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Let $\mathcal{C}$ be a class of subsets of $\Omega $, $A\subset \Omega $, $\mathcal{F}=\sigma(\mathcal{C})$, and $\mathcal{G}=\{ B \in \mathcal{F}:B\cap A\in \sigma_A(\mathcal{C}\cap A)\}$.

How can I prove that $\mathcal{G}$ is a sigma field?

For example: $B\in \mathcal{G}\iff (B\cap A \in \sigma_A(\mathcal{C}\cap A)) \land (B\in \mathcal{F}) \Rightarrow (B^c\cup A^c \in \sigma_A(\mathcal{C}\cap A)) \land (B^c\in \mathcal{F}) $, but what I would want is $(B^c\cap A \in \sigma_A(\mathcal{C}\cap A)) \land (B^c\in \mathcal{F}) $

I'm trying to use the good sets principle to prove that $\sigma_A(\mathcal{C}\cap A)=\mathcal{F}\cap A$.

Any help would be appreciated.

An old man in the sea.
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  • What exactly do you mean by $\mathcal{F} \cap A$? If you mean ${E \cap A : E \in \mathcal{F}}$, then note that this isn't (necessarily) a $\sigma$-algebra since $\Omega \notin \mathcal{F}$ unless $A = \Omega$. This means that you can't have $\sigma(\mathcal{C} \cap A) = \mathcal{F} \cap A$. – Marcus M May 31 '17 at 16:59
  • @MarcusM thanks for your interest. But since $\mathcal{F}$ is a sigma algebra, $\Omega$ should belong to it, no? – An old man in the sea. May 31 '17 at 17:25
  • Sure, but $\Omega \cap A = A$ which is not $\Omega$. – Marcus M May 31 '17 at 21:55
  • @MarcusM You're right. I've edited my question, so that make clear that the sigma algebra is of subsets of $A$ over $\mathcal{C}\cap A$ – An old man in the sea. May 31 '17 at 22:04

1 Answers1

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HINT: Let $\mathcal{B} = \{ B \in \mathcal{F} : B \cap A \in \sigma_A(\mathcal{C} \cap A)\}$, and show that $\mathcal B = \mathcal F$.

Note first that $\mathcal{C} \subset \mathcal{B}$. Since $\mathcal{F}$ and $\sigma_A(\mathcal{C} \cap A)$ are both $\sigma$-fields, it's straightforward to show that $\mathcal B$ is a $\sigma$-field. But since $\mathcal{B}$ is a $\sigma$-field with $\mathcal C \subset\mathcal{B}$, we must have $\mathcal{F} = \sigma(C) \subset \mathcal{B}$; we also have that $\mathcal B \subset \mathcal F$ by definition, thereby showing that $\mathcal B = \mathcal F$. Finally, by definition of $\mathcal B$, we have that $\mathcal B \cap A \subset \sigma_A(\mathcal C \cap A)$. Since $\mathcal B = \mathcal F$, this is shows $\mathcal F \cap A \subset \sigma_A(\mathcal C \cap A)$.

The reverse inclusion is immediate: since $\mathcal C \subset \mathcal F$, we have $\sigma_A(\mathcal C \cap A) \subset \sigma_A(\mathcal F \cap A) = \mathcal F \cap A$.

EDIT: How to show that $\mathcal{B}$ is a $\sigma$-field:

  • Since $\Omega \in \mathcal{F}$ and $\Omega \cap A = A \in \sigma_A(\mathcal{C} \cap A)$, we also have $\Omega \in \mathcal{B}$.

  • Let $E \in \mathcal{B}$. Then $E \in \mathcal{F} \implies \Omega \setminus E \in \mathcal{F}$. Similarly, $E \cap A \in \sigma_A(\mathcal{C} \cap A) \implies A \setminus (E \cap A) \in \sigma_A(\mathcal{C} \cap A)$. But $A \setminus (E \cap A) = \{x \in A : x \notin E\} = (\Omega \setminus E) \cap A$. Thus $\Omega \setminus E \in \mathcal{B}$.

  • Let $\{E_i\}_{i = 1}^\infty \subset \mathcal{B}$. Then $$\{E_i\}_{i=1}^\infty \subset \mathcal{F} \implies \bigcup\limits_{i} E_i \in \mathcal{F}.$$ Similarly, we have $$\{E_i \cap A\}_{i = 1}^\infty \subset \sigma_A(\mathcal{C} \cap A) \implies \bigcup\limits_{i} (E_i \cap A) \in \sigma_A(\mathcal{C} \cap A).$$ Since $\bigcup_i (E_i \cap A) = (\bigcup_i E_i) \cap A$, this implies that $\bigcup_i E_i \in \mathcal{B}$, thereby completing the proof.

Marcus M
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  • Marcus, thanks for your answer, but my whole doubt is precisely how to show that $\mathcal{B}$ is a sigma-field – An old man in the sea. Jun 01 '17 at 09:47
  • @Anoldmaninthesea. Ah, okay. I've added a bit that proves that. I think the trickiest part is the complement one; I'd recommend giving them all a shot before reading what I did, but I think showing $\Omega \in \mathcal{B}$ and that $\mathcal{B}$ is closed under countable unions is more doable. – Marcus M Jun 01 '17 at 14:27