Motivate why we may calculate $P[XGiven two independent random variables $X $ and $Y $ with continuous densities we know that there exists a regular conditional distribution $P[X<Y|Y=y] $, and futher that the regular conditional distribution has density $f_{X,Y}(x,y)/f_Y(y) $.
How do we motivate that $P[X<Y]=\int_\infty^\infty P[X<Y|Y=y]f_Y(y)dy$. This looks like a generalization of the discrete case where we may write $\{X<Y\}$ as the disjoint union $\bigcup _{k \in Z } \{X<k\} \cap \{Y=k \}$ and then use the countable additivity of $P $ to conclude that $$P[X<Y]=\sum_{k \in Z }P[\{X<k\} \cap \{Y=k \}] $$  (And of course for every $k$, $P[\{X<k\} \cap \{Y=k \}]=P[X<y|Y=k]P[Y=k]$ )
Any guidance would be appreciated!

Reading my own post two monts later I would like to add the following for any future viewer of this post.
To derive 
$$P[X<Y]=\int P[X<Y|Y=y] f_(y)dy$$ 
we do the following.
$P[X<Y]=\int P[X<Y|Y](\omega)P(d \omega )=\int P[X<Y|Y=y]P_Y(dy)=\int P[X<Y|Y=y]f_Y(y)dy$
where the first equality is the double expectation formula, the second the definition of conditional expectation of $1_{X<Y} $ given that  $Y=y $ [See Definition 4 on p. 262 in Shiryaev's Probability 1 (2016)] and the third equality the definition of a density, $P_Y=f_Y dy $. None of this uses the independence of $X $ and $Y $.
If we also assumes independence we may use the following result which may be found on p.263 in Shiryaev's Probability 1 (2016) to get $P[X<Y]=\int P[X<y] f_(y)dy$

If $X,Y $ are independent and $\phi $ is a $\mathcal B(R^2) $
  measurable function such that $E[|\phi(X,Y) |]<\infty $ then  $$\int E[\phi(X,Y)|Y=y]P_Y(dy)=\int E[\phi(X,y)]P_Y(dy)$$

We simply let $\phi(x,y)=1_{\{x<y\}}(x,y)$.
