Suppose $g:[0,1]^2\to R$ and $g$ can have finitely many discontinuities. $F$ is continuous and atomless c.d.f on $[0,1]$
$$\int_{[0,1]} g(x,y)dF(y)=1/2, \forall x$$
$$\int_{[0,1]} g(x,y)dF(x)=1/2, \forall y$$
Can we conclude that $g(\cdot)$ must be $1/2$ for all $(x,y) \in [0,1]^2$?
How about if $g$ is nondecreasing in $x$ and $y$ everywhere on $[0,1]$ for the same question?
No. We cannot conclude that $g \equiv 1/2$.
A counterexample is $F(x) = x$ and $$ g(x,y) = \frac{1}{2} (1+a -2a (x+y) + 4 a xy) ,$$ for $a \in [-1,1]$ and defined for $(x,y) \in [0,1]^2$. This is a continuous function which is nondecreasing in each coordinate.
The basic idea behind the counterexample is that it does not suffice to know only the marginal distributions of a bivariate random vector $(X,Y)$ to determine its joint distribution. The function $g$ can be seen as the density.
If $g$ has the additional requirement to be nondecreasing, then this simple counterexample is not working, though.