Intersection Sigma Algebra

Last updated Nov 1, 2022

Suppose $(X, \mathcal{M})$ is a Measure Space and let $A \subset X$. Then the collection $$\mathcal{M} \cap A := {E \cap A : E \in \mathcal{M}}$$ is the Intersection Sigma Algebra on $A$.

1. We should check that $\mathcal{M} \cap A$ is indeed a Sigma Algebra:

   Proof: We check the criteria for being a Sigma Algebra

   1. Since $X \in \mathcal{M}$ and $X \cap A = A$, we have that $A \in \mathcal{M} \cap A$. $\checkmark$
   2. Suppose $F \in \mathcal{M} \cap A$. Then there exists $E \in \mathcal{M}$ so $F = E \cap A$. Thus $$\begin{align*} A \setminus F &= A \cap (E \cap A)^{C}\\ &= A \cap (E^{C} \cup A^{C})\\ &=(A \cap E^{C}) \cup (A \cap A^{C})\\ &=E^{C} \cap A \in \mathcal{M} \cap A \end{align*}$$ since $E^{C} \in \mathcal{M}$. $\checkmark$
   3. Suppose $(F_n){n=1}^{\infty} \subset \mathcal{M} \cap A$. Then there exists $(E_n){n=1}^{\infty} \subset \mathcal{M}$ so that $F_{n} = E_{n} \cap A$ $\forall n \in \mathbb{N}$. Then $$\bigcup\limits_{n \in \mathbb{N}} F_{n} = \bigcup\limits_{n \in \mathbb{N}} (E_{n} \cap A) = A \cap \bigcup\limits_{n \in \mathbb{N}} E_{n} \in \mathcal{M} \cap A$$ since $\bigcup\limits_{n \in \mathbb{N}} E_{n} \in \mathcal{M}$. $\checkmark$ $\blacksquare$