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Let $\mathcal{C} \subset \mathcal{P}(\Omega) $ be a set of sets and $U \subset \mathcal{\Omega}$ be a set. Let $\mathcal{G}=\{A: A \cap U \in \sigma(\mathcal{C} \cap U)\}$. How to prove that $\Omega \in \mathcal{G}$?

PS: $\mathcal{C} \cap U =\{C \cap U: C \in \mathcal{C}\}$.

My try: to prove $\Omega \in \mathcal{G}$, we need to prove $\Omega \cap U = U \in \sigma(\mathcal{C} \cap U)$. But how to prove $U \in \sigma(\mathcal{C} \cap U)$? If we let $\mathcal{C} = \{\emptyset\} $, then we have $U \in \{\Omega,\emptyset \}$, which is impossible. How to explain this? Thanks!

Note: this problem is a substep of proof of the equation $\sigma(\mathcal{C} \cap U)=\sigma(\mathcal{C}) \cap U$. Therefore it is true for sure. The original proposition is Proposition E.1.5. at the bottom of this page. theanalysisofdata.com/probability/E_1.html.

Jonas Lionel
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  • Yes. Thx for reminding me. – Jonas Lionel May 13 '22 at 13:17
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    You are correct in stating that in general it cannot be deduced that $\Omega\in\mathcal G$. – Vera May 13 '22 at 13:31
  • Thanks for your comment. This is the original link that describes the proposition. Please refer to Proposition E.1.5. at the bottom of this page. http://theanalysisofdata.com/probability/E_1.html. I believe this prop is correct because it is mentioned in several books. – Jonas Lionel May 13 '22 at 13:45
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    Let $\iota:U\to\Omega$ denote the inclusion. Then $$\sigma\left(\iota^{-1}\left(\mathcal{C}\right)\right)=\iota^{-1}\left(\sigma\left(\mathcal{C}\right)\right)$$ Here the LHS equals $\sigma\left(\left{ C\cap U\mid C\in\mathcal{C}\right} \right)$ and the RHS equals $\left{ A\cap U\mid A\in\sigma\left(\mathcal{C}\right)\right} $. For a proof of this see here. Actually this also works if we are not dealing with an inclusion. – Vera May 13 '22 at 13:49
  • Thanks! I will go back on this once I learn the concept of inclusion. – Jonas Lionel May 14 '22 at 02:00
  • Thanks again for your comment. I think I figure it out. Your inclusion mapping is wonderful and concise. Below is my derivation. Under the inclusion, we have $f^{-1}(B) = B\cap U $ where $B \subset \Omega$ and $f^{-1}(\mathcal{C}) = \mathcal{C} \cap U$ where $\mathcal{C} \subset P(\Omega)$. Since for any mapping we have $f^{-1}(\sigma(\mathcal{C})) = \sigma(f^{-1}(\mathcal{C}))$, it can be readily to get $\sigma(\mathcal{C} \cap U)=\sigma(\mathcal{C}) \cap U$. So sweet! – Jonas Lionel May 14 '22 at 02:34
  • Glad to help. The equality $f^{-1}(\sigma(\mathcal C))=\sigma(f^{-1}(\mathcal C))$ is a very nice tool. Btw, it is also true that $f^{-1}(\tau(\mathcal C))=\tau(f^{-1}(\mathcal C))$ where $\tau(\mathcal C)$ stands for the topology generated by $\mathcal C$. Proof of that is in the same line as the proof for $\sigma$-algebras. – Vera May 15 '22 at 15:07

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