Results 1 to 3 of 3

Math Help - Borel-Sigma Algebras

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    46

    Borel-Sigma Algebras

    Let f: R --> R be a continuous function, and let \mathcal{A} = {A C R: there exists B \epsilon \mathcal{B} with A=f^{-1}(B) where \mathcal{B} are the Borel subsets of the range space R. Show that \mathcal{A} C \mathcal{B}, the Borel subsets of the domain space R.

    Okay, so I know f being continuous means that f^{-1}(v) is open for all open v in R.

    I'm not exactly sure what I need to show here. Is it just that \mathcal{A} is open in R? Then wouldn't that show it is in B? And wouldn't this be true because f inverse is open? So, all the A's in \mathcal{A} would be open. Am I on the right track? Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Focus's Avatar
    Joined
    Aug 2009
    Posts
    228
    I am somewhat confused as to what you are asking. I assume you want to show that \mathcal{A} is a subset of the Borel sigma algebra of the domain space.

    Take any A \in \mathcal{A}, then there exists a B in the Borel sigma algebra of the range, such that A=f^{-1}(B). As B is a Borel set, it is the intersection of some open sets (countable union of open sets is also open, all we are missing is the countable intersection), so B=\bigcap G_n. Now A=f^{-1}\left(\bigcap G_n \right)=\bigcap f^{-1}(G_n), and as f is cts, each f^{-1}(G_n) is open in the domain. Hence A is a Borel set.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Focus View Post
    As B is a Borel set, it is the intersection of some open sets (countable union of open sets is also open, all we are missing is the countable intersection), so B=\bigcap G_n.
    You definitely can't say that. Borel subsets can be much more complicated than intersections of open sets.

    The Borel \sigma-algebra is defined as the smallest \sigma-algebra containing the open subsets. That's the only property we can use to prove that a set is in the Borel \sigma-algebra. Hence the proof can't be direct. We know that f^{-1}(O)\in\mathcal{B} for any open subset O (because f^{-1}(O) is open, too). And we want to generalize this to any borelian subset instead of just open subsets.

    The proof goes this way: let

    \mathcal{C}=\{B\in\mathcal{B}|f^{-1}(B)\in\mathcal{B}\}.

    Then \mathcal{C} is a subset of \mathcal{B}, it contains the open subsets, and it is easy (but you really should write it down) to prove that \mathcal{C} is a \sigma-algebra. Therefore, by definition of the borelian \sigma-algebra, \mathcal{B}\subset\mathcal{C}, hence \mathcal{C}=\mathcal{B}. This is exactly what we needed: for any B\in\mathcal{B}, f^{-1}(B)\in\mathcal{B}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 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
  2. How many sigma algebras?
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 19th 2010, 01:46 PM
  3. Borel Sigma Algebras
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: October 4th 2009, 01:24 PM
  4. Sigma Algebras
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: June 9th 2008, 04:31 PM
  5. sigma algebras
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 20th 2008, 06:37 PM

Search Tags


/mathhelpforum @mathhelpforum