Let $f\in {L}^{1}({\mathbb{R}}^{2})$ with respect to the Lebesgue measure $m\times m$ on ${\mathbb{R}}^{2}$. Prove that if

${\iint}_{{\mathbb{R}}^{2}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0,$

then there exits a square ${S}_{a,b}=\{(x,y)\mid a\le x\le a+1,b\le y\le b+1\}$, such that

${\iint}_{{S}_{a,b}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0.$

I tried to show that the integral

${\iint}_{[a,a+x]\times [b,b+y]}f(s,t)\phantom{\rule{thinmathspace}{0ex}}ds\phantom{\rule{thinmathspace}{0ex}}dt$

is absolutely continuous by Fubini's Theorem and Fundamental Theorem. And by the countable additivity of integration, I proved the integral on the whole plane is still A.C. However, I could not directly apply a theorem like the IVT for the single variable functions.

Is there any theorem for the two-dimensional case?

${\iint}_{{\mathbb{R}}^{2}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0,$

then there exits a square ${S}_{a,b}=\{(x,y)\mid a\le x\le a+1,b\le y\le b+1\}$, such that

${\iint}_{{S}_{a,b}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0.$

I tried to show that the integral

${\iint}_{[a,a+x]\times [b,b+y]}f(s,t)\phantom{\rule{thinmathspace}{0ex}}ds\phantom{\rule{thinmathspace}{0ex}}dt$

is absolutely continuous by Fubini's Theorem and Fundamental Theorem. And by the countable additivity of integration, I proved the integral on the whole plane is still A.C. However, I could not directly apply a theorem like the IVT for the single variable functions.

Is there any theorem for the two-dimensional case?