Dynkin System Contains Empty Set

Theorem

Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.


Then the empty set $\O$ is an element of $\DD$.


Proof

As $\DD$ is a Dynkin system, $X \in \DD$.

By Set Difference with Self is Empty Set, $X \setminus X = \O$.

Hence, by property $(2)$ of a Dynkin system, $\O = X \setminus X \in \DD$.

$\blacksquare$


Sources

  • 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $5.2$, $\S 5$: Problem $1$
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