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Math Help - evaluating a lebesgue integral

  1. #1
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    evaluating a lebesgue integral

    I'm stuck on a textbook exercise: given f is lbesgue integrable on \mathbb{R}^n, evalate

    $\lim_{|h|\to\infty} \int_{\mathbb{R}^n} |f(x+h)+f(x)|\, {dx$


    The related section proves that if f is lebesgue integrable on \mathbb{R}^n (ie. f is in L(\mathbb{R}^n)) then $\lim_{|h|\to 0} \int_{\mathbb{R}^n} |f(x+h)-f(x)|\, {dx} = 0$ I'm failing to see how this might to be used (if at all) for the above.


    Thanks for your help
    Last edited by jacobi; November 15th 2009 at 01:27 PM.
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  2. #2
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    Quote Originally Posted by jacobi View Post
    I'm stuck on a textbook exercise: given f is lebesgue integrable on \mathbb{R}^n, compute

    \lim_{|h|\to\infty} \int_{\mathbb{R}^n} |f(x+h)+f(x)|\, {dx}.
    Here's an intuitive answer, it's up to you to convert it into a rigorous argument!

    If f is integrable then it tends to 0 at infinity. So there is a bounded region of \mathbb{R}^n, call it K, outside which |f| is very small. Choose h to be larger than the diameter of K. Inside K, |f(x+h)| will be very small, so |f(x+h)+f(x)| is approximately |f(x)|. Let L = {k–h : k ∈ K} (so L is the translate of K through –h). Inside L, |f(x)| will be very small, so |f(x+h)+f(x)| will be approximately |f(x+h)|. Outside K∪L, both |f(x)| and |f(x+h| are small, hence so is |f(x+h)+f(x)|.

    Thus the integral of |f(x+h)+f(x)| will be approximately the sum of the integrals over K and L, each of which is approximately equal to the integral of |f|. In the limit, it follows that \lim_{|h|\to\infty} \int_{\mathbb{R}^n} |f(x+h)+f(x)|\, {dx} = 2\!\!\int_{\mathbb{R}^n} |f(x)|\, {dx}.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Opalg View Post
    If f is integrable then it tends to 0 at infinity.
    I don't believe this is necessarily true. There are some weird and very contrived functions that are unbounded but still Riemann integrable on the real line, such as:

    f(x)=\left\{\begin{array}{lr}0:&0\leq x<1\\-4n^7(x-n)(x-(n+n^{-3})):&x\in[n,n+n^{-3}]\\0:& otherwise\end{array}\right\} for n\in\mathbb{N}.

    The maximum value of f in the interval [n,n+n^{-3}] is f\left(n+\frac{n^{-3}}{2}\right)=n, so it's unbounded.

    \int_0^{\infty} f(x)\,dx=\frac{2}{3}\sum_{n=1}^{\infty}\frac{1}{n^  2}=\frac{\pi^2}{9}
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    Quote Originally Posted by redsoxfan325 View Post
    I don't believe this is necessarily true. There are some weird and very contrived functions that are unbounded but still Riemann integrable on the real line, such as:

    f(x)=\left\{\begin{array}{lr}0:&0\leq x<1\\-4n^7(x-n)(x-(n+n^{-3})):&x\in[n,n+n^{-3}]\\0:& otherwise\end{array}\right\} for n\in\mathbb{N}.

    The maximum value of f in the interval [n,n+n^{-3}] is f\left(n+\frac{n^{-3}}{2}\right)=n, so it's unbounded.

    \int_0^{\infty} f(x)\,dx=\frac{2}{3}\sum_{n=1}^{\infty}\frac{1}{n^  2}=\frac{\pi^2}{9}
    Easier still f(x)=n  \ if \ x\in [n,n+n^{-3}], using the same arguments it's improper Riemann integrable. Also Lebesgue integrable, but notice that the measure of the intervals in which f is not zero tends to zero ie. we can make the measure of A=\{ y\in \mathbb{R} : f(y)\neq 0 \} arbitrarily small outside some compact set, so the argument still works (I think).
    Last edited by Jose27; November 13th 2009 at 05:29 PM.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Touché. I had to come up with an unbounded integrable function as a homework exercise last year, and the one I posted was what I came up with. I can't believe I went to all that trouble when I could have just used the example you came up with. [smacks head]
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  6. #6
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    Quote Originally Posted by redsoxfan325 View Post
    I don't believe this is necessarily true. There are some weird and very contrived functions that are unbounded but still Riemann integrable on the real line
    Well I did say I was only giving an intuitive answer. But you're right, the set where |f(x)| > \varepsilon need not be bounded. What I should have said is that there is a bounded set outside which the integral is small: \int_{\mathbb{R}^n\setminus K} |f(x)|\, {dx} < \varepsilon. You can then use the triangle inequality |f(x)| - |f(x+h)| \leqslant |f(x+h)+f(x)| \leqslant |f(x)| + |f(x+h)| to show that \int_{K} |f(x+h)+f(x)|\, {dx} is approximately \int_{\mathbb{R}^n} |f(x)|\, {dx}, and similarly for the integral over L. The answer \lim_{|h|\to\infty} \int_{\mathbb{R}^n} |f(x+h)+f(x)|\, {dx} = 2\!\!\int_{\mathbb{R}^n} |f(x)|\, {dx} still stands.

    Notice that the question specifies the Lebesgue integral. This is essential, because the proof relies on the fact that if f is integrable then so is |f|. That is built into the definition of the Lebesgue integral, but does not hold for the Riemann integral.
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    Quote Originally Posted by Opalg View Post
    What I should have said is that there is a bounded set outside which the integral is small: \int_{\mathbb{R}^n\setminus K} |f(x)|\, {dx} < \varepsilon.
    Thanks; your explanation was very helpful. It's just this step (probably the most obvious) I don't see. I know the integral of |f| is finite, but how does this imply the above? (That is, for any epsilon>0 there is a bounded set K such that ...)
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  8. #8
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    Quote Originally Posted by jacobi View Post
    Quote Originally Posted by Opalg View Post
    What I should have said is that there is a bounded set outside which the integral is small: \int_{\mathbb{R}^n\setminus K} |f(x)|\, {dx} < \varepsilon.
    Thanks; your explanation was very helpful. It's just this step (probably the most obvious) I don't see. I know the integral of |f| is finite, but how does this imply the above? (That is, for any epsilon>0 there is a bounded set K such that ...)
    Define a sequence of functions f_n(x) = \begin{cases}|f(x)|&\text{if $|x|\leqslant n$, }\\ 0&\text{if $|x|>n$.}\end{cases} Then f_n(x) increases to |f(x)| as n\to\infty, so by the Monotone Convergence Theorem \int f_n\to \int|f|. Thus for n large enough, the integral of |f| over the region outside the ball of radius n can be made as small as we want.
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