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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 $\mathcal{F}$?

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As this is very easy, what you have attempted? – Jonas Teuwen Jan 23 '11 at 20:41
@Jonas T: This is my first touch with sigma algebras and I'm mainly having trouble fully comprehending the question and taking the first step. – tsiki Jan 23 '11 at 20:50
I don't understand. If you're up to the point where you see $\sigma$-algebras then you should be able to check the definitions on such a trivial example. – Jonas Teuwen Jan 23 '11 at 21:18
That makes two confused people then. – tsiki May 2 '13 at 22:08
up vote 0 down vote accepted

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 $B$.

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Thanks, I'm pretty sure I got it now – tsiki Jan 23 '11 at 21:20
@Theo: I would like you to read the comments and what caused this exchange between Didier and me. I made a mistake of wrongly interpreting over which set the op wanted to check it was a $\sigma$-algebra. I acknowledged the mistake and decided to edit it. Didier started an unwarranted comment which triggered me. I am totally fine with down-voting my answer but the comment(s) left should not be provocative. – user17762 Aug 18 '11 at 10:10
@Didier: As a show of good spirit and respect, I have now edited and changed my answer and removed all my comments. – user17762 Aug 18 '11 at 10:29

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.

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