# Math Help - evaluating a lebesgue integral

1. ## 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.

2. Originally Posted by jacobi
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 $\lim_{|h|\to\infty} \int_{\mathbb{R}^n} |f(x+h)+f(x)|\, {dx} = 2\!\!\int_{\mathbb{R}^n} |f(x)|\, {dx}$.

3. Originally Posted by Opalg
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.

$\int_0^{\infty} f(x)\,dx=\frac{2}{3}\sum_{n=1}^{\infty}\frac{1}{n^ 2}=\frac{\pi^2}{9}$

4. Originally Posted by redsoxfan325
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).

5. 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]

6. Originally Posted by redsoxfan325
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.

7. Originally Posted by Opalg
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 ...)

8. Originally Posted by jacobi
Originally Posted by Opalg
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.