# Sigma Algebra

Let W be a sample space with a countable number of outcomes. Let A be the set of elementary events (an elementary event is a set containing only one outcome from the sample space). Prove that the sigma-algebra generated by A is the set of all subsets.

A sigma algebra es defines as a set which contains the empty set, any set and its complement and infinite unios of such subsets from the sample space.
The case here is that I came up with the fact that must prove that a sigma-algebra generated by a set of elementary events must have asmany elements as 2^W, where W is the sample space where we generated the set A of elementary events and the sigma-algebra from A.

I thought it could be proven by induction.

matt grime
Homework Helper
No, you prove it by showing that any element in the power set can be constructed using those operations on the collection of elementary events. Note a sigma algebra is closed under countable union, not infinite union.

(demonstrating two sets have equal cardinality says nothing about them being equal, necessarily)

That is true. So it is close by finite unions and by DeMorgan's Law it is also by finite intersection, right?
Then if |W| = n, and A = {set of elementary events from sample space W}, a sigma-algebra generated by A would have 2^W elements.

matt grime
Homework Helper
You cannot argue by cardinality, it is a useless tactic, since the same argument you've just used states that the rationals and the naturals are the same subset of R.

Let X be some subset of W.

Write X as a countable union of sets that are definitely in the sigma algebra (you do get that countable sets may be infinite too?)

I think I could say this:
Suppose we have a countable set:
W = {a1, a2, ..., an}
And a set of elementary events:
F = { {a1}, {a2}, ..., {an} }

Then the sigma-algebra generated by F, say Sigma-Algebra(F)=2^W, where 2^W is the set of all subsets.

Let A belongs to 2^W, (let A be contained in W), and
A = {b1, b2, ..., bn} then A = U{bn} from n=1, infinity

Then the sigma-algebra(F) has all the possible subsets and their complement in all possible combinations, thus we can say that: Sigma-algebra(F) = 2^W

I feel I am still missing something...

matt grime
Homework Helper
Notationally that is wrong.

Let S be any subset of W. We must simply show that S is in B(A) the sigma algebra generated by A.

Since S =U{s}, where the union is taken over s in S, and such is a countable union, since S is a subset of a countable set, it follows that S is in the sigma algebra generated by A.

End of proof.

Your sets have terminal elements, last ones, for some reason. They need not.

Oh, I see. Thank you very much for your explanation.