Showing that $P(X_1>X_2) = \int_{0}^1 P(X_1>X_2 | X_2=x) f_{X_2}(x) dx$ I am going through this post in trying to prove the probabilistic interpretation of the AUC for a ROC Curve (for a classifier):

The AUC for a ROC curve is the the probability of the classifier scoring a randomly chosen positive point higher than a randomly chosen negative one.

I was able to follow till the end, but am a little stuck on getting the final result. So far I have the following
\begin{align*}
 {AUC} &= \int_0^1 tpr(fpr) d(fpr)\\
 &= \int_0^1 tpr(fpr(s))d(fpr(s))\\
 &= \int_{+\infty}^{-\infty} tpr(s)fpr'(s)ds\\
 &= \int_{+\infty}^{-\infty} \left(1-F_1(s)\right) \left(-f_{-1}(s)\right)ds\\
 &= \int_{-\infty}^{+\infty} (1-F_1(s)) (f_{-1}(s))ds \\
 &= \int_{-\infty}^{+\infty} Pr(s<X_{1}) (f_{-1}(s))ds \\
 &= \int_{-\infty}^{+\infty} Pr(X_{-1}<X_{1}\ |\ X_{-1}=s) (f_{-1}(s))ds \\
 &= \int_{-\infty}^{+\infty} Pr(X_{-1}<X_{1}, X_{-1}=s ) ds \\
 &= Pr(X_{-1}<X_{1} ).
%  &= \int_{-\infty}^{+\infty} P(S>s) (f_{-1}(s))ds \\
\end{align*}
I am struggling in my confidence that
$$\int_{-\infty}^{+\infty} Pr(X_{-1}<X_{1}\ |\ X_{-1}=s) (f_{-1}(s))ds = \int_{-\infty}^{+\infty} Pr(X_{-1}<X_{1}, X_{-1}=s ) ds$$
is correct. I found the result in this post, but cannot find it elsewhere for continuous random variables. Is this a standard definition without proof? I have not seen it before and I can't find anything about conditional probabilities that makes me get it, could I get some help on that part?
Thanks
 A: It's easier to see if everything  is discrete, so you have on the left
$$\sum_{s=-\infty}^\infty P[X_{-1}<X_1 | X_{-1}=s] P[X_{-1}=s]$$
and on the right
$$\sum_{s=-\infty}^\infty P[X_{-1}<X_1 \cap X_{-1}=s] $$
so for each $s$ it's just the equality $P(A|B)P(B)=P(A\cap B)$
In the continuous case, probabilities with equalities turn into densities times $ds$, and sums into integrals.
A: It's probably easiest to use the law of iterated expectations. To do so we can write the probability as the expectation of an indicator:
$$ \mathbb{P}(X_1 > X_2) = \mathbb{E}[ \mathbb{I}\{X_1 > X_2 \} ]$$
The law of iterated expectations states that
\begin{align*}
 \mathbb{E}[ \mathbb{I}\{X_1 > X_2 \} ] &= \mathbb{E}[ \mathbb{E}[\mathbb{I}\{X_1 > X_2 \} \mid X_2]] \\
&= \mathbb{E}[ \mathbb{P}(X_1 > X_2 \mid X_2)]
\end{align*}
The second line substitutes the definition of conditional probability as the expectation of an indicator function (similar to what we did before).
Finally, if $X_2$ has a continuous distribution, we can put the results together to show that
$$ \mathbb{P}(X_1 > X_2)  = \mathbb{E}[ \mathbb{P}(X_1 > X_2 \mid X_2)] = \int \mathbb{P}(X_1 > X_2 \mid X_2 = x_2) f(x_2) dx_2 $$
Continuity of $X_2$ is only used for the second equality, which substitutes the integral definition of the expectation. I like this proof because you don't have to take a stance on whether $X_1$ or $X_2$ are discrete/continuous until the last step.
