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Math Help - Prove measurability

  1. #1
    Junior Member TheProphet's Avatar
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    Prove measurability

    Let  f = (f_1, f_2): \Omega \to E \times F .
    Show that  f : (\Omega,\mathcal{A}) \to (E \times F, \mathcal{E} \otimes \mathcal{F}) is measurable if and only if f_1 is measurable from  (\Omega,\mathcal{A}) to  (E, \mathcal{E}) and f_2 is measurable from (\Omega,\mathcal{A}) to (F,\mathcal{F}).

    Attempted solution.

    " \Rightarrow"
     \mathcal{E} \times \mathcal{F} \subset \mathcal{E} \otimes \mathcal{F} . Set  \mathcal{B} = \{ C \in \mathcal{E} \otimes \mathcal{F} : C = \Lambda \times F, \Lambda \in \mathcal{E} \} . Then  \mathcal{B} \subset \mathcal{E} \otimes \mathcal{F} , so  f^{-1}(\mathcal{B}) \subset \mathcal{A} . But  f^{-1}(\Lambda \times F) = \{ \omega \in \Omega : f_{1}(\omega) \in \Lambda \} . Similarily,  f^{-1}(E \times \Gamma) = \{ \omega \in \Omega : f_{2}(\omega) \in \Gamma\} .

    "  \Leftarrow "

     \{ \omega : f \in A \times B, A \in \mathcal{E}, B \in \mathcal{F} \} =
     \{ \omega : f_{1}(\omega) \in A, f_{2}(\omega) \in B, A \in \mathcal{E}, B \in \mathcal{F} \} , which is in  \mathcal{A} .
    Since f^{-1}(\mathcal{E} \times \mathcal{F}) \subset \mathcal{A} and  \sigma(\mathcal{E} \times \mathcal{F}) = \mathcal{E} \otimes \mathcal{F} , it follows that  f^{-1}(\mathcal{E} \otimes \mathcal{F}) \subset \mathcal{A} .

    Appreciate any input/corrections!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Prove measurability

    Quote Originally Posted by TheProphet View Post
    Let  f = (f_1, f_2): \Omega \to E \times F .
    Show that  f : (\Omega,\mathcal{A}) \to (E \times F, \mathcal{E} \otimes \mathcal{F}) is measurable if and only if f_1 is measurable from  (\Omega,\mathcal{A}) to  (E, \mathcal{E}) and f_2 is measurable from (\Omega,\mathcal{A}) to (F,\mathcal{F}).

    Attempted solution.

    " \Rightarrow"
     \mathcal{E} \times \mathcal{F} \subset \mathcal{E} \otimes \mathcal{F} . Set  \mathcal{B} = \{ C \in \mathcal{E} \otimes \mathcal{F} : C = \Lambda \times F, \Lambda \in \mathcal{E} \} . Then  \mathcal{B} \subset \mathcal{E} \otimes \mathcal{F} , so  f^{-1}(\mathcal{B}) \subset \mathcal{A} . But  f^{-1}(\Lambda \times F) = \{ \omega \in \Omega : f_{1}(\omega) \in \Lambda \} . Similarily,  f^{-1}(E \times \Gamma) = \{ \omega \in \Omega : f_{2}(\omega) \in \Gamma\} .

    "  \Leftarrow "

     \{ \omega : f \in A \times B, A \in \mathcal{E}, B \in \mathcal{F} \} =
     \{ \omega : f_{1}(\omega) \in A, f_{2}(\omega) \in B, A \in \mathcal{E}, B \in \mathcal{F} \} , which is in  \mathcal{A} .
    Since f^{-1}(\mathcal{E} \times \mathcal{F}) \subset \mathcal{A} and  \sigma(\mathcal{E} \times \mathcal{F}) = \mathcal{E} \otimes \mathcal{F} , it follows that  f^{-1}(\mathcal{E} \otimes \mathcal{F}) \subset \mathcal{A} .

    Appreciate any input/corrections!
    Whoah, overusage of fancy letters. What is the tensor product? Just the usual product measure?
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  3. #3
    Junior Member TheProphet's Avatar
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    Re: Prove measurability

    Quote Originally Posted by Drexel28 View Post
    Whoah, overusage of fancy letters. What is the tensor product? Just the usual product measure?
    This was not my intention, I used the same the letters my book uses =)

    Yeah, its the usual product measure.  \mathcal{E} \otimes \mathcal{F} is the \sigma-algebra generate by  \mathcal{E} \times \mathcal{F} ,
    where  \mathcal{E} and  \mathcal{F} are  \sigma-algebras on  E and  F recpectively.
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