I remarked that it may not be possible for F to be all subsets of Ω. Here is an example to illustrate the point.
Let Ω be the set of all points on the circumference of a unit circle. Define a equivalence relation on points in Ω such that x∼y if the angle between them is a rational fraction of π. This partitions Ω 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 Aq. Clearly, A and Aq are disjoint, and it ought to be that P(Aq)=P(A). Now taking the union over all rationals, Ω=⋃qAq is a union of a countable number of disjoint sets (since every point in Ω is some rational angle away from some point in A). But it is impossible to satisfy the probability axioms, which would require P(⋃qAq)=1=∑qP(Aq), and also have P(A)=P(Aq) for all q. We are forced to conclude that no probability measure P can cope with a subset of Ω like A.
Let Ω be the set of all points on the circumference of a unit circle. Define a equivalence relation on points in Ω such that x∼y if the angle between them is a rational fraction of π. This partitions Ω 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 Aq. Clearly, A and Aq are disjoint, and it ought to be that P(Aq)=P(A). Now taking the union over all rationals, Ω=⋃qAq is a union of a countable number of disjoint sets (since every point in Ω is some rational angle away from some point in A). But it is impossible to satisfy the probability axioms, which would require P(⋃qAq)=1=∑qP(Aq), and also have P(A)=P(Aq) for all q. We are forced to conclude that no probability measure P can cope with a subset of Ω like A.
Sam · 583 weeks ago
Richard Weber 35p · 583 weeks ago
I admit that this is not an argument that follows directly from the probability axioms I-III.
Notice that construction of A requires the axiom of choice.