Dynkin System Contains Empty Set
Jump to navigation
Jump to search
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 $1$