$\int_0^{x_0} \int_0^{y_0} F(x, y)\,dy\,dx= 0$, $F = 0$ almost everywhere? Suppose $F$ is real-valued and integrable with respect to $2$-dimensional Lebesgue measure on $[0, 1]^2$ and$$\int_0^{x_0} \int_0^{y_0} F(x, y)\,dy\,dx= 0$$for all $x_0 \in [0, 1]$ and $y_0 \in [0, 1]$. Does it follow that $F = 0$ almost everywhere?

Define the following signed measure
\begin{align}
\nu(A) = \int_A F\ d\mu
\end{align}
which one could check is absolutely continuous with respect to the Lebesgue measure $d\mu=dxdy$ since $\mu(A) = 0$ implies $\nu(A)=0$. 
Since $\nu(A) = 0$ on all rectangles of $[0, 1]^2$, then, by extension, we see that $\nu(A) = 0$ for all Borel subsets of $[0, 1]^2$, i.e. effectively $\nu$ is the zero measure. 
Next, by Radon-Nikodym theorem, we know there exists a unique measurable function $F$ such that 
\begin{align}
\frac{d\nu}{d\mu}= F.  
\end{align}
Then it follows $F=0$ a.e. since both generate the same zero measure. 
