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I am having some problems joining sigma algebras. So I have: $$\textit{K} =\left \{ A\cap B: A\in \sigma (D),B\in \sigma (E) \right \}$$ and I need to show $$\sigma(K)= \sigma (D,E)$$

What I've done so far

I intend to do this by showing $$\sigma(K)\subseteq \sigma (D,E)$$ and $$\sigma(K)\supseteq\sigma (D,E)$$

The first part is easy enough, my problem is $\sigma(K)\supseteq\sigma (D,E)$ as I'm not sure where to begin.

I've started by saying by definition $\sigma(D,E)= \sigma(\sigma(D)\cup\sigma(E))$ and $\sigma(\sigma(D)\cup\sigma(E))=\sigma(\sigma(D)\cap\sigma(E))$ by De Morgan and $\sigma(D)\cap\sigma(E)=K$, but I think that is wrong as $\sigma(D)\cap\sigma(E)$ may be a sigma algebra and thus its smaller than $ \sigma (A,B)$.

I've also considered something like $\sigma(K^{c})$ to somehow argue that $\sigma(\sigma(D)\cup\sigma(E))$ is a subset of that but intuitively it feels like what I did with the De Morgan above.

Thanks

EDIT: sorry for the confusing notation everyone, everything has been changed.

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3  
Confusing notation, $A \in \sigma(A)$ ... perhaps there are two different letters $A$ intended here? –  GEdgar Aug 16 '12 at 15:39
    
Thanks for changing notation. Now explain what the letters mean. There is some large set $X$, and we have sigma-algebras on it? Is $D$ a subset of $X$, so that $\sigma(D)$ is a very small sigma-algebra? Or what? –  GEdgar Aug 19 '12 at 16:57

2 Answers 2

Notice that $K$ contains $A$ and $B$. The definition of $\sigma(A,B)$ would be $\sigma(A\cup B)$, I think. Not that it makes for much of a difference, but still...

I think you're overthinking it. I believe that the first part is actually harder (if only a little bit).

Also, you might want to use different symbols to denote the algebras and their elements (\mathcal might help you).

If you're still stuck, see the further hint:

If we have a set $G$ and a sigma-algebra $\Sigma$, then if $G\subseteq \Sigma$, then $\sigma(G)\subseteq \Sigma$. Thus, if for some $G_1,G_2$ we have that $\sigma(G_1)\supseteq G_2$ and $\sigma(G_2)\supseteq G_1$, then $\sigma(G_1)=\sigma(G_2)$.

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Thanks, I realised I can see how I was overthinking. Jus ta misc question would it be correct to say: $$\sigma(\sigma(A)\cup\sigma(B))=\sigma(\sigma(D)\cap\sigma(E))$$? –  Pk.yd Aug 19 '12 at 14:34
    
@Pk.yd: what are $A,B$? If they are elements of $\sigma(D),\sigma(E)$, then not really. The only sensible way to interpret $\sigma(A)$ I can think of in this case would be as $\sigma(\{ A\})$, which, except for trivial cases, is something much less rich than $\sigma(D)$ –  tomasz Aug 19 '12 at 18:26

Your definition of $K$ has no sense: you should say for instance $K=\sigma(X\cap Y, X \in\sigma(A), Y\in\sigma(B))$ (because $A \in \sigma(A)$ is puzzling).

The first inclusion $\sigma(K) \subset \sigma(A,B)$ is easy: it follows from the inclusions $\sigma(A) \subset \sigma(A,B)$ and $\sigma(B) \subset \sigma(A,B)$ and from the stability of a $\sigma$-field under intersection.

The second inclusion $\sigma(K) \supset \sigma(A,B)$ is easy too: clearly $\sigma(K)$ is a $\sigma$-algebra containing $\sigma(A)$ and $\sigma(B)$, therefore it contains the smallest $\sigma$-algebra containing $\sigma(A)$ and $\sigma(B)$, which is nothing but $\sigma(A,B)$.

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ahh forgot about about your very last point. Thanks. –  Pk.yd Aug 19 '12 at 14:35

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