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Sum of two closed sets are measurable

  1. Mar 25, 2012 #1
    I tried very long time to show that
    For closed subset A,B of R^d, A+B is measurable.

    A little bit of hint says that it's better to show that A+B is F-simga set...
    It seems also difficult for me as well...

    Could you give some ideas for problems?
     
  2. jcsd
  3. Mar 25, 2012 #2
    Try showing that if A and B are compact, then A+B is also compact. Then in the general case where A and B are closed, can you see how to write A+B as the countable union of compact sets?
     
  4. Mar 25, 2012 #3
    hmm.. I thought that if A,B is closed, then A+B is not necessarily closed...
     
  5. Mar 25, 2012 #4
    That's true, but you're not being asked to prove that A+B is closed, you're being asked to prove that it's a countable union of closed sets. That's why you start with the case where A+B is compact. If A and B are compact instead of just closed, then A+B really is compact. Then you can use the fact that every closed set is a countable union of compact sets to prove that A+B is F_σ
     
  6. Mar 25, 2012 #5
    Aha. I got your idea!
    I'm now wondering how I can show the fact that compactness is closed under the set addition..

    Is it better to use open covering argument?
     
  7. Mar 25, 2012 #6
    Actually it's probably faster to use the fact that a subset of R^n is compact iff every sequence has a convergent subsequence.
     
  8. Mar 25, 2012 #7
    hmm.. I'm not sure It is actually true that every seq in A+B has a convergent subseq.

    assuring the exsitence of conv subseq in A and B does not implies so as well A+B... Isn't it?
     
  9. Mar 25, 2012 #8
    A sequence in A+B is of the form [itex]\{a_n + b_n\}_{n\in \mathbb{N}}[/itex], where [itex]a_n\in A[/itex] and [itex]b_n\in B[/itex]. Can you find a subsequence such that [itex]a_{n_k}[/itex] converges and [itex]b_{n_k}[/itex] converges? What does that say about [itex]a_{n_k} + b_{n_k}[/itex]?
     
  10. Mar 25, 2012 #9
    I got it. Thank you very much!
     
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