Show that then also $ \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n}$ is a ring and an $ \sigma $ algebra Let $ \Omega $ be a set.
(a) Let $ \left(\mathcal{A}_{n}\right)_{n \in \mathbb{N}} $ rings on $ \Omega $ such that $ \mathcal{A}_{n} \subset \mathcal{A}_{n+1} $ for all $ n \in \mathbb{N} $. Show that then also $ \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n}$ is a ring.
(b) Now let the $ \mathcal{A}_{n} $ in (a) be even $ \sigma $ algebras. Then is $ \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $ an $ \sigma $ algebra?
Attempt:
The three properties of a ring are to be shown:
(1) Show: $ \varnothing \in \bigcup_{n \in \mathbb{N}}  \mathcal{A}_{n} $
Since $\mathcal{A}_{n}$ are rings, the following applies: $\varnothing \in \mathcal{A}_{n}, \forall n \in \mathbb{N} $.
$ \Rightarrow \varnothing \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $.
(2) Show $ A, B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} \Rightarrow  A\backslash B \in \bigcup_{n \in N} \mathcal{A}_{n} $.
Be $ A, B \in \bigcup_{n \in \mathbb{N}} A_{n} $, i.e there is a largest index i and j with:
$ A \in \mathcal{A}_{i}, B \in \mathcal{A}_{j}, i,j \in \mathbb{N} $.
Case 1: $ i=j $, also $ \mathcal{A}_{i}= \mathcal{A}_{j}$
$$ \Rightarrow A, B \in \mathcal{A}_{i} \underset{(R 2)}{\Rightarrow} A \backslash B \in \mathcal{A}_{i}  \Rightarrow A \backslash B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $$
Case 2: $ i<j $, also $ \mathcal{A}_{i} \subset \mathcal{A}_{j}$
$$ \Rightarrow A, B \in \mathcal{A}_{j} \underset{(R 2)}{\Rightarrow} A \backslash B \in \mathcal{A}_{j}  \Rightarrow A \backslash B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $$
Case 3: $ i>j $, also $ \mathcal{A}_{j} \subset \mathcal{A}_{i}$
$$ \Rightarrow A, B \in \mathcal{A}_{i} \underset{(R 2)}{\Rightarrow} A \backslash B \in \mathcal{A}_{i}  \Rightarrow A \backslash B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $$
(3) Show $ A, B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} \Rightarrow  A \cup B \in \bigcup_{n \in N} \mathcal{A}_{n} $.
Be $ A, B \in \bigcup_{n \in \mathbb{N}} A_{n} $, i.e there is a largest index i and j with:
$ A \in \mathcal{A}_{i}, B \in \mathcal{A}_{j}, i,j \in \mathbb{N} $.
Case 1: $ i=j $, also $ \mathcal{A}_{i}= \mathcal{A}_{j}$
$$ \Rightarrow A, B \in \mathcal{A}_{i} \underset{(R 3)}{\Rightarrow} A \cup B \in \mathcal{A}_{i}  \Rightarrow A \cup B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $$
Case 2: $ i<j $, also $ \mathcal{A}_{i} \subset \mathcal{A}_{j}$
$$ \Rightarrow A, B \in \mathcal{A}_{j} \underset{(R 3)}{\Rightarrow} A \cup B \in \mathcal{A}_{j}  \Rightarrow A \cup B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $$
Case 3: $ i>j $, also $ \mathcal{A}_{j} \subset \mathcal{A}_{i}$
$$ \Rightarrow A, B \in \mathcal{A}_{i} \underset{(R 3)}{\Rightarrow} A \cup B \in \mathcal{A}_{i}  \Rightarrow A \cup B \in \bigcup_{n \in \mathbb{N}} \mathcal{A}_{n} $$
Can someone check if this proof fits for a)?
And what about b)? I don't know if this statement would be true or false. What do you think and why?
 A: Your proof of (a) is correct.
The item (b) is not true. Here is a sample counter-example.
Let $\Omega =\Bbb N$. For each $n \in \Bbb N$, let $\mathcal{A}_n$ be the  $\sigma$- algebra generated by the subsets $\{0\}, ... \{n\}$.
For instance: $\mathcal{A}_0$ is the $\sigma$- algebra generated by $\{0\}$, that is  $\mathcal{A}_0= \{\emptyset, \{0\}, \{0\}^c, \Bbb N\}$; $\mathcal{A}_1$ is the $\sigma$- algebra generated by $\{0\}$ and $\{1\}$ , that is  $\mathcal{A}_1= \{\emptyset, \{0\}, \{1\}, \{0,1\}, \{0\}^c, \{1\}^c, \{0,1\}^c, \Bbb N\}$.
Clearly, $ \mathcal{A}_{n} \subset \mathcal{A}_{n+1} $ for all $ n \in \Bbb N $. Note also that, for each $n \in \Bbb N$ and each $A \in \mathcal{A}_{n}$, $A$ is either a finite set or a co-finite set (its complement is a finite set). In particular, for any $A \in \bigcup_{n \in \Bbb N} \mathcal{A}_{n}$, $A$ is either a finite set or a co-finite set .
Now, for each $n \in \Bbb N$, let $C_n =\{2n\}$. It follows immediately that $C_n \in \mathcal{A}_{2n}$ and so, $C_n \in \bigcup_{n \in \Bbb N} \mathcal{A}_{n}$. However $\{0,2,4,...\}= \bigcup_{n \in \Bbb N} C_n$ is not in $\bigcup_{n \in \Bbb N} \mathcal{A}_{n}$, because $\bigcup_{n \in \Bbb N} \mathcal{A}_{n}$ only has finite or co-finite sets.
So  $\bigcup_{n \in \Bbb N} \mathcal{A}_{n}$ is not closed by countable unions.
