# Union of increasing sigma-algebras is not sigma-algebra

Tags:
1. Jan 28, 2015

### A.Magnus

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.

Last edited: Jan 28, 2015
2. Jan 28, 2015

### Stephen Tashi

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.

3. Jan 30, 2015

### A.Magnus

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.