I remarked that it may not be possible for $\mathscr{F}$ to be all subsets of $\Omega$. Here is an example to illustrate the point.
Let $\Omega$ be the set of all points on the circumference of a unit circle. Define a equivalence relation on points in $\Omega$ such that $x\sim y$ if the angle between them is a rational fraction of $\pi$. This partitions $\Omega$ into equivalence classes (in fact an uncountable number of them). Now pick one point from each equivalence class and call the set of these points $A$. If we shift every point in $A$ around the circumference of the circle by the same rational angle $q$ we get a new set, say $A_q$. Clearly, $A$ and $A_q$ are disjoint, and it ought to be that $P(A_q)=P(A)$. Now taking the union over all rationals, $\Omega=\bigcup_q A_q$ is a union of a countable number of disjoint sets (since every point in $\Omega$ is some rational angle away from some point in $A$). But it is impossible to satisfy the probability axioms, which would require $P(\bigcup_q A_q)=1=\sum_q P(A_q)$, and also have $P(A)=P(A_q)$ for all $q$. We are forced to conclude that no probability measure $P$ can cope with a subset of $\Omega$ like $A$.
Let $\Omega$ be the set of all points on the circumference of a unit circle. Define a equivalence relation on points in $\Omega$ such that $x\sim y$ if the angle between them is a rational fraction of $\pi$. This partitions $\Omega$ into equivalence classes (in fact an uncountable number of them). Now pick one point from each equivalence class and call the set of these points $A$. If we shift every point in $A$ around the circumference of the circle by the same rational angle $q$ we get a new set, say $A_q$. Clearly, $A$ and $A_q$ are disjoint, and it ought to be that $P(A_q)=P(A)$. Now taking the union over all rationals, $\Omega=\bigcup_q A_q$ is a union of a countable number of disjoint sets (since every point in $\Omega$ is some rational angle away from some point in $A$). But it is impossible to satisfy the probability axioms, which would require $P(\bigcup_q A_q)=1=\sum_q P(A_q)$, and also have $P(A)=P(A_q)$ for all $q$. We are forced to conclude that no probability measure $P$ can cope with a subset of $\Omega$ like $A$.
