Sigma algebras on [0,1]

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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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Are singletons in A?? Is [itex]\mathbb{Q}[/itex] in A?
 
  • #3
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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.
 
  • #4
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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.
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.


That I assume implies, all the rationals in [0,1] are also in A.
Why??
 
  • #5
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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].
 
  • #6
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3,282
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].
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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