^*)^{-1}\{0\}$ then $S'=\{\cup_{y\in S(n)}B_y:n\in \omega\}\subset m^{-1}\{0\}, $ so $m^*(\cup S)=m(\cup S')=0.$


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