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I am looking for an example of a collection of sets $\mathcal{F}$ where $\Omega=\mathbb{R}$, $\mathbb{R}\in \mathcal{F}$, $A^c\in\mathcal{F}$ for all $A\in\mathcal{F}$, but the countable union property does not hold, thereby ensuring $\mathcal{F}$ is not a sigma algebra.

For natural numbers, a standard example is $\mathcal{F}=\{A \subseteq\mathbb{N}\mid \text{$A$ or $A^c$ has a finite number of elements}\}$. How do we extend this example to reals?

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Also I think you mean $\mathcal{F}=\{A \in\mathcal P(\mathbb{N})\mid \text{$A$ or $A^c$ has a finite number of elements}\}$ in general this works for any set besides $\Bbb N$ as long as it is not finite. –  user10444 Feb 9 at 13:55
    
@user10444 $\{\emptyset,[0,1],\mathbb R\}$ is not closed under complement (and is in fact closed under countable union) –  Hagen von Eitzen Feb 9 at 13:57
    
@Hagen Yes it is Not a $\sigma$-algebra, I apparently missed the fact it has to be an algebra. I will delete my first comment. –  user10444 Feb 9 at 13:58
    
@user10444: Yes, I meant subset. Thanks for pointing out the error. –  Sultan Feb 9 at 14:00

2 Answers 2

Let $$\mathcal{G} := \{(a,b]; -\infty \leq a \leq b \leq \infty\}$$ and $$\mathcal{F} := \{A \subseteq \mathbb{R}; \exists A_1,\ldots,A_n \in \mathcal{G}: A = \bigcup_{k=1}^n A_k\},$$ i.e. $\mathcal{F}$ consists of all finite unions of the intervals contained in $\mathcal{G}$. (We use the convention $(a,\infty] := (a,\infty)$.)

It is not difficult to see that $\mathbb{R} \in \mathcal{F}$ and $A^c \in \mathcal{F}$ for any $A \in \mathcal{F}$. On the other hand, $\mathcal{F}$ is not a $\sigma$-algebra. Consider for example the sets $A_n := (1-1/n,1] \in \mathcal{F}$. Then $$\bigcap_{n \geq 1} A_n = \{1\} \notin \mathcal{F}$$

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Let's start from what we want. $\mathcal{F}$ should not be closed under countable unions. For this it suffices to have $A, B \in \mathcal{F}$ with $A \cup B \notin \mathcal{F}$. $\mathcal{F}$ should be closed under complement, so let's throw in $A^c$ and $B^c$. And we need $\mathbb R$, and hence $\emptyset$, too. Does the closed-under-complement requirement force us to add any more sets? No. So let's put $A=\{0\}$ and $B=\{1\}$ and we get $$\{\emptyset, \{0\}, \{1\}, \mathbb R \setminus \{0\}, \mathbb R \setminus \{1\}, \mathbb R\}$$ It is easy to check that this works.

If you want the lack of closure under countable union to be witnessed by countably infinitely many distinct sets, you can take $\{n\}$ and its complement for every natural number $n$ instead of just $0$ and $1$.

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