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

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

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

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

    3 - But if $\displaystyle A_1,A_2,... \in \beta^*$, is it true that $\displaystyle A^* = \bigcup_{i \in N} A_i \in \beta^*$? 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 $\displaystyle \Omega=\{1,2,3,4,5,6\}$, $\displaystyle \beta_1=\{\emptyset, \{1\}, \{2,3,4,5,6\}, \Omega\}$, and $\displaystyle \beta_2=\{\emptyset, \{2\}, \{1,3,4,5,6\},\Omega\}$.

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

    I think that you will have better luck with intersections.
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