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Math Help - measure theory question

  1. #1
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    measure theory question

    A* is a σ-algebra on a set Ω*, and let f be a function from a set Ω to Ω*. prove that A={f^(-1)(E*):E* is an element of A*} is a σ-algebra on Ω. (σ-algebra is said to be generated by f).

    i'm not sure how to do this question, can somebody help me?
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  2. #2
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    Given that

    A^* is a \sigma- algebra on \Omega^*

    it follows that

    {\Omega^*}\in{A^*}

    If f^{-1} exists, it follows that f is bijective.

    Thus f^{-1}(\Omega^*)=\Omega

    Therefore \Omega\in{A}.

    Similarly, given that

    \forall{E^*}\in{A^*},{({E^*})^c}\in{A^*}

    it follows that

    f^{-1}{({E^*})^c}\in{A}

    Then you have left only to show:

    (iii) \forall infinite sets {(A_i)}\in{A},   \cup{A_i}\in{A}

    (iv) \forall infinite sets {(A_i)}\in{A},   \cap{A_i}\in{A}


    Can I ask which book your university is using for Measure Theory?
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  3. #3
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    TXGirl, the use of notation, f^{ - 1}, is not the inverse of a function.
    It is confusing notation. That is exactly why many of us prefer a different notation.
    f:A \mapsto B\quad  \wedge C \subseteq B\quad  \Rightarrow \quad \overleftarrow f (C) = f^{ - 1} [C] = \left\{ {x \in A:f(x) \in C} \right\}.
    In other words, we now have a mapping \overleftarrow f :P(B) \mapsto P(A)\,\& \,\overleftarrow f (B) = A.
    The second property comes from the way mappings are defined.
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  4. #4
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    Quote Originally Posted by Plato View Post
    TXGirl, the use of notation, f^{ - 1}, is not the inverse of a function.
    It is confusing notation. That is exactly why many of us prefer a different notation.
    f:A \mapsto B\quad  \wedge C \subseteq B\quad  \Rightarrow \quad \overleftarrow f (C) = f^{ - 1} [C] = \left\{ {x \in A:f(x) \in C} \right\}.
    In other words, we now have a mapping \overleftarrow f :P(B) \mapsto P(A)\,\& \,\overleftarrow f (B) = A.
    The second property comes from the way mappings are defined.

    Ok, so the goal then seems to be to show that

    {A}:= \left\{ ({x}\in{\Omega}:f(x)\in{E^*}):{E^*}\in{A^*} \right\} is a \sigma-algebra on \Omega

    So your first step is:

    (i) Show \Omega\in{A}

    You have that f(\Omega)\subseteq {\Omega^*}\in{A^*} the latter inclusion following from the fact that {A^*} is a \sigma-algebra on {\Omega^*}

    Thus it follows that {\Omega}\in{A}

    (ii) Show \forall{X}\in{A}, {X^c}\in{A}

    If X \in A \quad  \Rightarrow \quad f(X)\in{A^*}
    \quad  \Rightarrow \quad {(f(X))^c}\in{A^*}
    \quad  \Rightarrow \quad {f(X^c)}\in{A^*}
    \quad  \Rightarrow \quad {X^c}\in{A}

    (iii) Show then that \forall infinite sets {(A_i)}\in{A},   \cup{A_i}\in{A}

    (iv) Finally show that \forall infinite sets {(A_i)}\in{A},   \cap{A_i}\in{A}

    **Note: {X^c} denotes the complement of {X} in the relevant space.
    Last edited by TXGirl; May 17th 2008 at 10:04 AM.
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  5. #5
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    Here is the first part: \Omega ^*  \in A^*  \Rightarrow \quad \Omega  = \overleftarrow f \left( {\Omega ^* } \right) \in A.

    If we have \overleftarrow f \left( {E^* } \right) \in A \Rightarrow \quad \Omega \backslash \overleftarrow f \left( {E^* } \right) = \overleftarrow f \left( {\Omega ^* \backslash E^* } \right) \in A.

    If we have a countable collection \left\{ {\overleftarrow f \left( {E_\alpha ^* } \right)} \right\} \Rightarrow \quad \overleftarrow f \left( { \cup E_\alpha ^* } \right) =  \cup \left( {\overleftarrow f \left( {E_\alpha ^* } \right)} \right) \in A.
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  6. #6
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    one question, f with a backward dash means f^(-1)?
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  7. #7
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    ahh sorry just saw it. many thanks
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  8. #8
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    Quote Originally Posted by TXGirl View Post
    Ok, so the goal then seems to be to show that

    {A}:= \left\{ ({x}\in{\Omega}:f(x)\in{E^*}):{E^*}\in{A^*} \right\} is a \sigma-algebra on \Omega

    So your first step is:

    (i) Show \Omega\in{A}

    You have that f(\Omega)\subseteq {\Omega^*}\in{A^*} the latter inclusion following from the fact that {A^*} is a \sigma-algebra on {\Omega^*}

    Thus it follows that {\Omega}\in{A}

    (ii) Show \forall{X}\in{A}, {X^c}\in{A}

    If X \in A \quad  \Rightarrow \quad f(X)\in{A^*}
    \quad  \Rightarrow \quad {(f(X))^c}\in{A^*}
    \quad  \Rightarrow \quad {f(X^c)}\in{A^*}
    \quad  \Rightarrow \quad {X^c}\in{A}

    (iii) Show then that \forall infinite sets {(A_i)}\in{A},   \cup{A_i}\in{A}

    (iv) Finally show that \forall infinite sets {(A_i)}\in{A},   \cap{A_i}\in{A}

    **Note: {X^c} denotes the complement of {X} in the relevant space.

    hi, how would you do part (iii). would you please help?
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