Simplify $\int_{\mathbb{R}^2} f(x,y) dy dx$ assuming $\int_a^\infty \int_{-\infty}^a f(x,y) =\int_{-\infty}^a \int_a^\infty f(x,y) =0$ for every $a$ If we assume
\begin{equation}
\int_{a}^\infty \int_{-\infty}^a f(x,y) dy dx = \int_{-\infty}^a \int_a^\infty f(x,y) dy dx = 0, \quad \forall a \in \mathbb{R}
\end{equation}
what can be said about the integral
\begin{equation}
\int_{\mathbb{R}^2} f(x,y) dy dx?
\end{equation}
With the assumption we find
\begin{equation*}
\int_{\mathbb{R}^2} f(x,y) dy dx = \int_a^\infty \int_a^\infty f(x,y) dy dx + \int_{-\infty}^a \int_{-\infty}^a f(x,y) dy dx, \quad \forall a \in \mathbb{R}.
\end{equation*}
Is there any further simplification that can be done with the assumption?
 A: 
The function $f$ might not be integrable on $\mathbb R^2$.

To see this, first choose any integrable function $g$ on the real line, such that $g(t)=0$ for every $t\leqslant1$, and $$\int_1^\infty g(t)dt=0$$ and define $$f(x,y)=\frac{g(y-x)}{y-x}$$ Then $f(x,y)=0$ for every $y\leqslant x$ hence, for every $a$, $$\int_a^\infty \int_{-\infty}^a f(x,y) dy dx=0$$ On the other hand, for every $a$, $$\int_a^\infty f(x,y) dy=\int_a^\infty \frac{g(y-x)}{y-x} dy =\int_{a-x}^\infty \frac{g(t)}t dt=\int_{\max(1,a-x)}^\infty \frac{g(t)}t dt$$ hence $$\int_{-\infty}^a \int_a^\infty f(x,y) dy dx =\int_{-\infty}^a\int_{\max(1,a-x)}^\infty \frac{g(t)}t dtdx=I+J$$ where $$I=\int_{a-1}^a\int_{\max(1,a-x)}^\infty \frac{g(t)}t dtdx=\int_{a-1}^a\int_1^\infty \frac{g(t)}t dtdx=\int_1^\infty \frac{g(t)}t dt$$ and $$J=\int_{-\infty}^{a-1}\int_{\max(1,a-x)}^\infty \frac{g(t)}t dtdx=\int_{-\infty}^{a-1}\int_{a-x}^\infty \frac{g(t)}t dtdx=\int_1^\infty\int_{a-t}^{a-1}\frac{g(t)}t dxdt$$ that is, $$J=\int_1^\infty(t-1)\frac{g(t)}t dt$$ hence $$I+J=\int_1^\infty g(t) dt=0$$ which shows that $$\int_{-\infty}^a \int_a^\infty f(x,y) dy dx = 0$$
Finally, $$\iint_{\mathbb R^2}|f(x,y)|\,dxdy=\int_\mathbb R\int_{x+1}^\infty\frac{|g(y-x)|}{y-x}dydx=\int_\mathbb R\int_1^\infty\frac{|g(t)|}tdtdx=+\infty$$ for every function $g$ not almost everywhere zero.
