Why is $\mathbb{P}(F\geqslant G) = \int_{\mathbb{R}} \mathbb{P}(F \geqslant g | G=g) \, D_G(g) \text{d}g$? For random variables $F,G$  I have problems with understanding the equation
$$\mathbb{P}(F \geqslant G) = \int_{\mathbb{R}} \mathbb{P}(F \geqslant g | G=g) \, D_G(g) \text{d}g, $$
where $D_G$ is the density function of $G$. 
Why is this correct? I know that we can write 
$\mathbb{P}(F \geqslant G)=  \int_{\mathbb{R}}  \mathbb{P}(F \geqslant G)\, D_G(g) \text{d}g $ but I am not sure how to deal with the conditional probability.
 A: Do you know about conditional expectation? For random variable $X$ and continuous random variable $Y$ jointly defined on the same probability space, "$\mathbb{P}(X \geqslant Y)=  \int_{\mathbb{R}}  \mathbb{P}( X\geqslant y)\, f_Y(y) \text{d}y$" is true if the random variables in question are independent; it does not hold in general. More generally, using the indicator function ${\bf 1}_{X\geqslant Y}$,
$$
P(X\geqslant Y) = \mathbb E[{\bf 1}_{X\geqslant Y}]  = \mathbb E[\mathbb E[{\bf 1}_{X\geqslant Y}\vert Y]] = \mathbb E[P(X\geqslant Y\vert Y)] \\= \int_{\mathbb{R}}  \mathbb{P}( X\geqslant y\ \vert\ Y=y)\, f_Y(y) \text{d}y
\tag{1}$$
The first 3 equalities follow by definition. The last, because $Y$ is a continuous random variable.

For example, as indicated in the comments, for jointly continous random variables $X$ and $Y$, we define the function $P(X\geqslant y\vert Y=y):=\int^{\infty}_{-\infty}{\bf 1}_{x\geqslant y}\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\text dx$ and, indeed rather trivially, see that
$$
P(X\geqslant Y) = \int^{\infty}_{-\infty}P(X\geqslant y\vert Y=y)f_{Y}(y)\text dy = \int^{\infty}_{-\infty}\Bigl(\int^{\infty}_{-\infty}{\bf 1}_{x\geqslant y}\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\text dx\Bigr)f_{Y}(y)\text dy = \\\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}{\bf 1}_{x\geqslant y}f_{X,Y}(x,y)\text dx\text dy,\\
$$
where the far rhs, by definition of the joint density $f_{X,Y}(x, y)$, must equal the far lhs. The important point to note here is that, in a rigorous sense, this example is unsurprisingly consistent and works because the relationships and entities in (1) are already well-defined.
