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Math Help - lebesgue measurable function

  1. #1
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    lebesgue measurable function

    Q:
    Prove that if  f : \mathbb{R} \longrightarrow [o,\infty] and  f^{-1}((r,\infty]) \in M for each  r \in \mathbb{Q} , then  f is lebesgue measurable. (  M is the  \sigma -algebra of Lebesgue measurable sets)

    Attempt:
    I need to show that for any set  A \in \mathcal{B}_{[0,\infty]} ,  f^{-1}(A) \in M (is this correct?). So I take a set  A \in \mathcal{B}_{[0,\infty]} . By definition of \mathcal{B}_{[0,\infty]} ,  A is an open interval (is this correct?). Now, i want to write  A as a union of sets of the form  (r,\infty] for each  r \in \mathbb{Q} but im sturggling to do this! Is this the right method? Can you help me? Thank you
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  2. #2
    Moo
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    Hello,

    I need to show that for any set A \in \mathcal{B}_{[0,\infty]}, f^{-1}(A) \in M (is this correct?).
    I think so, because the Lebesgue measure is a measure over the Borelian sigma-algebra of \mathbb{R}.

    So I take a set A \in \mathcal{B}_{[0,\infty]}. By definition of \mathcal{B}_{[0,\infty]}, A is an open interval (is this correct?).
    No, we just know that \mathcal{B}_{[0,\infty)} is generated by the open intervals in [0,\infty)
    In other words, it's generated by \{(a,\infty) \mid a\in \mathbb{R}_+\}
    But an element of this will not necessarily be an open set.


    It may be useful to use the fact that \mathbb{Q} is dense in \mathbb{R} (same goes for the positive numbers only).
    Thus, \forall a \in\mathbb{R}_+, there exists a (decreasing) sequence (a_n)_{n\geq 0}, where \forall n \in\mathbb{N} ~,~ a_n \in \mathbb{Q}_+ such that \lim_{n\uparrow \infty} a_n=a
    And we can write : (a,\infty)=\bigcup_{n\geq 0} [a_n,\infty)

    But one can prove that f^{-1}\left(\bigcup_{j\in J} A_j\right)=\bigcup_{j\in J} f^{-1}(A_j) (where J is a set without any restriction on it). It's one of Haussdorff formulas.

    So you'll have f^{-1}((a,\infty))\in M, \forall a\in \mathbb{R}_+

    --------------------
    Now the second part (to finish) isn't easier imo.

    There's a lemma that we called "Transport lemma", but I didn't find an English page of it... Here is the French one : Lemme de transport - Wikipédia
    ******transcription******
    X,Y 2 sets. f : X -> Y. \mathcal{E}\subset \mathcal{P}(Y) ( = the power set of Y = the family of all subsets of Y).
    Then f^{-1}(\sigma(\mathcal{E}))=\sigma(f^{-1}(\mathcal{E}))

    Here we have \mathcal{E}=\{(a,\infty) \mid a\in\mathbb{R}_+\}

    But we know that f^{-1}(\mathcal{E})\in M
    Hence, since M is a sigma-algebra, we have the (red) inclusion f^{-1}(\sigma(\mathcal{E}))=\sigma(f^{-1}(\mathcal{E}))~{\color{red}\subset}~ M

    And thus for any element A in \sigma(\mathcal{E})=\mathcal{B}_{[0,\infty)}, f^{-1}(A)\in M



    Whew... Does it look clear ? And I'm pretty sure there's a more elegant way to prove it, but I'm not an expert...
    Well, you have a method, and maybe it will give you guidelines for building your own proof...
    (as you can see, I'm very self-confident ^^)
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