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Math Help - Measurable function in intersection set

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    Measurable function in intersection set

    If X = A \cup B where  A,B \in \mathbb {M} , the  \sigma -Algebra of X, then a function f on X is measurable iff f is measurable on A and on B.

    Proof so far.

    Assume that f:X \rightarrow Y , where  \mathbb {M} , \mathbb {N} are the set of all  \sigma -Algebra of X and Y, respectively.

    First, assume that f is measurable on X, that is, I have f^{-1}(E) \in \mathbb {M} \ \ \ \ \forall E \in \mathbb {N}

    Consider the function f:A \rightarrow Y , I need to show that  f^{-1}(E) \in \mathbb {M}_A \ \ \ \forall E \in \mathbb {N} , where  \mathbb {M}_A is the a  \sigma -Algebra of A.

    I know that has to be true, but I'm just having a hard time trying to write up the sigma algebra of A in X, any hints?

    Thank you.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    If X = A \cup B where  A,B \in \mathbb {M} , the  \sigma -Algebra of X, then a function f on X is measurable iff f is measurable on A and on B.

    Proof so far.

    Assume that f:X \rightarrow Y , where  \mathbb {M} , \mathbb {N} are the set of all  \sigma -Algebra of X and Y, respectively.

    First, assume that f is measurable on X, that is, I have f^{-1}(E) \in \mathbb {M} \ \ \ \ \forall E \in \mathbb {N}

    Consider the function f:A \rightarrow Y , I need to show that  f^{-1}(E) \in \mathbb {M}_A \ \ \ \forall E \in \mathbb {N} , where  \mathbb {M}_A is the a  \sigma -Algebra of A.

    I know that has to be true, but I'm just having a hard time trying to write up the sigma algebra of A in X, any hints?
    If f is measurable on A then f^{-1}(E)\cap A \in \mathbb{M}. If f is measurable on B then f^{-1}(E)\cap B \in \mathbb{M}. It then follows that f^{-1}(E) = f^{-1}(E)\cap (A\cup B) = (f^{-1}(E)\cap A)\cup(f^{-1}(E)\cap B) \in \mathbb{M}.
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