Sigma-Algebra of Countable Sets
Theorem
Let $X$ be a set.
Let $\Sigma$ be the set of countable and co-countable subsets of $X$.
Then $\Sigma$ is a $\sigma$-algebra.
Proof
Let us verify in turn the axioms of a $\sigma$-algebra.
Axiom $(1)$
By Relative Complement with Self is Empty Set:
- $\relcomp X X = \O$
In particular, it is countable.
Hence $X$ is co-countable, and so $X \in \Sigma$.
$\Box$
Axiom $(2)$
The relative complement of a countable set is by definition co-countable.
The converse holds by Relative Complement of Relative Complement.
Hence:
- $E \in \Sigma \implies \relcomp X E \in \Sigma$
$\Box$
Axiom $(3)$
Let $\family {E_n}_{n \mathop \in \N} \in \Sigma$ be a family of elements of $\Sigma$.
Suppose that all the $E_n$ are countable.
Then by Countable Union of Countable Sets is Countable, $\ds \bigcup_{n \mathop \in \N} E_n$ is also countable.
Hence in $\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$.
Now suppose that at least one $E_n$ is only co-countable.
Then by Set Complement inverts Subsets and Set is Subset of Union: General Result:
- $\ds \relcomp X {\bigcup_{n \mathop \in \N} E_n} \subseteq \relcomp X {E_n}$
By definition of co-countable, the latter is countable.
Thus, by Subset of Countably Infinite Set is Countable, it follows that $\ds \bigcup_{n \mathop \in \N} E_n$ is co-countable.
Hence:
- $\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$
Hence the result, from Proof by Cases.
$\blacksquare$
Sources
- 1984: Gerald B. Folland: Real Analysis: Modern Techniques and their Applications : $\S 1.2$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $3.3 \ \text{(v)}$