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Measure of borel set minus open <e

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    We have a metric space [tex] X=\cup X_k [/tex] where [itex] X_k\subset X_{k+1}[/itex] and each [itex] X_k[/itex] is open. Show that for any Borel set E, there is an open set U such that [itex] \mu (U-E)<\epsilon [/itex]. (Its supposed to be "U \ E".)


    2. Relevant equations

    [itex] \mu [/itex] is a measure, so probably the important thing is countable subadditivity.
    A borel set is a set generated by countable union, countable intersection, and relative complement of open sets.

    3. The attempt at a solution

    I know that if I have an open set I can intersect it with an [itex] X_k[/itex] and still have an open set... In this way I believe I can chop up any open set to countable pieces. But how can I get the difference in measure to be less than [itex] \epsilon [/itex] ?
    The only solutions to a problem like this that I have seen are in the context of Lebesgue measure and [itex]\mathbb{R}^n [/itex], but I cannot use this context. I must prove it in a general metric space with a general measure. Also, if anyone can recommend a book that has a good treatment of general measures instead of focusing on Lebesgue, I would appreciate the suggestion. So far I have Royden and Folland.
     
  2. jcsd
  3. Sep 22, 2013 #2
    Going back to set notation A/B= A[itex]\cap[/itex] B* (with * denoting the compliment).

    This should make it easier to use like like countable subadditivity and finite intersections for measure.
     
  4. Sep 22, 2013 #3
    Okay, good point, I like that. But... I still don't know how to get an open set (or countably many open sets) tightly wrapped around E.
     
  5. Sep 22, 2013 #4

    pasmith

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    You're working in a metric space so you should consider looking at a collection (or more than one collection) of open balls.
     
  6. Sep 22, 2013 #5
    I'm just guessing because I haven't actually thought this problem out in detail but....

    If we let E be generated by some of the Xk 's. I would think for a general borel set it doesn't have to be the X's but I'm not sue if the logic would flow nicely otherwise.


    Then E= [itex]\cup[/itex]some k's Xk.
    So E*= ([itex]\cup[/itex]k Xk)* = [itex]\cap[/itex] Xk*

    and U/E is simply U[itex]\cap[/itex]([itex]\cap[/itex]k Xk) so this leaves you with a countable number of intersections.

    And then you can try and work it from here. I think you may have to assume at some point that X has finite measure; not sure though.
     
  7. Sep 22, 2013 #6
    But what if E is not a union of Xk's? What if E is only parts of the Xks?

    Right, I forgot to mention [itex] \mu (X_k)<\infty [/itex]
     
  8. Sep 22, 2013 #7
    Yeah, but how would I get only countably many open balls?
     
  9. Sep 22, 2013 #8
    Yeah, generating from the Xk 's is kind of nice but isn't really true.

    But that is ok because if your generating sets are Ji then each Ji [itex]\subset[/itex] [itex]\cup[/itex] for some k's Xk.

    And then you could use that the fact that if A[itex]\subset[/itex] B then μ(A)≤ μ(B).

    I think that should be enough to get the ball rolling.

    The goal, in my mind, is to find what the measure of set U has to be and so that you can try and construct it.

    Also, a trivial answer is U = ∅. Cheap, but finding a solution sort of proves that a solution has to exist.
     
    Last edited: Sep 22, 2013
  10. Sep 23, 2013 #9
    Holy crap I solved it. :surprised
     
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