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Math Help - almost everywhere

  1. #1
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    almost everywhere

    Let f be nonnegative measurable function. Show \int f = 0 implies f=0 almost everywhere.

    let E=\{x:f>0\}. Then E=\bigcup E_n where E_n=\{x:f>1/n \}.
    now 0= \int f \geq \int f1_{E_n} \geq \frac{1}{n}\int 1_{E_n} = \frac{1}{n} mE_n.
    for each n, 1/n \neq 0 so mE_n=0.

    Here is my question. i am not sure if i can say lim_{n \rightarrow \infty} mE_n = 0. can i say that?
    Last edited by PRLM; December 24th 2009 at 09:06 PM. Reason: fixing a sentence that didnt make sense.
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  2. #2
    Senior Member Shanks's Avatar
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    Of course.
    Notice that {E_n} is an increasing sequence converges to E.
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  3. #3
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    a.e

    i can see that for each n mE_n=0 and i guess it makes sense that lim_{n \rightarrow \infty} mE_n=0. but i dont know how to show that. would u get me started on it?
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  4. #4
    Senior Member Shanks's Avatar
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    Since the sequence \{E_n\}\text{ is an increasing sequence converging to }E,mE_n=0, \forall n.we obtain
    mE=lim_{n \to \infty} mE_n=lim_{n\to\infty}0=0
    that is what we want.
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