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Math Help - a special measurable function

  1. #1
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    a special measurable function

    Hi,

    I have a bdd Lebesgue measurable function h on \mathbb{R} such that h(x)=h(x+1) a.e. on \mathbb{R}. Define h_k(x)=h(kx).

    How do I prove the following:

    (i) \lim_{k\rightarrow\infty}\int_a^bh_k(x)dx=(b-a)\int_0^1h(x)dx

    (ii) If h isn't constant a.e. then there will be no subseq that will converge m-a.e. on \[a,b\].
    -----------------------------------------------------------
    My attempts:

    (i) For this one I'm thinking that the condition on h implies that h is constant a.e., say h=C a.e. so
    \lim_{k\rightarrow\infty}\int_a^bh_k(x)dx=C\int_a^  b1dx=m((a,b))C=(b-a)\int_0^1h(x)dx.
    But if that is true... the the \lim_k is of no use.

    (ii) Here I'm thinking a contradiction, i.e. assuming that there is such a sub.seq. but I get it to work...
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  2. #2
    Super Member girdav's Avatar
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    Re: a special measurable function

    What happens if h(x)=\sin (2\pi x)?
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  3. #3
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    Re: a special measurable function

    Quote Originally Posted by mgarson View Post
    Hi,

    I have a bdd Lebesgue measurable function h on \mathbb{R} such that h(x)=h(x+1) a.e. on \mathbb{R}. Define h_k(x)=h(kx).

    How do I prove the following:

    (i) \lim_{k\rightarrow\infty}\int_a^bh_k(x)dx=(b-a)\int_0^1h(x)dx

    (ii) If h isn't constant a.e. then there will be no subseq that will converge m-a.e. on \[a,b\].
    -----------------------------------------------------------
    My attempts:

    (i) For this one I'm thinking that the condition on h implies that h is constant a.e., say h=C a.e.
    The condition on h says that it is periodic, with period 1. That certainly does not imply that h is constant a.e. For example, you could have h(x) = \sin(2\pi x).

    To prove (i), you could start by using the periodicity of h to show that the integral of h over any interval of length 1 is equal to the integral from 0 to 1. Then to find \int_a^bh_k(x)\,dx, divide the interval [a,b] into subintervals of length 1/k and make the substitution y=kx to show that \int h(kx)\,dx over each of these subintervals is equal to \frac1k\int_0^1h(x)dx. The only snag is that ba may not be an exact multiple of 1/k, so there may be a little subinterval left over at the end. As k\to\infty, the contribution of the integral over that subinterval will become negligible.

    For (ii), I would start as you suggest, suppose that there is a subsequence of the functions h_k that converges a.e. on [a,b] to a limit function f. Use one of the Lebesgue convergence theorems, together with the result of (i), to deduce that for each u\in(a,b], \int_a^uf(x)\,dx = (u-a)\int_0^1h(x)dx. Then differentiate both sides to deduce that f is constant a.e. You still have to deduce that this implies that h is constant a.e., and I must admit I haven't thought why that follows. (But I think it should.)
    Last edited by Opalg; September 4th 2011 at 11:44 PM. Reason: corrected error
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  4. #4
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    Re: a special measurable function

    Ok, so knowing that f=C, where C is some constant, should imply that f is constatnt a.e.
    So, \int_0^1h(x)dx=C which then gives me h_k-C=0 a.e. and therefore also h-C=0 a.e.

    All I'm a bit concerned about is the follwoing equality:
    Quote Originally Posted by Opalg View Post
    \int_a^uf(x)\,dx = (u-a)\int_0^1h(x)dx.
    what were using here is (i), i.e \lim_{n\rightarrow\infty}\int_a^uh_{k_n}(x)dx=(u-a)\int_0^1h(x)dx. But if the same proof as in (i) is to hold we need to have k_n\rightarrow \infty as n\rightarrow\infty. Does this really need to be true? What if k_n\rightarrow A<\infty as n\rightarrow\infty?
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