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Math Help - Choosing a sigma algebra such that a function is measurable

  1. #1
    Member Last_Singularity's Avatar
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    Choosing a sigma algebra such that a function is measurable

    I am using the following definition of "measurable." Suppose we have a set \Omega with an associated sigma-algebra F of subsets. A function f: \Omega \rightarrow R is measurable w.r.t. F if: for all c \in R, the set \{\omega \in \Omega | f(\omega) < c\} = f^{-1}((-\infty,c)) belongs to F.

    So I choose \Omega = R (the reals) and let the function f be the identity function f(x) = x. I hope that this innocent bijective function is measurable. And for all c, we would have: f^{-1}((-\infty,c)) = (-\infty,c).

    But "measurable" is defined in the context of "with respect to" a sigma-algebra, right? So this requires me to choose a sigma-algebra, no?

    Would choosing the trivial sigma-algebra F= \{\Omega,\emptyset\} = \{R,\emptyset\} work?

    What other possible sigma-algebras F of \Omega could we choose to keep f measurable w.r.t. F?

    Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Last_Singularity View Post
    Would choosing the trivial sigma-algebra F= \{\Omega,\emptyset\} = \{R,\emptyset\} work?

    It does not work because

    f^{-1}(-\infty,c)=}(-\infty,c)\not\in F

    What other possible sigma-algebras F of \Omega could we choose to keep f measurable w.r.t. F?

    Any sigma-algebra which contains to the sigma-algebra generated by the family:

    S=\{(-\infty,c):c\in\mathbb{R}\}



    Fernando Revilla
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  3. #3
    Member Last_Singularity's Avatar
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    Doesn't a sigma-algebra need to be closed under complements? As in, if F is a sigma-algebra and A \in F, then A^C = \Omega - A \in F.

    In this case, if I let c be given, A=(-\infty,c_A) creates A^C = R  - (-\infty,c_A) = [c_A,\infty), which is not in F = \{(-\infty,c): c \in R\}, right?

    I must be mis-understanding the definition...
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  4. #4
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    Quote Originally Posted by Last_Singularity View Post
    Doesn't a sigma-algebra need to be closed under complements? As in, if F is a sigma-algebra and A \in F, then A^C = \Omega - A \in F.

    In this case, if I let c be given, A=(-\infty,c_A) creates A^C = R  - (-\infty,c_A) = [c_A,\infty), which is not in F = \{(-\infty,c): c \in R\}, right?

    I must be mis-understanding the definition...
    The \sigma-algebra generated by a family \mathcal{F} is defined as the smallest \sigma-algebra which contains \mathcal{F}.
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