Let $f\in L^1(\mathbb{R}^2)$ with respect to the Lebesgue measure $m\times m$ on $\mathbb{R}^2$. Prove that if $$\iint_{\mathbb{R}^2} f(x,y)dxdy=0$$ then there exists a square $S_{a,b}=\{(x,y)\,|\, a\le x\le a+1,\; b\le y\le b+1\}$, such that $$ \iint_{S_{a,b}}f(x,y)dxdy=0$$
My try: The union of all squares $S_{a+i,b+j}$ ,$i,j \in \mathbb{Z}$, covers the whole $\mathbb{R^2}$. Fix some enumeration of the set $ \{ S_{a+i,b+j} \}_{i,j} $ and denote it by $ \{ S^{a,b}_{n} \}^{\infty}_{n=1}$ . We can epress $\mathbb{R}^2$ as the disjoint union $$ \mathbb{R^2}=\bigsqcup^{\infty}_{n =1} S^{a,b}_{n}$$ Therefore $$\iint_{\mathbb{R}^2} f(x,y)dxdy=\sum^{\infty}_{n=1} \iint_{S^{a,b}_{n}}f(x,y)dxdy=0$$ because the series converges, $\displaystyle\lim_{n \to \infty} \iint_{S^{a,b}_{n}}f(x,y)dxdy =0 $ for any enumeration. But I can not find a rigorous way to show that over one of the squares the integral vanishes, because we can not consider a particular pattern in which positive and negative integrals cancel out each other. Should I fix an enumeration in which the $1\times 1$ squares are adjacent ?
Hints\suggestions are more appreciated than answers. Thank you in advance.