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Thread: 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 $\displaystyle \Omega$ with an associated sigma-algebra $\displaystyle F$ of subsets. A function $\displaystyle f: \Omega \rightarrow R$ is measurable w.r.t. $\displaystyle F$ if: for all $\displaystyle c \in R$, the set $\displaystyle \{\omega \in \Omega | f(\omega) < c\} = f^{-1}((-\infty,c))$ belongs to $\displaystyle F$.

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

    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 $\displaystyle F= \{\Omega,\emptyset\} = \{R,\emptyset\}$ work?

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

    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 $\displaystyle F= \{\Omega,\emptyset\} = \{R,\emptyset\}$ work?

    It does not work because

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

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

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

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



    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 $\displaystyle F$ is a sigma-algebra and $\displaystyle A \in F$, then $\displaystyle A^C = \Omega - A \in F$.

    In this case, if I let $\displaystyle c$ be given, $\displaystyle A=(-\infty,c_A)$ creates $\displaystyle A^C = R - (-\infty,c_A) = [c_A,\infty)$, which is not in $\displaystyle F = \{(-\infty,c): c \in R\}$, 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 $\displaystyle F$ is a sigma-algebra and $\displaystyle A \in F$, then $\displaystyle A^C = \Omega - A \in F$.

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

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