If all projections on the axis are integrable, do the iterated integrals exist?

Question

Suppose $f(x,y) : [0,1] \times [0,1] \to \mathbb{R}_{\ge 0}$ is a function, not necessarily continuous, nor necessarily Riemann integrable on $[0,1] \times[0,1]$. By existence of an integral here we will also admit also the case in which it is unbounded.

Moreover assume that:

• for each $x \in [0,1]$ the Riemann integral $\int_0^1 f(x,y) dy$ exists;
• for each $y \in [0,1]$ the Riemann integral $\int_0^1 f(x,y) dx$ exists.

Does this implies that the Riemann integral of $F(x) = \int_0^1 f(x,y) dy$ and $G(y) = \int_0^1 f(x,y) dx$ exist?

The latter are nothing else than the iterated integrals of $f(x,y)$, i.e. $\int_0^1 F(x)dx = \int_0^1 \int_0^1 f(x,y) dy dx$, $\int_0^1 G(y)dy = \int_0^1 \int_0^1 f(x,y) dxdy$.

Observe that I cannot apply Fubini's theorem since I do not know whether $f(x,y)$ is integrable in $[0,1] \times [0,1]$. As far as I know Tonelli's theorem only apply to Lebesgue integrals.

Attempt 1: proving that the answer is yes

I have been trying to prove this by contraddiction. First suppose that $f$ is bounded and that, for example, $F(x)$ is non-Riemann integrable. Then its set of discontinuity points $X^*$ must have non-zero Lebesgue measure. I would like to show that this implies that at least for one $y^*$ $f(x, y^*)$ must have a non-zero measure set of discontinuity points, therefore $\int_0^1 f(x,y^*) dx$ does not exist. However, my attempts have failed so far because I do not know how to link the discontinuities of $F(x)$ and those of $f(x,y)$ for $x$ or $y$ fixed.

Attempt 2: finding a counter-example

I know that there are sets $S$ (such as the Sierpinsky set) that have non-zero measure on the plane but such that every projection on the axis has zero measure. However this is not enough, because, for example, taking the characteristic function of $S$ as a counter-example I always end up with iterated integrals that exist and are zero. I also experimented a bit with two-variables version of Thomae's function, but also in that case I always end up with iterated integrals that exist.

Answer 1

No, it being integrable along every slice doesn’t necessarily mean the slices are Reiman integrable.

Here’s a counterexample. For rational $x=a/b \in [0,1]$, let $f(a/b,y)=b 1_{[1/b,2/b]}$. For irrational $x$, let $f(x,y)=0$.

Then, for any $y$, $y$ is only in the range $[1/b,2/b]$ for a finite number of integers $b$ (from $1/y$ to $2/y$) each of which is the denominator of a finite number of rationals within $[0,1]$, so for fixed $y$, we have $f(x,y)$ is nonzero a finite number of times, so $G(y)=\int^1_0 f(x,y)dx=0$ exists.

Going the other way, for any rational $x=a/b$, we have that $F(a/b)=b\times(2/b-1/b)=1$. For any irrational it is $0$, so $F=1_\mathbb{Q}$.

However, despite $F$ and $G$ existing, $F$ is not Reimann integrable (since on every open set, it is both $0$ and $1$, so the lower integral is always $0$ and the upper is always $1$ and so they don’t converge).

If you really want both $F$ and $G$ to exist and be nonintegrable you can consider $h(x,y)=f(x,y)+f(y,x)$.

Answer 2

Let's consider the function $$ f(x,y) = \frac{1}{x} \mathbb{I}_{\exp(-1/y)<x} \mathbb{I}_{y \neq 0} \mathbb{I}_{x \neq 0}. $$

Then the projections are clearly Riemann integrable as they only have finitely many discontinuities. Moreover, they are integrable:

$$ \begin{aligned} x \geq 0: &\int_0^1 f(x,y) dy = \frac{1}{x} \int_0^1 \mathbb{I}_{\exp(-1/y)<x} dy \leq \frac{1}{x} \\ x = 0: &\int_0^1 f(x,y) dy = 0 \\ y \geq 0: &\int_0^1 f(x,y) dx = \int_{\exp(-1/y)}^1 \frac{1}{x} dx = \ln(1) - \ln(\exp(-1/y)) = \frac{1}{y} \\ y=0: &\int_0^1 f(x,y) dx=0 \end{aligned} $$

However, the iterated integral diverges:

$$\int_0^1 \int_0^1 f(x,y) dx dy = \int_0^1 \frac{1}{y} dy = \infty. $$