Atom of Countably Generated Sigma-Algebra is Measurable

Theorem

Let $X$ be a set.

Let $\Sigma$ be a countably generated $\sigma$-algebra of $X$.

Then:

$\ds \forall x \in X : \quad \sqbrk x_\Sigma \in \Sigma$

where $\sqbrk x_\Sigma$ denotes the atom.

Proof

By assumption, there is a countable collection $\GG \subseteq \Sigma$ such that:

$\ds \Sigma = \map \sigma \GG$

Let $x \in X$.

It suffices to show:

$\ds \sqbrk x_\Sigma = \bigcap_{ x \in A \in \GG } A \cap \bigcap_{ x \not \in A \in \GG} X \setminus A$

$\subseteq$ is clear.

We need to show $\supseteq$.

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