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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.

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2 Answers 2

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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 $\cap_{n \in \mathbb{N}}A^c_n\in \Sigma$.

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    $\begingroup$ I understand it now. Thanks for your answer! $\endgroup$ Commented Nov 6, 2021 at 11:10
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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 !

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