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Math Help - Measurable Functions

  1. #1
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    Measurable Functions

    Let (f_n) be a sequence of real-valued measurable functions on [a,b]. If (f_n) converges to a function f almost everywhere on [a,b], prove that there exists a sequence (E_n) of measurable subsets of [a,b] such that
    m([a,b] \setminus \cup E_n) = 0
    and (f_n) converges uniformly to f on each E_n.

    How do I even start this question?

    Thanks in advance.
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  2. #2
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    Re: Measurable Functions

    If (f_n) converges almost everywhere to f, then there is a zero set of points on which it does not converge. A zero set has certain properties. For one thing, it has measure zero. So, if \[a,b\] \setminus \cup E_n = Z\text{ where }Z\text{ is your zero set}, then you are guaranteed that the measure will be zero. So, let's construct E_1. Since the zero set is a subset of \[a,b\], by the well ordering principle of real numbers, there exists a greatest lower bound of Z. Can you find a really small interval E_1 that will include a, but not a_1? Will that interval be measurable? How do you know? Will (f_n) converge to f uniformly on that interval? Again, how do you know?
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  3. #3
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    Re: Measurable Functions

    Is a_1 a point in the set Z or just part of the partition of [0,1]?
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  4. #4
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    Re: Measurable Functions

    It is a point in the set Z.
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  5. #5
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    Re: Measurable Functions

    Alright I managed to solve it. It's actually applying Egorov's Theorem to each E_n and intersect [a,b]\E_n for all n to get m([a,b]\(\cup E_n)) = 0 and f_n converges uniformly to f on each E_n.

    Thanks.
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