# Measurable cardinals: non-trivial two-valued measures

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.

• How about $A\in \mathcal{U}\iff \mu (A) = 1$ ? – Max Jan 10 at 22:58
• @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. – bof Jan 11 at 0:48
• What properties do you have to prove to show that $\kappa$ is measurable? Which ones have you done, and which are you stuck on? – bof Jan 11 at 0:55
• 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? – bof Jan 11 at 0:56
• 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. – 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 $$\omega and $$A=\{A_x:x such that $$m(\cup A)=1.$$

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

Let $$B=\{A_x\setminus \cup_{y Then $$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

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 $$j so this contradicts the minimality of $$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 $$|A| so $$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|

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 $$m^*(\cup S)=m(\cup S')=0.$$