## Measurable Spaces – Problem (40/365)

The past few problems looked at how probability works when we have an infinite sample space. It didn’t cover how one can actually assign probabilities to such spaces. That will be the next task. Before that, the book covers the topic of $\sigma$-algebras which form the algebra of events on top of which we can assign a measure.

Given a set $\Omega$, and a set of subsets $\mathcal{A}$, we say that $\mathcal{A}$ is an algebra if $\Omega \in \mathcal{A}$ and is closed under unions and complementation. A $\sigma$-algebra adds to that the requirement that it also be closed under countable unions. The pair $(\Omega, \mathcal{A})$ is called a measureable space.

Let $\mathcal{A}_1, \mathcal{A}_2$ be $\sigma$-algebras of $\Omega$. Are the following systems of sets $\sigma$-algebras?

$\displaystyle \mathcal{A}_1 \cap \mathcal{A}_2 = \{ A : A \in \mathcal{A}_1 \text{ and } A \in \mathcal{A}_2 \} \\ \mathcal{A}_1 \cup \mathcal{A}_2 = \{ A : A \in \mathcal{A}_1 \text{ or } A \in \mathcal{A}_2 \}$

The intersection of $\sigma$-algebras is also a $\sigma$-algebra because $\Omega \in \mathcal{A}_1 \cap \mathcal{A}_2$, and $A_1 \cup A_2 \cup \dots \in$ in the intersection is contained in both $\mathcal{A}_1$ and $\mathcal{A}_2$.

However, the union of $\sigma$-algebras is not always a $\sigma$-algebra. For instance, let $\mathcal{A} = \{ A, \bar{A}, \emptyset, \Omega_1 \}$, $\mathcal{B} = \{ B, \bar{B}, \emptyset, \Omega_2 \}$, then their union does not contain $A \cup B$.

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