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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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    Confusing notation, $A \in \sigma(A)$ ... perhaps there are two different letters $A$ intended here?2012-08-16
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    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?2012-08-19

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