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Math Help - Little question - measure theory

  1. #1
    Member mabruka's Avatar
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    Little question - measure theory

    I had this little thought bothering me:

    Let (X,\mathcal A,m) be a finite measure space , two functions f,g, and \mathcal B a sub \sigma - algebra.

    If f=g a.e. and g is \mathcal B-measurable,

    is f \mathcal B-measurable too?

    Im thinking yes but im not sure how to prove it.


    Any ideas?

    thnx
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  2. #2
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    Quote Originally Posted by mabruka View Post
    I had this little thought bothering me:

    Let (X,\mathcal A,m) be a finite measure space , two functions f,g, and \mathcal B a sub \sigma - algebra.

    If f=g a.e. and g is \mathcal B-measurable,

    is f \mathcal B-measurable too?

    Im thinking yes but im not sure how to prove it.


    Any ideas?

    thnx
    Yes it is

    Let O be an open set

     g^{-1}(O) = {  x ;  g(x) \in O } = [{  x ;  f(x) \in O } \ {  x ;  f(x) \neq g(x) }]  \cup {  x ;  g(x) \in O &  f(x) \neq g(x) }

    But everyone of these sets is measurable (see why?), so then  g^{-1}(O) is too. Thus g is measurable.
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  3. #3
    Member mabruka's Avatar
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    You reversed the roles of f and g but it is ok, lets do it your way .

    i was considering that decomposition of the inverse mapping but got stuck in the following part.


    Lets call E the exceptional set where f and g differ , m(E)=0 and E=\{x: f(x)\not =g(x)\}

    First question:

    The left side of f^{-1}(O) is

    f^{-1}(O)\cap E^c =g^{-1}(O)\cap E which is obviously in \mathcal A but i am not sure why it must be in \mathcal B since \mathcal B \subset \mathcal A

    The right side

    f^{-1}(O)\cap E \subset E ... and thus
    m(f^{-1}(O)\cap E \subset E)=0 but i dont know what to do next..


    thank you southprkfan1
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  4. #4
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    Quote Originally Posted by mabruka View Post
    You reversed the roles of f and g but it is ok, lets do it your way .

    i was considering that decomposition of the inverse mapping but got stuck in the following part.


    Lets call E the exceptional set where f and g differ , m(E)=0 and E=\{x: f(x)\not =g(x)\}

    First question:

    The left side of f^{-1}(O) is

    f^{-1}(O)\cap E^c =g^{-1}(O)\cap E which is obviously in \mathcal A but i am not sure why it must be in \mathcal B since \mathcal B \subset \mathcal A

    The right side

    f^{-1}(O)\cap E \subset E ... and thus
    m(f^{-1}(O)\cap E \subset E)=0 but i dont know what to do next..


    thank you southprkfan1
    My Bad. I made 2 mistakes. That one, and more importantly, I didn't notice you had 2 different - algebras.

    ... Let me get back to you, sorry
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  5. #5
    Member mabruka's Avatar
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    Well it turns out that even when considering only one sigma algebra like you did the statement may not be true if the measure space (X,\mathcal F, m) is not complete in the sense that


    (X,\mathcal F, m) is complete if the all the null sets are measurable.
    \mathcal N = \{A\subset X: \exists B\in \mathcal F \text{ such that } A\subset B, m(B)=0 \}\subset \mathcal F


    So well, lets add the " (X,\mathcal F, m) complete" to the hypothesis.

    The case when we have only one sigma algebra is true.
    The other one remains yet unclear.


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  6. #6
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    Quote Originally Posted by mabruka View Post
    Well it turns out that even when considering only one sigma algebra like you did the statement may not be true if the measure space (X,\mathcal F, m) is not complete in the sense that


    (X,\mathcal F, m) is complete if the all the null sets are measurable.
    \mathcal N = \{A\subset X: \exists B\in \mathcal F \text{ such that } A\subset B, m(B)=0 \}\subset \mathcal F


    So well, lets add the " (X,\mathcal F, m) complete" to the hypothesis.

    The case when we have only one sigma algebra is true.
    The other one remains yet unclear.


    I'll be honest, I'm completely stuck .

    I had never of a measure space being complete. I was taught that all zero sets are measurable, period. Very interesting.
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  7. #7
    Member mabruka's Avatar
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    Well the completeness is just a technical aspect that most of the time is overlooked because if you have a measure m over \mathcal F_1 which is not complete you can always extend it to a measure M over a bigger sigma algebra \mathcal F_2

    satisfying that M|_{\mathcal F_1} =m


    This is done using Carathedory's Extension theorem. In fact if the initial measure is finite then the extension is unique.
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