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Math Help - Measurable Function and Sequence of Real Numbers

  1. #1
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    Measurable Function and Sequence of Real Numbers

    Let E be a Lebesgue measurable subset of \mathbb{R} with m(E) < \infty and let (f_n) be a sequence of real-valued Lebesgue measurable functions on E. Prove or disprove that there exists a sequence (\alpha_n) of positive real numbers such that lim_{n \to \infty} \alpha_n f_n = 0 almost everywhere on E.

    I suppose the statement is true? So far can't think of any counterexample. If it's true, how do I go about starting the proof?

    Thanks in advance.
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  2. #2
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    Re: Measurable Function and Sequence of Real Numbers

    Quote Originally Posted by Markeur View Post
    Let E be a Lebesgue measurable subset of \mathbb{R} with m(E) < \infty and let (f_n) be a sequence of real-valued Lebesgue measurable functions on E. Prove or disprove that there exists a sequence (\alpha_n) of positive real numbers such that lim_{n \to \infty} \alpha_n f_n = 0 almost everywhere on E.

    I suppose the statement is true? So far can't think of any counterexample. If it's true, how do I go about starting the proof?

    Thanks in advance.
    For fixed n, the union of the sets S_k = \{x\in E:|f_n(x)|>k\}\ \ (k\in\mathbb{N}) is the whole of E. So there exists K such that m(S_K)>m(E)-2^{-n}. Let \alpha_n = 1/(2^nK). Then |\alpha_nf_n(x)|<2^{-n} except on a subset of E of measure less than 2^{-n}.

    If you now let n vary, you should be able to show that given N, there is a subset E_N of E with measure less than 2^{-N+1} with the property that if x\in E\setminus E_N then |\alpha_nf_n(x)|<2^{-n} for all n\geqslant N. It follows that \alpha_nf_n(x)\to0 except on the null set \textstyle\bigcap_N E_N.
    Last edited by Opalg; September 27th 2011 at 11:16 AM.
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    Re: Measurable Function and Sequence of Real Numbers

    Hey thanks very much! Anyway I would like to know why we need to establish S_k = {x \in E : |f_n(x)| > k}. Is it because (f_n) is a sequence of real-valued Lebesgue measurable functions without any other condition given?

    Thanks in advance.
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  4. #4
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    Re: Measurable Function and Sequence of Real Numbers

    Quote Originally Posted by Markeur View Post
    Hey thanks very much! Anyway I would like to know why we need to establish S_k = {x \in E : |f_n(x)| > k}. Is it because (f_n) is a sequence of real-valued Lebesgue measurable functions without any other condition given?
    If you knew that the functions were bounded then the result would be easy to prove. But a measurable function, even on a set of finite measure, need not be bounded. So the idea is to use the measurability condition to find a "large" subset of E on which f_n is bounded. But the details of the proof are quite delicate, and I have had to edit my previous comment a couple of times to get an argument that (I think!) works.
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