Suppose I have a CDF of the form:
$$F_{XY}(x,y)=\begin{cases} g(x,y) & x\le \frac {1}{2} , \ y\le \frac {1}{2}\mathstrut \mathstrut \\[5pt] g(x, \frac 12\mathstrut ) & x\le \frac 12, \ y >\frac 12\mathstrut \mathstrut \\[5pt] g(\frac 12,y\mathstrut ) & x>\frac 12 , \ y\le \frac 12\mathstrut \mathstrut \\[5pt] g(\frac 12, \frac 12\mathstrut \mathstrut ) & x>\frac 12, \ y > \frac 12\mathstrut \mathstrut \end{cases}$$
If I differentiate to obtain the density function, I would get $$f_{XY}(x,y)=\begin{cases} g_{xy}(x,y) & x\le \frac 12 , \ y \le \frac 12\mathstrut \mathstrut \\[5pt] 0 & \text{otherwise} \end{cases}.$$
Of course, there is nothing insightful about differentiating constants resulting in $0$. But I was wondering if there was intuition behind why, when the joint CDF depends on only one of the variables (e.g. only $x$ or only $y$), the density is $0$ in that region?
Look at the definition of cdf: $F_{X,Y}(x,y)=\mathbb P(X\leq x, Y\leq y)$. Let us consider the case when in some region this function depends on $y$ only and does not depend on $x$. Take two points $(a,y)$ and $(b,y)$ with $a<b$, from this region. We supposed that $F_{X,Y}(a,y) = F_{X,Y}(b,y)$.
Using the definition, the probability that a pair of r.v.'s belongs to a halh of strip $a<X\leq b$, $Y\leq y$ is $$ \mathbb P(a<X\leq b, Y\le y) = F_{X,Y}(b,y) - F_{X,Y}(a,y) = 0. $$ What does it mean? The probability to see a pair $(X,Y)$ inside the region where cdf does not depend on $x$, is zero. Clearly, pdf inside this region is zero too since only zero pdf can provide zero probability being integrated over the whole half of strip.