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?

  • $\begingroup$ 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? $\endgroup$
    – coffeemath
    Mar 8, 2021 at 20:18
  • $\begingroup$ @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. $\endgroup$
    – user866761
    Mar 8, 2021 at 20:24

1 Answer 1


Since your proposed sigma algebra consists of the various unions of a specific list of pairwise disjoint sets, it is obvious anyway that if you take a countable union of those it will be in your proposed sigma algebra, merely because that proposed algebra was already defined to be the collection of all subsets of the power set of the basic collection of pairwise disjoint sets.

It still seems there might be a simpler way to say the above; the more I look at it my version seems overly complicated.

  • 1
    $\begingroup$ 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. $\endgroup$
    – user866761
    Mar 8, 2021 at 20:44

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