$\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A$

measure-theory

Suppose $\mathcal{C}$ is a class of subsets of $\Omega$. Let $A$ be a non-empty subset of $\Omega$. Show that
$$\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A,\text{ a }\sigma\text{-field of subsets of }A$$
where, $\mathcal{C}\cap A:=\{X\cap A:X\in\mathcal{C}\}$.

Proving

  1. $\sigma(\mathcal{C})\cap A$ is a $\sigma$-field of subsets of $A$
  2. $\sigma(\mathcal{C})\cap A$ contains $\mathcal{C}\cap A$

is easy. I also need to prove

  1. For any field containing $\mathcal{C}\cap A, F(\mathcal{C}\cap A)$ (say), $F(\mathcal{C}\cap A)\supseteq\sigma(\mathcal{C})\cap A$.

However for any $B\in\sigma(\mathcal{C})\cap A$, I am not getting anywhere. Hints please?

Best Answer

Important: Note that in order to prove $\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A$, we must consider $\sigma(\mathcal{C}\cap A)$ as the $\sigma$-field generated in $A$, that means, a $\sigma$-field of subsets of $A$.

You already know:

  1. $\sigma(\mathcal{C})\cap A$ is a $\sigma$-field of subsets of $A$
  2. $\sigma(\mathcal{C})\cap A$ contains $\mathcal{C}\cap A$

From 1 and 2, it follows immediately that $\sigma(\mathcal{C}\cap A) \subseteq\sigma(\mathcal{C})\cap A$, because $\sigma(\mathcal{C}\cap A)$ is the intersection of all $\sigma$-fields containing $\mathcal{C}\cap A$.

Now, we must prove that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$. Let us define $$\mathcal{K}=\{S + T : S \in \sigma(\mathcal{C}\cap A) \text{ and } T \in \sigma(\mathcal{C}\cap A^c)\} $$ where $+$ indicates disjoint union. It is to see that $\mathcal{K}$ is a $\sigma$-field of subsets of $\Omega$ and $\mathcal{C} \subseteq \mathcal{K}$. It follows immediately that $\sigma(\mathcal{C}) \subseteq \mathcal{K}$. So we have $$ \sigma(\mathcal{C}) \cap A \subseteq \mathcal{K} \cap A = \sigma(\mathcal{C}\cap A)$$ So we have proved that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$. So, we have $\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A$.

Remark: Another way to prove that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$ is to consider $$ \mathcal{H} = \{ S \subseteq \Omega : S \cap A \in \sigma(\mathcal{C}\cap A)\}$$ It is easy to prove that $\mathcal{H}$ is a $\sigma$-field of subsets of $\Omega$ and $\mathcal{C} \subseteq \mathcal{H}$. It follows immediately that $\sigma(\mathcal{C}) \subseteq \mathcal{H}$. So we have $$ \sigma(\mathcal{C}) \cap A \subseteq \mathcal{H} \cap A \subseteq \sigma(\mathcal{C}\cap A)$$ So we have proved that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$.

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