Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.