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It seems to be obvious, but how to give a formal proof?

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The power set $2^\Omega$ is such a $\sigma$-field... –  Nate Eldredge May 1 '12 at 23:21
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...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 1 '12 at 23:23
    
@t.b.: Beat me to it! –  Brian M. Scott May 1 '12 at 23:23
    
See also this thread for a lengthy discussion. –  t.b. May 1 '12 at 23:33
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1 Answer

up vote 2 down vote accepted

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".

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