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Sigma algebras on [0,1]

  1. Dec 7, 2012 #1
    This topic came up while studying measures on sub-intervals of [0,1]. The collection of all intervals in [0,1] is a semi-algebra, say J. Now from finite disjoint union of members of J lets say we form a set A.

    I was able to prove that A is an algebra, since for any C,D ε A, C[itex]\cap[/itex]D and C[itex]^{c}[/itex] belong to A.

    I'm not able to understand why A isn't a σ-algebra. Can anyone please outline a proof or give me a counter-argument.
  2. jcsd
  3. Dec 7, 2012 #2
    Are singletons in A?? Is [itex]\mathbb{Q}[/itex] in A?
  4. Dec 7, 2012 #3
    Yeah singletons are in A. Since singletons of the form (a,a) exist in J, the single union of such elements should exist in A. That I assume implies, all the rationals in [0,1] are also in A.
  5. Dec 7, 2012 #4
    Isn't [itex](a,a)=\emptyset[/itex]?? Singletons would rather come from [itex][a,a]=\{a\}[/itex]. But ok, let's assume that the singletons are in A.

  6. Dec 7, 2012 #5
    Yeah sorry, that's [a,a].

    We have [0,r)[itex]\cup[/itex](r,1] in A, if r is any rational in [0,1]. Therefore the element {r} ε A. So I think we could generalize this to all r ε Q[itex]\cap[/itex][0,1].
  7. Dec 7, 2012 #6
    Yeah, I agree that every {r} is in A. But why does that imply that [itex]\mathbb{Q}\in A[/itex]. Is it even true??
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