TITLE: A question about sigma-algebras and generators
QUESTION [8 upvotes]: 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

REPLY [7 votes]: 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,
$$
  \sigma(\mathcal{G}) \cap A
  \subset
  \sigma_A(\mathcal{G} \cap A).
$$