Clarification from old post: Union of sigma-algebras is non sigma-algebra

I have been working on slightly different problem from one posted back in 2013 here. I followed closely the hints given by @martini there, but nevertheless I still got stuck. I am retyping the question here not to duplicate but for your convenience:

Suppose $\mathscr A_1 \subset \mathscr A_2 \subset \ldots$ are $\sigma$-algebras of subsets of set $X$. Give example of $\bigcup_{i=1}^{\infty} \mathscr A_i$ that is non $\sigma$-algebra.
(The 2013 old post did not have the phrase in bold.)

And to that question @martini suggested to generate an example using natural numbers $\mathbb N$ and $F_n = \{\{1\}, \{2\}, \ldots \{n\}\}$, which I think makes sense and is relevant for my question. Here are what I have gone so far:

(1) Let $F_n = \{\{1\}, \{2\}, \ldots \{n\}\}$, let $\sigma(F_n)$ be its $\sigma$-algebra, and let make this example simple by making $n=2$ only
(2) If $F_1:=\{1\}$, then $\sigma(F_1)=\{\emptyset,\{1\},\{1\}^c,\mathbb N \}$
(3) If $F_2:=\{\{1\},\{2\}\}$, then $\sigma(F_2)=\{\emptyset,\{1\},\{1\}^c,\{2\},\{2\}^c \{1,2\},\{1,2\}^c,\mathbb N\}$
(4) Here $\sigma(F_1) \cup \sigma(F_2)=\sigma(F_2)$, and $\sigma(F_2)$ is a $\sigma$-algebra. Since I am looking for non $\sigma$-algebra, therefore this is not the example I have been looking for.

I thinks I have been misunderstanding some concepts from the beginning, but what are they? Thank you for your time and help.

NOTE: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To find out if this posting is a duplicate, I did my due diligence by asking opinions from experienced users here, and prior to that I had tried to post the question outside but got only lukewarm response.

I think you were on the right track, you just needed to make a small observation: consider the set of even integers in the union, it's not in any particular algebra, but it is in the union of all of them.

As for what your misunderstanding was: you stopped at the case where $n = 2$, you needed to consider he union over all of them.

More formally,

Let $X = \mathbb N$ and define $\mathcal A_n$ be the set of all the subsets of $\{1, 2, \ldots, n\}$ and their complements. Notice that $\mathbb E = \{2, 4, 6, \ldots\} \notin \mathcal A_n$ for any $n$, so $\mathbb E \notin \bigcup_{k = 1}^\infty \mathcal A_k$ but $\{2k\} \in \mathcal A_{2k}$ for each $k \in \mathbb N$. Hence, we have a countable collection of subsets of $\bigcup_{k = 1}^\infty \mathcal A_k$ whose union is not in $\bigcup_{k = 1}^\infty \mathcal A_k$. Conclude that $\bigcup_{k = 1}^\infty \mathcal A_k$ is not a $\sigma$-algebra.

The algebra $\bigcup_{k = 1}^\infty \mathcal A_k$ is the algebra of subsets of $X$ whose elements are either finite or cofinite, so any infinite set with infinite complement will not be in $\bigcup_{k = 1}^\infty \mathcal A_k$ since it is neither finite nor cofinite.

• Thank you for your patient explanation! Now I can rest my case. Thanks again! – Amanda.M Jan 29 '15 at 18:11

Let $\mathscr A_i$ be the Lebesgue $\sigma-$algebra over $[0,i)$. We have now a sequence of monoton classes. Their union, however is not a $\sigma-$algebra because $[0,+\infty)$ is not included. On the other hand all the sets of the form $[0,i)$ are included. As a result their union should be included, but it is not.

I can see that the old post answers the question... (a different way though) You can ignore my answer.