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Math Help - Measurable Functions (2)

  1. #1
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    Measurable Functions (2)

    Let E be a measurable subset of \mathbb{R} with m(E) < \infty and let f be a real-valued measurable function on E. Prove that for any \epsilon > 0, there exists a bounded measurable function g on E such that m(\{x \in E : f(x) \not= g(x)\}) < \epsilon.

    Again, how do I go about doing it? Thanks in advance.
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  2. #2
    Senior Member Tinyboss's Avatar
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    Re: Measurable Functions (2)

    Okay, so what's the main issue here? If you just let g:=f, then the measure of the set where they differ is certainly less than \epsilon. But of course, then g may not be bounded as required. So try defining g_n(x):=\mathrm{min}(f(x),n), and looking at what happens to m(\{x\mid f(x)\ne g_n(x)\}) as n goes to infinity.
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  3. #3
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    Re: Measurable Functions (2)

    Okay managed to solved it. Though I used g(x) = f(x) if x lies in A, g(x) = 0 otherwise.

    Thanks anyway.
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