Sigma-Algebra Contains Empty Set

Theorem

Let $X$ be a set.

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): $3.2 \ \text{(i)}$
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