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Math Help - Is a countable union/intersection of sigma-algebra also a sigma-algebra?

  1. #1
    Member Last_Singularity's Avatar
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    Is a countable union/intersection of sigma-algebra also a sigma-algebra?

    Suppose that I have a set \Omega over which I define a collection of \sigma-algebra: \{\beta_i\}; that is, each \beta_1,\beta_2,... is a \sigma-algebra of \Omega. Is the countable union of those also a \sigma-algebra? That is, is \bigcup_{i \in N} \beta_i = \beta^* a \sigma-algebra over \Omega? How about countable intersection?

    1 - It seems pretty clear that \Omega and \emptyset are contained in \beta^*.

    2 - It is also clear that if A \in \beta^*, then A^C \in \beta^*

    So the first two conditions of a \sigma-algebra have been fulfilled: \beta^* contains the full sample space / empty set and it is closed under complements.

    3 - But if A_1,A_2,... \in \beta^*, is it true that A^* = \bigcup_{i \in N} A_i \in \beta^*? I am unable to prove this fact. Yet I cannot find a counterexample.

    A counterexample or hint of proof would be very appreciated - thanks.
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  2. #2
    Senior Member roninpro's Avatar
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    For a counterexample, take \Omega=\{1,2,3,4,5,6\}, \beta_1=\{\emptyset, \{1\}, \{2,3,4,5,6\}, \Omega\}, and \beta_2=\{\emptyset, \{2\}, \{1,3,4,5,6\},\Omega\}.

    The set \beta_1\bigcup \beta_2 is not a \sigma-algebra because \{1\}\bigcup \{2\}=\{1,2\} is not in \beta_1\bigcup \beta_2.

    I think that you will have better luck with intersections.
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