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Thread: 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

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

    it follows that

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

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

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

    Therefore $\displaystyle \Omega\in{A}$.

    Similarly, given that

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

    it follows that

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

    Then you have left only to show:

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

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


    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, $\displaystyle f^{ - 1}$, is not the inverse of a function.
    It is confusing notation. That is exactly why many of us prefer a different notation.
    $\displaystyle f:A \mapsto B\quad \wedge C \subseteq B\quad \Rightarrow \quad \overleftarrow f (C) = f^{ - 1} [C] = \left\{ {x \in A:f(x) \in C} \right\}$.
    In other words, we now have a mapping $\displaystyle \overleftarrow f :P(B) \mapsto P(A)\,\& \,\overleftarrow f (B) = A$.
    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, $\displaystyle f^{ - 1}$, is not the inverse of a function.
    It is confusing notation. That is exactly why many of us prefer a different notation.
    $\displaystyle f:A \mapsto B\quad \wedge C \subseteq B\quad \Rightarrow \quad \overleftarrow f (C) = f^{ - 1} [C] = \left\{ {x \in A:f(x) \in C} \right\}$.
    In other words, we now have a mapping $\displaystyle \overleftarrow f :P(B) \mapsto P(A)\,\& \,\overleftarrow f (B) = A$.
    The second property comes from the way mappings are defined.

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

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

    So your first step is:

    (i) Show $\displaystyle \Omega\in{A}$

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

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

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

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

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

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

    **Note: $\displaystyle {X^c}$ denotes the complement of $\displaystyle {X}$ 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: $\displaystyle \Omega ^* \in A^* \Rightarrow \quad \Omega = \overleftarrow f \left( {\Omega ^* } \right) \in A$.

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

    If we have a countable collection $\displaystyle \left\{ {\overleftarrow f \left( {E_\alpha ^* } \right)} \right\} \Rightarrow \quad \overleftarrow f \left( { \cup E_\alpha ^* } \right) = \cup \left( {\overleftarrow f \left( {E_\alpha ^* } \right)} \right) \in A$.
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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

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

    So your first step is:

    (i) Show $\displaystyle \Omega\in{A}$

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

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

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

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

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

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

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

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