I know that there exist no measures (i.e. no non-negative, $\sigma$-additive, extended real-valued function) on a free $\sigma$-algebra with $\mathbb{N}$ generators. Where can I find a proof of that?
Measure on free Boolean sigma-algebras addresses the issue, but I don't understand the argument, especially what $x_A$ represents.
Thanks
$x_A$ is defined as the meet of all $x_n, n \in A$ and all $\lnot x_n$ where $n \notin A$. Here $A$ is a subset of $\Bbb N$ and $\{x_n\mid n \in \Bbb N\}$ is a countable set of generators for the Boolean $\sigma$-algebra. These meets exist as we have a $\sigma$-algebra… They are not $0$ as we have a free algebra.
The answer then argues that $x_A \land x_B = 0$ whenever $A \neq B$ in $\mathscr{P}(\Bbb N)$, which is correct and shows that the set $P:=\{x_A: A \subseteq \Bbb N\}$ is a pairwise disjoint family of size continuum (so uncountable and of uncountable cofinality) and this cannot happen in any measure algebra (as these are ccc: any set of pairwise disjoint elements is at most countable). So a lot of algebras cannot be made into a measure algebra. I assume (as is common in this field) that $x \neq 0$ will imply $\mu(x) >0$ so $\mu$ is strictly positive.
The argument to form $P$ is used also to show that an infinite $\sigma$-algebra cannot be countable. It’s pretty standard.