Sigma algebra proof problem

Question

Let $\mathcal{F}$ be a $\sigma$-algebra of subsets of $\Omega$ and $B \in \mathcal{F}$. Show that $E = \{A \cap B: A \in \mathcal{F}\}$ is a $\sigma$-algebra of subsets of $B$. Is it still true when $B$ is a subset of $\Omega$ that does not belong to $\mathcal{F}$?

Answer 1

We are given that $\mathcal{F}$ is a $\sigma$-algebra of subsets of $\Omega$. Let $B \subset \Omega$. ($B$ may (or) may not be in $\mathcal{F}$)

Let $\mathcal{E}_B = \{A \cap B : A \in \mathcal{F} \}$. We want to show that $\mathcal{E}_B$ is a $\sigma$-algebra of the subsets of $B$.

1. $B \in \mathcal{E}_B$. This is so since $\Omega \in \mathcal{F}$ and hence $B = \Omega \cap B \in \mathcal{E}_B$
2. If $E_1 \in \mathcal{E}_B$, then $E_1 = A_1 \cap B$ for some $A_1 \in \mathcal{F}$. Now $E_1^c$ relative to $B$ is $E_1^c \cap B = \left(A_1 \cap B \right)^c \cap B = \left(A_1^c \cup B^c \right) \cap B = \left(A_1^c \cap B \right) \cup \left(B_1^c \cap B \right) = \left(A_1^c \cap B \right) \cup \emptyset = \left(A_1^c \cap B \right)$ Hence, we have $E_1^c$ relative to $B$ is $A_1^c \cap B$ and since $\mathcal{F}$ is a $\sigma$-algebra, we have $A_1^c \in \mathcal{F}$ and hence $E_1^c \in \mathcal{E}_B$
3. If we have a sequence of sets $\{E_k\}_{k=1,2,\ldots}^{\infty} \in \mathcal{E}_B$, then each $E_k = A_k \cap B$ for some $A_k \in \mathcal{F}$ and since $\mathcal{F}$ is a $\sigma$-algebra, we have $\displaystyle \cup_{k=1}^{\infty} A_k \in \mathcal{F}$. Hence, $\displaystyle \cup_{k=1}^{\infty} E_k = \displaystyle \cup_{k=1}^{\infty} \left(A_k \cap B \right) = \displaystyle \left(\cup_{k=1}^{\infty} A_k \right) \cap B$ and since $\left(\cup_{k=1}^{\infty} A_k \right) \in \mathcal{F}$ we have $\left(\cup_{k=1}^{\infty} A_k \right) \cap B \in \mathcal{F} \implies \displaystyle \cup_{k=1}^{\infty} E_k \in \mathcal{E}_B$

Hence, $\mathcal{E}_B$ is a $\sigma$-algebra of the subsets of $B$.

Answer 2

Both questions can be checked by the definition of the $\sigma$-algebra. For the example by @Sivaram (when $B\notin{\cal F}$), ${\cal E} = \{\emptyset, B\}$, which is still a (trivial) $\sigma$-algebra.