Let $\mu$ be a measure on $S$ such that:
- $\mu\left(\emptyset\right)=0$ and $\mu(S)=1$
- if $X\subseteq Y$, then $\mu(X)\leq\mu(Y)$
- $\mu\left(\{a\}\right)=0$ for all $a\in S$
- if $X_n$, $n=0,1,2,\ldots$, are pairwise disjoint, then $\displaystyle\mu\left(\bigcup \limits_{n=0}^\infty X_n\right)=\sum \limits_{n=0}^\infty\mu(X_n)$
- $\mu$ is two-valued, i.e. $\mu(X)=1$ or $\mu(X)=0$ for all $X$
Let $U=\{X\subseteq S\colon\mu(X)=1\}$. $U$ is a non principal ultrafilter on $S$, and this implies a weaker axiom of choice.
I'm not sure where I used it, but I guess it's in that measure. Can you show me an example of such a measure? Alternatively, can you tell me where I need $\sf AC$ if it's not in the measure?
Edit: Let $S$ be uncountable and $\mu\left(X\right)=0$ if $X$ is finite or countable. It seems to me that this measure can be construct without any assumption about $\sf AC$. And this leads to a nonprincipal ultrafilter, which in fact does require some weak form of $\sf AC$? I still don't understand..
Think about how such a $\mu$ would be created. We know its value for finite sets. Now we go through the infinite sets on at a time. The first set is $A$. Decide $\mu(A) = 1$. Then necessarily $\mu(A^c) = 0$. Now go to the next set, call it $B$. If $B$ is in the sigma algebra generated by the finite sets and $A$, then the value is known. Otherwise set $\mu(B) = 1$, and look at the next infinite set.
You will have to make these choices again and again. In fact you will have to make these choices a transfinite number of times. Since you have to make infinitely many choices, the axiom of choice has to be invoked.
Is it possible that there exists a clever argument to find a $\mu$ in a more constructive manner? The only way to show this cannot be done is to prove there exists models of set theory in which such a $\mu$ doesn't exist. And I believe that this has been done.