Show that a function is not integrable but an iterated integral exist

Question

So, I have the funcion

$f(x,y) = \begin{cases} 0 & \text{if }x \text{ irrational} \\ 2y & \text{if } x \text{ rational}\ \end{cases} $

Defined in $R=[0,1]\times[0,1]$

I know that $f$ is not Riemann-integrable since the value of the lower/upper Darboux' sums depends on the choice of the sample points.

I just don't understand why should the iterated integral exist:

$\int_{0}^{1}[\int_{0}^{1}f(x,y)dy]dx$

But it does, and I don't know what its value should be.

Answer 1

Firstly, be a little careful here. The function is not Riemann integrable, but not because different Darboux sum values can be achieved with different partitions. It is not integrable because the supremal lower Darboux sum disagrees with the infimal upper Darboux sum. That is, there is always a positive gap between lower Darboux sums and upper Darboux sums that cannot be mader arbitrarily small.

As for the iterated integral, just evaluate it inside out. Consider, for fixed $x \in [0, 1]$, the integral $$\int_0^1 f(x,y) \, \mathrm{d}y = \begin{cases} \int_0^1 2y \, \mathrm{d}y & \text{if } x \in \mathbb{Q} \\ \int_0^1 0 \, \mathrm{d}y & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}\end{cases} = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}\end{cases}.$$ This function is not Riemann integrable, so the iterated integral doesn't make sense.

If you want an example of an iterated integral that does make sense, even though the function is not integrable, try $$f(x, y) = \begin{cases} \sin(y) & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}\end{cases}.$$ over $[-\pi, \pi] \times [-\pi, \pi]$.