It seems to be obvious, but how to give a formal proof?
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3The power set $2^\Omega$ is such a $\sigma$-field... – Nate Eldredge May 01 '12 at 23:21
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3...and in case you look for the smallest one containing $A$, show that an arbitrary intersection of $\sigma$-fields is a $\sigma$-field. – t.b. May 01 '12 at 23:23
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@t.b.: Beat me to it! – Brian M. Scott May 01 '12 at 23:23
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See also this thread for a lengthy discussion. – t.b. May 01 '12 at 23:33
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The usual assertion is "there is a smallest $\sigma$-field containing $A$".
For this, you show two things:
1) that an arbitrary intersection of $\sigma$-fields is a $\sigma$-field;
2) that there exists a $\sigma$-field containing $A$ (for this you can take the power set of $\Omega$).
The two assertions together guarantee that the intersection of all $\sigma$-fields containing $A$ is non-empty. It is easy to verify that this has to be the smallest such set.
This reasoning works for many different algebraic structures (because one can verify 1) and 2)). It is usually called "the $\sigma$-field (or group, or ring, or algebra, or field, or vector space, etc., etc.) generated by the set".
Martin Argerami
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