# $\mathcal{A}:= \left\{\bigcup_{i \in I} A_i : I \subseteq \mathbb{N} \right\}$ sigma-algebra?

Let $$\Omega$$ be a set $$(A_n)_{n \in \mathbb{N}}$$ a sequence of subsets of $$\Omega$$ such that $$A_m \cap A_n = \emptyset$$ whenever $$m \neq n$$ and $$\bigcup_{n\in \mathbb{N}}A_n = \Omega$$. Consider the family $$\mathcal{A}$$ of subsets of $$\Omega$$ defined by:

$$\mathcal{A}:= \left\{\bigcup_{i \in I} A_i : I \subseteq \mathbb{N} \right\}$$.

To show: $$\mathcal{A}$$ is a $$\sigma$$-algebra.

My attempt:

• Choose $$I = \mathbb{N}$$ then $$\bigcup_{i \in \mathbb{N}} A_i \in \mathcal{A}$$ and by assumption $$\bigcup_{i \in \mathbb{N}} A_i = \Omega$$

• Let $$A\in \mathcal{A}$$ then $$A=\bigcup_{i \in \lambda} A_i$$ for $$\lambda \subset \mathbb{N}$$. Now $$A^c = (\bigcup_{i \in \lambda} A_i)^c = \bigcup_{j\in \mathbb{N}\setminus \lambda}A_j$$. This is true because $$A_m \cap A_n = \emptyset$$ whenever $$m \neq n$$. But since ($$\mathbb{N} \setminus \lambda) \subset \mathbb{N} \implies A^c \in \mathcal{A}$$

• Let $$A_{I_n} \in \mathcal{A}$$. Then $$A_{I_n} = \bigcup_{i \in I_n}A_i.$$ But then $$\bigcup_{n \in \mathbb{N}} A_{I_n} = \bigcup_{i \in \mathbb{N}}A_i$$ which is by assumption in $$\mathcal{A}$$.

Is that correct?

• Why is the subscript $n$ there twice in the notation $(A_n)_n$? Also does $I_n$ mean :the identity function on $n$ (regarding $n$ as nonnegative or ositive integers up to $n$)? Or is $I_n$ simply the integers in the latter? Mar 8, 2021 at 20:18
• @coffeemath $(A_n)_{n \in \mathbb{N}}$ I think this clarifies it. A sequence $A_n \forall n \in \mathbb{N}$. $I_n$ should be here just a subset of $\mathbb{N}$ but depends on $n$ since I'll have to show that it is stable under countable union and therefore I'm iterating with this $n$ over the natural numbers.
– user866761
Mar 8, 2021 at 20:24

• After reading your answer it became clear for me why it is obvious. I think we were just given this exercise to get familiar with the definition of $\sigma$-algebras and prove it theoretically.