Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
@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

1 Answer 1

up vote 3 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, $$ \sigma(\mathcal{G}) \cap A \subset \sigma_A(\mathcal{G} \cap A). $$

share|improve this answer
    
+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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.