Lebesgue integrable and improper Riemann integrable 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?
 A: 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.

Theorem
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.
