Solutions to Probability Essentials - Chapter 2

Probability

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Problem 2.17 Suppose that $\Omega$ is an infinite set (countable or not), and let $\mathcal{A}$ be the family of all subsets which are either finite or have a finite complement. Show that $\mathcal{A}$ is an algebra, but not a $\sigma$-algebra.

Proof:
Firstly, we need to show that $\mathcal{A}$ is an algebra. Let's check the properties one by one:
1. It's trivial to see that $\emptyset\in\mathcal{A}$ and $\Omega\in\mathcal{A}$.
2. Suppose that $A\in\mathcal{A}$, by the definition of $\mathcal{A}$, $A$ is either finite or have a finite complement, then $A^c$ is either finite or have a finite complement. Thus $A\in\mathcal{A}\Rightarrow A^c\in\mathcal{A}$.
3. Suppose that $A_1,A_2\in\mathcal{A}$, it's sufficient to show that $A_1\cap A_2\in\mathcal{A}$ and $A_1\cup A_2\in\mathcal{A}$. There are three cases: (1) Both $A_1$ and $A_2$ are finite, then $A_1\cap A_2$, $A_1\cup A_2$ are finite. (2) Both $A_1$ and $A_2$ have finite complements, then $(A_1\cup A_2)^c=A_1^c\cap A_2^c$ is finite, $(A_1\cap A_2)^c=A_1^c\cup A_2^c$ is finite. (3) One of $A_1$ and $A_2$ is finite, WLOG, say $A_1$ is finite, $A_2^c$ is finite. Clearly, $A_1\cap A_2$ is finite, and $(A_1\cup A_2)^c=A_1^c\cap A_2^c$, since $A_2^c$ is finite, $(A_1\cup A_2)^c$ is finite.
Hence, $\mathcal{A}$ is closed under finite unions and finite intersections, and we have shown that $\mathcal{A}$ is an algebra.
Secondly, we need to show that $\mathcal{A}$ is not closed under countable unions and intersections.
Let $C\subset\Omega$ be a countable infinite subset, with $C^c$ be infinite(we can always do this, no matter $\Omega$ is countable or not). Because $C$ is countable, $C=\{c_i|i\in\mathbb{N}^*\}$. Both $C$ and $C^c$ are infinite, so $C\notin\mathcal{A}$. Now construct a sequence $\{A_i\}$ such that $A_i=\{c_i\}$. Each $A_i$ contains only one element, $A_i$ is finite and $A_i^c$ is infinite. So $A_i\in\mathcal{A},\forall i\in\mathbb{N}^*$, but $\cup_{i=1}^\infty A_i=C\notin\mathcal{A}$.
Hence $\mathcal{A}$ is not closed under countable unions and intersections.