Let's take $f:[0,1] \to \mathbb{R}$ as a measurable and not Lebesgue integrable function. Is that true that the function $g$ defined by the formula $g(x,y) := f(x) - f(y)$ is not Lebesgue integrable on $[0,1] \times [0,1]$?
I am sure it's true, because if $f$ is not Lebesgue integrable, then, there must be some place in $[0,1]$ where it "blows up" to $-\infty$ or $\infty$. Because of that, it must be true that $g(x,y)$ is also not Lebesgue integrable since those places where $f$"blows up" cannot cancel each other because we would have $\infty - \infty$ and that can't be interpreted in any way.
I think that this intuition must be correct, but I don't know how to show that. I didn't find any elegant way of doing so while looking on that forum.
Edit:
Let's assume that $g$ is Lebesgue integrable on $[0, 1] × [0, 1]$. Then from Fubini's theorem, we have that:$$ \int \int_{[0,1][0,1]} g(x,y) \ \lambda_2(x,y) = \int \int_{[0,1][0,1]} f(x) - f(y) \ \lambda_2(x,y) = \int_0^1 \int_0^1 \left[ f(x) - f(y) \right] \ dx dy = $$
$$= \int_0^1 f(x) \ dx - \int_0^1 f(y) \ dy = \infty - \infty$$
Is that enough to say that we got to a contradiction and conclude that $g$ is not Lebesgue integrable on $[0, 1] × [0, 1]$?