Showing that $\mathbb{P}(U - V > 0) = \frac{2}{3}$ for dependent Uniform r.v. with given joint DF 
Assume $U, V$ to be $Unif[0,1]$ variables with the following joint distribution function $F$: 
$$F(u,v)=\left\{ \begin{array}{ll} u, &0\leq u \leq v/2 \leq 1/2\\
v/2, & 0 \leq v/2 \leq u < 1 - v/2 \\
u + v - 1, & 1/2 \leq 1 - v/2 \leq u \leq 1 \end{array} \right.$$
I'm trying to show that $\mathbb{P}( V - U > 0 ) = \frac{2}{3} $

In other words,  $\mathbb{P}(U \leq V) = \frac{2}{3}$. 
On a side note, it can be shown that the support of $F$ lies through the following points:  $(0,0)(\frac{1}{2}, 1)(1, 0)$. Since the mass here is assumed to be uniform, the easiest way here is to calculate the mass above $v = u$ line, which results the wanted $\frac{2}{3}$.
My question: what would be the technical way of evaluating this probability? That is, assuming we don't know how does the support of $F$ look like? 
From what I've gathered (please correct me if I'm wrong), one way would be through conditional probabilities, as in 
$$ \begin{align*}\mathbb{P}(U \leq V) &= \int_{0}^{1}\mathbb{P}( U \leq z | V = z)f_{V}(z)dz \\
&=\int_{0}^{1} \mathbb{P}( U \leq z,  V = z)dz \\
&=\int_{0}^{1} \left( \int_0^{z} \mathbb{P}( U = x,  V = z) dx \right ) dz 
\end{align*}$$
But this does not seem convenient, since the $\mathbb{P}(U = u, V = v)$ expression does not seem trivial to derive from the known $F$ (or is it?). 
Could we condition on some other event, allowing for $\{ V \leq z \}$ condition?
 A: I'm not sure what you have in mind by saying "...assuming we don't know how does the support of $F$ look like..."
Basically when we are given such a CDF and we see that it's mixed derivative $\partial^2/ \partial u \partial v$ is zero almost everywhere, we WILL go find out that the density is concentrated on the boundaries of the regions: the two line segments joining the points $(0,0)(\frac{1}{2}, 1)(1, 0)$.
It is not like in a homework that we are just given this fact. In the natural course of solving this problem we deduce that the density is the 2-dim Dirac delta `wall' along the region-boundaries (what you call the support of $F$). This is not cheating or handwaving but the legit approach.
In particular, when you take the 1st derivative of $dF/du$ you will have uniform distributions that take $v$ as a parameter. Formally, uniform distribution consists of the Heaviside step function (or more often in statistics people use the indicator function).
The 2nd derivative (differentiating a step function) will give you $\delta(\cdot)$ the Dirac delta function, and BTW the first derivative of the Dirac delta function $\delta'(\cdot)$. Please consult wiki or Eq.(10) of the Wolfram page to see how $\delta'(\cdot)$ works.
So if you want the technical way of doing this, first get familiar with taking the step function and its derivatives. Basically this is low level functional analysis where you treat the `support' of $F$ still as the entire unit square, and you treat a concentrated mass (be it a point mass in 1-dim, or a line of mass in 2-dim) just the same as any continuous and nicely behaving densities.
If you don't like it, please realize that you still have to face the same "issue" going with the conditional density.
