# Prove certain $\Sigma$ is a $\sigma$-algebra

I've been solving some problems from my Functional Analysis course and there's a part of this problem I'm not sure how to do. The problem goes like this:

Given $$\Sigma$$ a collection of subsets of a non-empty set $$X$$ that verifies:

1. $$X\in\Sigma$$
2. $$A,B\in\Sigma \implies B\setminus A\in\Sigma$$
3. $$A,B\in\Sigma, A\cap B=\emptyset \implies A\cup B\in\Sigma$$
4. $$(A_n)_{n\in\mathbb{N}}\subseteq\Sigma \text{ decreasing sequence} \implies \bigcap_{n=1}^\infty A_n\in\Sigma$$

Prove $$\Sigma$$ is a $$\sigma$$-algebra.

What I did is just prove the three different points of the $$\sigma$$-algebra definition:

• First I prove $$\emptyset\in\Sigma$$. This is easy since (1) says that $$X\in\Sigma$$ and using (2), $$X\setminus X=\emptyset\in\Sigma$$.

• Now to prove $$A\in\Sigma$$ implies $$A^c\in\Sigma$$, I use again (1) and (2), $$X\setminus A = A^c\in\Sigma$$.

These two properties were easy to prove, the problem is the last one:

• Now I need to prove that $$\Sigma$$ is closed under countable union, to be said, $$(A_n)_{n\in\mathbb{N}}\subseteq \Sigma \implies\bigcup A_n\in\Sigma$$. It's clear that I'll need to use hypothesis (3) and (4). From (4) I can conclude that $$\Sigma$$ is closed under countable unions of increasing sequences of sets, because $$\left[\bigcap_{n\in\mathbb{N}} A_n \right]^c = \bigcup_{n\in\mathbb{N}} A_n^c$$ and $$(A_n^c)$$ is an increasing sequence (because $$(A_n)$$ was a decreasing one). But as I said this is verified for increasing countable sequences, not in general. I guess now I must use hypothesis (3) to extend it to all countable sequences, but I don't see how.

How can I finish the prove of the third property? Is the work I did till that point correct? Any help or hint will be appreciated, thanks in advance.

Consider $$(A_n)_{n \in \mathbb{N}}\subset \Sigma$$. We must show that $$\cup_{n \in\mathbb{N}}A_n \in \Sigma$$. Define $$A:=\cup_{n \in\mathbb{N}}A_n$$, we have that $$A=(\cap_{n \in \mathbb{N}}A^c_n)^c$$. As $$\Sigma$$ is closed under complements, we just need to show that $$\cap_{n \in \mathbb{N}}A^c_n\in \Sigma$$ to come up with the conclusion. We need a decreasing sequence $$(B_n)_{n \in \mathbb{N}}\subset \Sigma$$ s.t. $$B_n\downarrow \cap_{n \in \mathbb{N}}A^c_n$$. This can be done as follows: $$B_n:=\bigcap_{k=1}^nA_{k}^c$$ By definition, $$(B_n)_{n \in \mathbb{N}}\subset \Sigma$$ and $$B_n \downarrow \cap_{n \in \mathbb{N}}A^c_n$$ so $$\cap_{n \in \mathbb{N}}A^c_n\in \Sigma$$.
First you note that for $$A_1,A_2\in\Sigma$$, $$A_1\cup A_2=(A_2\setminus A_1)\cup A_1\in \Sigma$$. Thus you have closure under finite union by induction. From there, pick any sequence $$(A_i)_i\subset \Sigma$$. Note that $$\cup_{i=1}^\infty A_i=\cup_{i=1}^\infty B_i$$ where $$B_i=\cup_{j=1}^i A_j$$ and $$(B_i)_i$$ is an increasing sequence of sets that you have shown to be included in $$\Sigma$$. Here note that $$B_i\in \Sigma$$ since we have closure under finite unions. You are done !