# Let f in L^1(R^2) with respect to the Lebesgue measure m*m on R^2. Prove that if int int_(R^2) f(x,y)dxdy=0

Let $f\in {L}^{1}\left({\mathbb{R}}^{2}\right)$ with respect to the Lebesgue measure $m×m$ on ${\mathbb{R}}^{2}$. Prove that if
${\iint }_{{\mathbb{R}}^{2}}f\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0,$
then there exits a square ${S}_{a,b}=\left\{\left(x,y\right)\mid a\le x\le a+1,b\le y\le b+1\right\}$, such that
${\iint }_{{S}_{a,b}}f\left(x,y\right)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0.$
I tried to show that the integral
${\iint }_{\left[a,a+x\right]×\left[b,b+y\right]}f\left(s,t\right)\phantom{\rule{thinmathspace}{0ex}}ds\phantom{\rule{thinmathspace}{0ex}}dt$
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?
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Zackary Galloway
Let $g\left(a,b\right)={\iint }_{{S}_{ab}}f\left(s,t\right)\phantom{\rule{thinmathspace}{0ex}}ds\phantom{\rule{thinmathspace}{0ex}}dt$. $g$ is a continuous real valued function on ${\mathbb{R}}^{2}$. If it is never zero it is always positive or always negative. [ Because its range is connected in $\mathbb{R}$]. If it is always positive then ${\iint }_{{\mathbb{R}}^{2}}f\left(s,t\right)\phantom{\rule{thinmathspace}{0ex}}ds\phantom{\rule{thinmathspace}{0ex}}dt>0$ because this integral is the sum of integrlas over ${S}_{nm}$ as $n,m$ vary over all integers.