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I need to get confirmation of my proof. If I've got it wrong, please just provide hints and not the full answer. The context here is to do with probability events as sets.

The question is:

Let $\mathcal{F}$ be a $\sigma$-field of subsets of $\Omega$ and suppose that $B \in \mathcal{F}$. Show that $\mathcal{G} = \{A \cap B : A \in \mathcal{F}\}$ is a $\sigma$-field of subsets of $B$.

  • Proof of: $\emptyset \in \mathcal{G}$.

Since $B^c \in \mathcal{F}$, set $A = B^c$. Therefore, $ A \cap B = \emptyset \in \mathcal{G}$.

  • Proof of: If $A_1 \cap B, A_2 \cap B, \ldots \in \mathcal{G}$ then $\bigcup_{i}^{\infty} A_{i} \cap B \in \mathcal{G}$.

$A \in \mathcal{F} \implies \bigcup_{i}^{\infty} A_{i} \in \mathcal{F}$. Therefore, $ \bigcup_{i}^{\infty} A_{i} \cap B \in \mathcal{G}$.

  • Proof of: $(A \cap B)^c \in \mathcal{G}$

$A^c, B^c \in \mathcal {F} \implies (A^c \cup B^c) \in \mathcal{F} \implies (A \cap B)^c \in \mathcal{G}$.

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The last part (stability under complement) should be slightly modified. What you need to show is that the relative complement of $A\cap B$ in $B$ (i.e. $B\cap (A\cap B)^c$) belongs to $\mathcal G$. But this is clear from what you did... –  Etienne Feb 18 at 22:26
    
Do you mean I should have stated: $$(A \cap B)^c \cap B = A^c \cap B \in \mathcal{G}$$ since $A^c \in \mathcal{F}$. –  I.K. Feb 18 at 22:42
    
Yes, this is what I meant. –  Etienne Feb 18 at 23:13

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