Here is a sketch of a proof.
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We want to show,
A n B is in sigma,
Let A', B' be complements.
We know that,
(A u B) is in Sigma,
So,
(A u B)' is in sigma,
So, (de Morga's law)
A' n B' is in sigma,
It completents are in sigma,
A n B is in sigma.
Mate, the problem is different!
I don't want to prove that A n B belongs to the sigma algebra!!That is easy. I just want to prove that A n B is ITSELF a sigma algebra of subsets of B, knowing that it belongs to the sigma algebra O!!
I think I have to prove that
1) empty set belongs to A n B.
2) B belongs to A n B.
I think but I am not sure!!!
Thak you anyway! I really appreciate that.
It strikes me as though there is a miss-statement the problem.
The set AnB is a single subset of B.
The collection {XnB: X is in S} can be shown to be a sigma algebra of subsets of B.
If XnB and YnB are such sets then (XnB)U(YnB)=(XUY)nB is such becsuse XUY is in S.
Consider XnB, both B’ and X’ are in S so X’UB’ is in S.
Because (XnB)’=X’UB’ we have closure by complement.
Similar arguments give countable unions.
Plato,
You are great!!!There was a mis-statement.Thanks very much. Only one more question. Let's suppose that this collection has just one set, let's say X n B. I can prove that this is a sigma algebra saying that X n B U empty_set = X n B which is of course in S.
Then closure by complement is the same.
So finally, what does this sigma algebra have in this particular case?
S= {empty_set, X n B, X'UB'}
Is this correct? Why can I call it sigma algebra of SUBSET of B? Where is B?
Thanks so much.
Andrea.
Probably I am wrong, but let's take a set A not empty. Then the Set S
S={empty_set, A} is the smallest sigma algebra. There is closure by complement and by union,isn't it.
Let's adapt that to our case.
If that collection C={X n B | X is in S} contains just one set and the empty set, what would be the sigma algebra?
Thanks a lot for your help! I've started studying it two days ago...I still get very confused!!!!