Sigma-Algebra Contains Empty Set

Theorem

Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.

Then $\O \in \Sigma$.

Proof

Axiom $(1)$ of a $\sigma$-algebra grants $X \in \Sigma$.

By axiom $(2)$ and Set Difference with Self is Empty Set, it follows that $\O = X \setminus X \in \Sigma$.

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $3.2 \ \text{(i)}$