By the definition of $\sigma$-algebra we have:
- The null set and the sample space are members of $F$.
- If an event $A$ is a member of $F$, then the complement of $A$ is a member of $F$
- If events $A_1$, $A_2$, … ,$A_n$ are members of F, then the event $A_1 \cup A_2 … \cup A_n$ is a member of $F$.
Let us consider two $\sigma$-algebras on the sample space $\Omega$, $F_1$ and $F_2$. Then is the intersection of the two collections a $\sigma$-algebras?
I understand that the answer is yes, but how do I formally prove this result?
For instance, the first property is easy to show. By definition, any any two sigma fields will contain the null set and the sample space, so the intersection between any two sigma fields will inevitably include both the null set and the sample space.
But how do I formally prove the claim?
Best Answer
Let $F_1,F_2$ be $\sigma$-algebras.
Let $A \in F_1 \cap F_2$, so $A\in F_1 \rightarrow \overline{A} \in F_1, A\in F_2\rightarrow \overline{A}\in F_2$ and we got that $\overline{A}\in F_1 \cap F_2$.
For 3, let $A_1, ...\in F_1 \cap F_2$. It means that $\forall i\in \mathbb{N}:A_i\in F_1\wedge A_i \in F_2$ so we have that $\bigcup_{i\in \mathbb{N}} {A_i} \in F_1,F_2$ as $F_1, F_2$ are both $\sigma$-algebras. Hence, $\bigcup_{i\in \mathbb{N}} {A_i}$.
Notice that I proved claim 3 for countably many sets and not for finitely many, we can prove the case of finitely many sets by setting $\phi=A_{n+1}=A_{n+2}=...$.