>>12926998First case is when A is countable, second case is when A^c is countable.
See
https://en.wikipedia.org/wiki/Probability_spaceAccording to the definition, Omega=R can be any set. Then, F must be a sigma algebra, i.e. a set of subsets of R which includes R, is closed under finite union, countable union and complements. Finally, P just needs to be a measure such that P(R)=1, and P(Omega)=0.
So that is all we need to show.
The first step is obvious, since R is clearly a set.
For the second step, all we need to show is (countable) additivity, since closure under complements is already given in the definition of F. To show closure under countable union, we just use the fact that a countable union of countable sets is countable.
Before showing the last step, first consider what do the elements of F look like? From the above step, we can see that every element of F is either countable or has countable complement. Then, by the definition of F, for any set A, the probability of A is either P(A)=0 or P(A)=1 depending on if A or its complement is countable. Then we can see that a countable collection of countable sets will always have zero probability, and this proves superadditivity for countable sets. Since 0 is the smallest measure, subadditivity is obvious, so now we consider the case involving sets whose complement is countable. It is obvious that any set which is disjoint to a set whose complement is countable must be countable, and thus, in a similar fashion to above, we can prove additivity in that case. This proves all the possible cases, and so we are done.
