TITLE: Can we put a probability measure on every $\sigma$-algebra?
QUESTION [16 upvotes]: The following question has puzzled me for some time:

Let $(\Omega,\Sigma)$ be a nonempty,
  measurable space. Does there
  necessarily exist a probability
  measure $\mu:\Sigma\to[0,1]$? 

If there exists a nonempty measurable set $A$ such that no nonempty subset of $A$ is measurable (an atom), we can simply let $\mu(B)=1$ if $A\subseteq B$ and $\mu(B)=0$ otherwise. So the problem is only interesting if the $\sigma$-algebra has not atoms. This rules out every countably generated $\sigma$-algebra. An example of a $\sigma$-algebra that has no atoms but supports a probability measure is $\{0,1\}^\kappa$ for $\kappa$ uncountable, which we can endow with the coin-flipping probability measure.

REPLY [6 votes]: The following is a (corollary of a) theorem of Sierpinskii from 1933:
If $\mu:\mathcal{P}(\Omega) \to [0,1]$ is a probability measure and $|\Omega|$ is smaller than the first weakly-inaccessible cardinal, then there must be a countable $A \subseteq \Omega$ such that $\mu(A)=1$.