Results 1 to 7 of 7

Math Help - sigma algebra problem

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    4

    sigma 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2006
    Posts
    4
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1
    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 XUB is in S.
    Because (XnB)=XUB we have closure by complement.
    Similar arguments give countable unions.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2006
    Posts
    4
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1
    I pray thee, think about what you have written!
    How can one set be a sigma algebra?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2006
    Posts
    4
    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!!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sigma-algebra
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 26th 2011, 12:03 PM
  2. Replies: 1
    Last Post: February 4th 2011, 08:39 AM
  3. Show intersection of sigma-algebras is again a sigma-algebra
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 20th 2010, 07:21 AM
  4. sigma-algebra
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: June 28th 2010, 10:19 PM
  5. sigma algebra and probability problem
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 19th 2008, 10:10 AM

Search Tags


/mathhelpforum @mathhelpforum