Union of increasing sigma-algebras is not sigma-algebra

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SUMMARY

The discussion centers on the properties of sigma-algebras, specifically addressing the union of increasing sigma-algebras and demonstrating that the union may not itself be a sigma-algebra. The example provided involves finite sigma-algebras on the set of natural numbers, where the sigma-algebras are defined as ##\sigma(F_n)##. The user initially concludes that the union of two sigma-algebras, ##\sigma(F_1)## and ##\sigma(F_2)##, remains a sigma-algebra, which contradicts the original query. The discussion highlights the need for careful consideration of the definitions and properties of sigma-algebras in this context.

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  • Understanding of sigma-algebras and their properties
  • Familiarity with the concept of countable unions in set theory
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A.Magnus
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I am working on a problem like this:

Suppose ##\mathscr A_1 \subset \mathscr A_2 \subset \ldots## are sigma-algebras consisting of subsets of a set ##X##. Give an example that ##\bigcup_{i=1}^{\infty} \mathscr A_i## is not sigma-algebra.

I was told to work along finite sigma-algebras on ##\mathbb N##, such that their union contains all singletons. For example, let ##F_n := \{\{1\},\{2\},…,\{n\}\}##, and then let ## \sigma (F_n)## be the sigma-algebra.

Here are what I got so far, making it as simple as possible to ##n = 2## only
(1) Let ##F_1 := \{1\}##, then ##\sigma(F_1) = \{ \emptyset, \{ 1\}, \{ 1\}^c, \mathbb N \}##
(2) Let ##F_2 := \{\{1\}, \{2\}\}##, then ##\sigma(F_2) = \{ \emptyset, \{ 1\}, \{ 1\}^c, \{ 2\}, \{ 2\}^c\ \{ 1, 2\}, \{ 1, 2\}^c ,\mathbb N \}##
(3) Here ##\sigma(F_1) \subset \sigma(F_2)##
(4) But here ##\sigma(F_1) \cup \sigma(F_2) = \sigma(F_2)##, which is a sigma-algebra. Therefore it looks like my example fails.

What was wrong with my analysis? If the above was hopelessly wrong, could you please give me another example? Thank you for your time and help.
 
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A.Magnus said:
I am working on a problem like this:

Suppose ##\mathscr A_1 \subset \mathscr A_2 \subset \ldots## are sigma-algebras consisting of subsets of a set ##X##. Give an example that ##\bigcup_{i=1}^{\infty} \mathscr A_i## is not sigma-algebra.

I was told to work along finite sigma-algebras on ##\mathbb N##, such that their union contains all singletons. For example, let ##F_n := \{\{1\},\{2\},…,\{n\}\}##, and then let ## \sigma (F_n)## be the sigma-algebra.

Unless the subset relation has a specialized meaning for sigma algebras, I think you are free to use different "whole spaces" for the different sigma algebras, as long as the "whole space" set for \mathscr A_i is a subset of the sets in \mathscr A_{i+1}.

Pick a sequence of sigma algebras with finite "whole"spaces" You can form an infinite set by taking a countable union of sets from the finite sets.
 
Stephen Tashi said:
Unless the subset relation has a specialized meaning for sigma algebras, I think you are free to use different "whole spaces" for the different sigma algebras, as long as the "whole space" set for \mathscr A_i is a subset of the sets in \mathscr A_{i+1}.

Pick a sequence of sigma algebras with finite "whole"spaces" You can form an infinite set by taking a countable union of sets from the finite sets.

I am sorry coming to you late. Let me take another look at the problem after your suggestion and get back with you. Thanks again.
 

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