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Let $ (\Omega, \mathcal{F}) $ be a measurable space. Let $ A, B \in \mathcal {F} $ with $ A \cap B = \emptyset $. Let $ \mathcal{A} \subset \mathcal{F} $ the smallest $ \sigma $-field containing $ A $ and not containing $ B $ and $ \mathcal {B} \subset \mathcal{F} $ the smallest $ \sigma $-field containing $ B $ and not containing $ A $.

Is it true that $ \mathcal{F} $ is the smallest sigma algebra containing $ \mathcal {A} \cup \mathcal {B}$? And if $|\Omega|=\infty$ ?

There is a counter example?

Thank´s.

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    $\begingroup$ It is possible that neither $\mathcal{A}$ nor $\mathcal{B}$ exist; e.g., if $A=\Omega-B$. $\endgroup$ – Arturo Magidin Feb 6 '12 at 19:32
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    $\begingroup$ I'm not sure I understand: if $A\neq \Omega\setminus B$, then $\mathcal A=\{\emptyset, A,\Omega\setminus A,\Omega\}$, $\mathcal B=\{\emptyset, B,\Omega\setminus B,\Omega\}$, and the smallest $\sigma$-field containing $\mathcal A\cup\mathcal B$ is finite. So taking $\mathcal F$ infinite we get a counter-example. $\endgroup$ – Davide Giraudo Feb 6 '12 at 19:33
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I'm not sure I understand: if $A\neq \Omega\setminus B$, then $\mathcal A=\{\emptyset,A,\Omega\setminus A,\Omega\}$, $\mathcal B=\{\emptyset,B,\Omega\setminus B,\Omega\}$, and the smallest $\sigma$-field containing $A\cup B$ is necessary finite. So taking $\mathcal F$ infinite we get a counter-example.

It's not even true when $\Omega$ is finite: take $\Omega:=\{a,b,c,d\}$, $\mathcal F=\mathcal P(\Omega)$, $ A=\{a\}$, $ B=\{b\}$. Then the smallest $\sigma$-algebra containing $\mathcal A\cup\mathcal B$ is $\{\emptyset,\{a\},\{b\},\{b,c,d\},\{a,c,d\},\{c,d\},\{a,b\},\Omega\}\neq \mathcal F$ since $\{c\}$ is not in the previous $\sigma$-field.

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    $\begingroup$ +1. One can even get counterexamples with $\Omega$ of size $3$ and $A$ and $B$ nonempty, and with $\Omega$ of size $2$ if empty sets are allowed... $\endgroup$ – Did Feb 6 '12 at 20:30

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