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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

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I don't understand the notation $\mathcal{G}\cap A$, since $\mathcal{G}$ is the set of subsets of $\Omega$ while $A$ is the subset of $\Omega$. – Jack Nov 17 '11 at 15:15
@Jack: I think that's why Si wrote "the family $\mathcal{G} \ \cap A$", to make clear that every set in $\mathcal{G}$ is being intersected with $A$, much like when we write $gH$ for a coset of a subgroup. – joriki Nov 17 '11 at 15:21
up vote 4 down vote accepted

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, $$ \sigma(\mathcal{G}) \cap A \subset \sigma_A(\mathcal{G} \cap A). $$

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+1, nice! To put it in words: It doesn't matter what happens outside $A$; the elements of $\mathcal G$ happen to have certain parts outside $A$, and these may interact in complicated ways to determine $\sigma_\Omega(\mathcal{G})$, but when we cut everything down to $A$ none of that matters; we might as well consider all possible extensions outside $A$ of the elements of $\sigma_A(\mathcal G\cap A)$ (as you did in $\mathcal B$); then the result is clearly a sigma algebra, and yet we don't get more than $\sigma_A(\mathcal G\cap A)$ when we cut it down to $A$. – joriki Nov 17 '11 at 15:57
Nice! That's what I was looking for... Cheers! – Mad Si Nov 17 '11 at 16:21

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