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!
 A: This is a special case of the very general "double expectation theorem" of conditional expectations:

Theorem.
  Let $(\Omega,\mathcal{F},P)$ be a probability space, $\mathcal{G} \subseteq \mathcal{F}$ a $\sigma$-algebra of events, and $Z$ a $\mathcal{F}$-measurable random variable such that either $Z$ is non-negative or $Z \in L^1(P)$.
  Then
  $$
E[Z] = E[E[Z \mid \mathcal{G}]].
$$

As a special case, using the fact that a probability of an event is the same as the expected value of the indicator function of that event, we obtain

Corollary.
  Let $(\Omega,\mathcal{F},P)$ be a probability space, $\mathcal{G} \subseteq \mathcal{F}$ a $\sigma$-algebra of events, and $A \in \mathcal{F}$ an event.
  Then
  $$
P(A) = E[P(A \mid \mathcal{G})].
$$

If we have a random variable $Y$, then conditioning on $Y$ means (by definition) conditioning on the $\sigma$-algebra $\mathcal{G} = \sigma(Y)$.
In particular,
$$
P(A)
= E[P(A \mid Y)]
$$
for any event $A$.
Even more specifically, in your case we can obtain the general formula
$$
P(X < Y)
= E[P(X < Y \mid Y)].
$$
If $X$ and $Y$ are both discrete, then this reduces to the usual formula:
$$
\begin{aligned}
P(X < Y)
&= E[P(X < Y \mid Y)] \\
&= \sum_y P(X < Y \mid Y = y) P(Y = y) \\
&= \sum_y P(X < Y, Y = y).
\end{aligned}
$$
Moreover, if $X$ and $Y$ are both absolutely continuous with joint density $f_{X, Y}$ and marginal densities $f_X$ and $f_Y$, respectively, then one can show that
$$
\begin{aligned}
P(X < Y)
&= E[P(X < Y \mid Y)] \\
&= \int_{\mathbb{R}} P(X < Y \mid Y = y) f_Y(y) \, dy.
\end{aligned}
$$
This is what you asked about, and it is because $P(X < Y \mid Y)$ is the $\sigma(Y)$-measurable random variable $g(Y)$, where $g : \mathbb{R} \to \mathbb{R}$ is the function given by
$$
g(y) = P(X < Y \mid Y = y) = \int_0^y f_{X \mid Y}(x \mid y) \, dx.
$$
Here $f_{X \mid Y}$ is the conditional density
$$
f_{X \mid Y}(x \mid y)
= \begin{cases} \frac{f_{X, Y}(x, y)}{f_Y(y)} &\text{if $f_Y(y) > 0$} \\
\text{undefined} & \text{otherwise}.
\end{cases}
$$
A: 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 $\phi(x,y)=1_{\{x<y\}}(x,y)$.
