Suppose you are given a set $ \Omega $ and a collection $ \mathcal{G} $ of subsets of $ \Omega $.
Assume further that $ A \subset \Omega $.
Now let $ \sigma_{\Omega} (\mathcal{G}) $ denote the smallest sigma-algebra on $ \Omega $ containing $ \mathcal{G} $, and let $ \sigma_{A}(\mathcal{G} \ \cap A) $ denote the smallest sigma-algebra on A containing the collection $ \mathcal{G} \ \cap A $.
Is it true that $ \sigma_{A}(\mathcal{G} \ \cap A) = \sigma_{\Omega} (\mathcal{G}) \cap A $ ?
The inclusion " $ \subset $ " is clear, since if $ \mathcal{H} $ is a sigma-algebra on $ \Omega $ containing $ \mathcal{G} $, then $ \mathcal{H} \cap A $ is a sigma-algebra on $ A $
containing $ \sigma_{A}(\mathcal{G} \ \cap A) $. But what about the other inclusion?
Thanks for your help!
Regards, Si
Best Answer
Let $$ \mathcal{B} = \left\{B \subset X | B \cap A \in \sigma_A(\mathcal{G} \cap A)\right\}. $$ Notice that $\mathcal{G} \subset \mathcal{B}$. It is easy to see that $\mathcal{B}$ is a $\sigma$-algebra. For example, if $B_j \in \mathcal{B}$, then $$ \left(\bigcup B_j\right) \cap A = \bigcup (B_j \cap A) \in \sigma_A(\mathcal{G} \cap A), $$ because $B_j \cap A \in \sigma_A(\mathcal{G} \cap A)$.
Therefore, since $\mathcal{B}$ is a $\sigma$-algebra containing $\mathcal{G}$, we can conclude that $\sigma(\mathcal{G}) \subset \mathcal{B}$. So, for every $B \in \sigma(\mathcal{G})$, $B \cap A \in \sigma_A(\mathcal{G} \cap A)$. In other words, $$ \sigma(\mathcal{G}) \cap A \subset \sigma_A(\mathcal{G} \cap A). $$