Sigma Algebras on [0,1]: Understanding Measures and Sub-Intervals

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Discussion Overview

The discussion revolves around the properties of a set A formed from intervals in [0,1] and whether it qualifies as a σ-algebra. Participants explore the implications of including singletons and rational numbers in A, as well as the definitions and characteristics of algebras and σ-algebras.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims to have proven that A is an algebra but questions why it is not a σ-algebra.
  • Another participant asks whether singletons are included in A and if the set of rational numbers, \mathbb{Q}, is also in A.
  • It is noted that singletons of the form (a,a) exist in the semi-algebra J, leading to the assumption that all rationals in [0,1] are included in A.
  • There is a clarification that (a,a) is equivalent to the empty set, and singletons should rather come from [a,a]=\{a\}.
  • Participants discuss the inclusion of intervals [0,r) and (r,1] for rational r, suggesting that this leads to the conclusion that {r} is in A.
  • There is a question about whether the inclusion of all singletons implies that \mathbb{Q} is in A, with uncertainty expressed regarding the truth of this implication.

Areas of Agreement / Disagreement

Participants generally agree that singletons are in A and that {r} for rational r is also in A. However, there is disagreement and uncertainty about whether this implies that \mathbb{Q} is in A, leaving the discussion unresolved.

Contextual Notes

Participants have not fully resolved the implications of including singletons and rational numbers in A, nor have they clarified the conditions under which A may or may not be a σ-algebra.

Old Monk
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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 let's say we form a set A.

I was able to prove that A is an algebra, since for any C,D ε A, C\capD and C^{c} 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.
 
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Are singletons in A?? Is \mathbb{Q} in A?
 
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.
 
Old Monk said:
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 (a,a)=\emptyset?? Singletons would rather come from [a,a]=\{a\}. 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??
 
Yeah sorry, that's [a,a].

We have [0,r)\cup(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\cap[0,1].
 
Old Monk said:
Yeah sorry, that's [a,a].

We have [0,r)\cup(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\cap[0,1].

Yeah, I agree that every {r} is in A. But why does that imply that \mathbb{Q}\in A. Is it even true??
 

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