Hi there,

I am stuck. Let's say that S is a sigma algebra of subsets of O. if A and B belongs to S, how can I prove that A intersection B is itself a sigma algebra of subsets of B?

I will appreciate your help!!!!

Andrea

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- October 18th 2006, 09:19 AMgarbinersigma algebra problem
Hi there,

I am stuck. Let's say that S is a sigma algebra of subsets of O. if A and B belongs to S, how can I prove that A intersection B is itself a sigma algebra of subsets of B?

I will appreciate your help!!!!

Andrea - October 18th 2006, 09:38 AMThePerfectHacker
Here is a sketch of a proof.

----

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. - October 18th 2006, 11:15 AMgarbiner
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. - October 18th 2006, 12:32 PMPlato
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. - October 18th 2006, 02:34 PMgarbiner
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. - October 18th 2006, 03:33 PMPlato
I pray thee, think about what you have written!

How can**one set**be a sigma algebra? - October 19th 2006, 02:12 AMgarbiner
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!!!!:D