I'm sitting here on a task, where I have to show that for the function:

$$f:(0,1]\times(0,1]\to\mathbb R, \quad f(x,y)=\frac{x-y}{(x+y)^3},$$

the double Riemann-integral is the double Lebesgue-integral:

$$\int_0^1 \int_0^1 f(x,y) \, dx \, dy= \int_{(0,1]} \int_{(0,1]} f(x,y) \, d\lambda_1(x) \, d\lambda_1(y),$$

and show that $f$ is not $\lambda_2$ integrable.

Can someone help me, please?

  • $\begingroup$ Someone proposed an edit that changed $(0,1]\times(0,1]$ to $(0,1]x(0,1]$ and left $->$ intact instead of changing it to $\to$, and changed the capital "L" in "Lebesgue" (a person's name) to a lower-case "L". I wonder if someone was testing us to see if we'd approve a bad edit? ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 22 '15 at 18:40
  • $\begingroup$ Isn't this really an iterated Lebesgue integral rather than a double Lebesgue integral? ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 22 '15 at 18:42
  • $\begingroup$ If the iterated integral has a finite value that is not $0$, then the iterated integral in the opposite order has minus that value, and by Fubini's theorem, that can only mean the double integral of the absolute value is $+\infty$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 22 '15 at 18:43
  • $\begingroup$ I finally finished my answer and my edit to the Q. I had forgotten the $->$ and written "Lebegue", and that was mostly all that was left to fix. I missed that proposed edit, or I would probably have rejected it. $\endgroup$ – MickG Dec 22 '15 at 19:02
  • $\begingroup$ Thank you for your answers. Now I'm trying to solve it! :) $\endgroup$ – serge Dec 22 '15 at 19:07

For the second part, you can argue by contradiction. If it were $\lambda_2$-integrabile, then by Fubini-Tonelli the integrals would swap. Yet you should be able to prove that swapping the integrals produces a change of sign.

Of course, that assumes you know that if a function is both Riemann- and Lebesgue-integrable the two integrals coincide, and that this is the case here, and so you can evaluate the double Lebesgue-integral as a Riemann-integral, which is question one.

So we state the theorem that answers question 1.


Let $X=[a,b]$ and $f:[a,b]\to\mathbb{R}$ be a Riemann- and Lebesgue-integrable function. Then: $$\int_Xf\mathrm{d}\lambda_1=\int_a^bf(x)\mathrm{d}x.$$ In other words, if the Lebesgue and the Riemann integrals of a function both exist, then they coincide.

Proof. By the Monotone convergence theorem, the integrals of any family of simple functions $\phi_n$ such that $\phi_n\uparrow f$ almost everywhere will converge to the integral of $f$. For example, if we partition $[a,b]$ into smaller and smaller intervals, and consider the functions that, for each partition, map a point to the minimum of the function $f$ on the interval where the point lies, we get $\phi_n$, a sequence of simple functions monotonely converging… to what? Well, if $f$ is continuous at some point, they will converge to $f$ in that point. There is a result, which I cannot prove, that states that $f$ is R-integrable iff the set of points of discontinuity has measure 0. Proof here, Theorem 3.4.2, pp. 24-25. With that, we have convergence a.e., so the integrals of $\phi_n$ converge to that of $f$. But those integrals are Riemann sums, so they also converge to the Riemann integral. But the limit of a sequence is unique, hence the result follows. $\square$

By continuity, your $f$ will be integrable in each single dimension, and if you calculate its primitive you will find the primitive is also integrable in $y$, in both the Riemann and Lebesgue senses, so the two double integrals exist, which completes question 1.

As for question 2, see paragraph 1.

| cite | improve this answer | |
  • $\begingroup$ OK, in my case the intervall is $(0,1]$ , i can write it in this way $[a,1]$ where $0<a<1$ is. So i can apply the Theorem. To show the existens of the impropper riemann integral, do I have to proof $\lim_{a\to 0}$ of the Riemann-integral to proof that this is equal to the Lebesgue-integral? $\endgroup$ – serge Dec 23 '15 at 10:10
  • $\begingroup$ AFAIK integrating over an open or closed or semi-open interval is exactly the same. So no need to use $a$ and take limits. What improper Riemann integral are you referring to? There seems not to be any improper integral here since when doing iterated integrals the singularity of the function at the origin is hidden away. And I see no reference to a double Riemann integral. $\endgroup$ – MickG Dec 23 '15 at 10:17
  • $\begingroup$ Yes I know what you mean, but I'm wondering because the task said to show the double improper Riemann-integral is equal...but I am seeing its not an improper-integral $\endgroup$ – serge Dec 23 '15 at 10:23
  • $\begingroup$ To show the two iterated integrals are equal, first concentrate on the internal one and notice how for all $y\in(0,1]$ $\int_0^1f(x,y)dx=\int_{(0,1]}f(x,y)d\lambda_1(x)$. So you reduce yourself to showing the equality of a single integral, the external one. Basically you are applying the theorem twice, first to the internal integral, then to the integral that is left after integrating the function in $x$ (or $d\lambda_1(x)$). $\endgroup$ – MickG Dec 23 '15 at 10:26
  • $\begingroup$ Now I have it, thank you. $\endgroup$ – serge Dec 23 '15 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.