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Math Help - Borel measurable function

  1. #1
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    Post Borel measurable function

    Hi there,
    I have a question on real analysis, I will write my try and could you please help me to solve it.
    we need to prove that f(x)=X^2 ,Q(x)= tanx, and J(x)= -sqrt(x) are borel measurable by the formal definition {f<= b}is in Borel, for all b in R.
    Thanks,
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by koko2009 View Post
    Hi there,
    I have a question on real analysis, I will write my try and could you please help me to solve it.
    we need to prove that f(x)=X^2 ,Q(x)= tanx, and J(x)= -sqrt(x) are borel measurable by the formal definition {f<= b}is in Borel, for all b in R.
    Thanks,
    What do you mean?
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  3. #3
    Moo
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    Quote Originally Posted by Drexel28 View Post
    What do you mean?
    What do you know about measure theory ?

    Any function f is Borel measurable if (with your definition) for any b in R, f^{-1}(]-\infty,b]) is a Borel set.

    In a more general way, since ]-\infty,b] is a closed set, it is sufficient for f to be continuous. Thus the pre-image of this set will be a Borel set, since the pre-image of a closed set under a continuous function is a closed set. (and a closed set is a Borel set)
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  4. #4
    Member mabruka's Avatar
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    Hi there,
    I have a question on real analysis, I will write my try and could you please help me to solve it.
    we need to prove that f(x)=X^2 ,Q(x)= tanx, and J(x)= -sqrt(x) are borel measurable by the formal definition {f<= b}is in Borel, for all b in R.
    Thanks,


    Fix b\in\mathbb R

    \{f\leq b\} =  \{x\in\mathbb R : f(x)\in((-\infty,b]\} = f^{-1}((-\infty,b])

    So as you pointed out, you need to see that f^{-1}((-\infty,b]) is a borel set.

    We have that f(x)=x^2 so

    f^{-1}((-\infty,b]) = \{x \in\mathbb R:-\infty<x^2\leq b \}<br />
    There are 3 cases:

    If b<0 then f^{-1}((-\infty,b]) =\emptyset

    If  b=0 then f^{-1}((-\infty,b]) =\{0\}

    If b\geq 0 then f^{-1}((-\infty,b]) =\{x: -\infty < x^2 \leq b \}=\{x: -\sqrt b  \leq x \leq\sqrt b \}=[-b,b]

    The first and last case obviously are borel sets (the empty set is in every \sigma-algebra, closed sets are borelian as well (their complement is open) ), what about the second one?

    Well, in \mathbb R with the usual metric, singletons are closed, so \{0\} is closed, therefore is borelian.

    Hence for every  b, \{f\leq b\} is in \mathcal B(\mathbb R) \Longrightarrow  f is measurable.


    You have to do this for every of your functions. Also as Moo already told you, if you already know (i bet you do) that the borel \sigma -algebra \mathcal B (\mathbb R) is the \sigma- algebra generated by all sets of the form :

    open (a,b)

    OR

    closed [a,b]

    OR

    psemi-open (a,b]

    OR
    psemi closed [a,b)


    (you can even use the above for rational end points still get the same sigma algebra)

    So, if \mathbb R,\mathcal B(\mathbb R) ) \longrightarrow (\mathbb R,\mathcal B(\mathbb R) )" alt=" h\mathbb R,\mathcal B(\mathbb R) ) \longrightarrow (\mathbb R,\mathcal B(\mathbb R) )" /> and you show that you have h^{-1}(K) is like any of the above type of sets, FOR EVERY K of any of the above, then h is measurable.
    Last edited by mabruka; March 11th 2010 at 03:38 PM.
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