Conditional probability explained? Let $F_A$ be the CDF to the random variable $A$ ( and $B$ another independet rv), how do we get that $P(A+B \le s) = \int_{\mathbb{R}} P(A+B \le s\mid A=x ) \, dF_A(x)$ (This is probably a Stieltjes-Integral) then? I have never seen that it is possible to integrate with respect to the CDF. Could anybody explain  what this means and how we get this?
If anything is unclear, please let me know. 
 A: It is a Riemann–Stieltjes integral.
Let's recall how $\mathbb E(C\mid A)$ is defined when $A$ and $C$ are Bernoulli-distributed  random variables and then look at random variables in general.
$$
\begin{array}{cc}
& C\\
A & \begin{array}{|c|cc|}
\hline
& 0 & 1 \\
\hline
0 & p & q \\
1 & r & s \\
\hline
\end{array}
\end{array}
$$
We have $p+q+r+s=1$. So
$$\mathbb E(C\mid A=0)= 0\cdot\Pr(C=0\mid A=0)+1\cdot\Pr(C=1\mid A=0)= \frac q {p+q}.$$
$$\mathbb E(C\mid A=1)= 0\cdot\Pr(C=0\mid A=1)+1\cdot\Pr(C=1\mid A=1)= \frac s {r+s}.$$
Therefore
$$
\mathbb E(A\mid C)=\begin{cases} q/(p+q) & \text{if }C=0, \\ s/(r+s) & \text{if }C=1. \end{cases}
$$
And then
$$
\begin{align}
\mathbb E(\mathbb E(C\mid A)) & = \dfrac q{p+q} \cdot\Pr(A=0) +\frac s {r+s}\cdot\Pr(A=1) \\[10pt]
& = \dfrac q{p+q} \cdot(p+q) +\frac s {r+s}\cdot(r+s) \\[10pt] & = q+s = \Pr(C=1) \\[10pt]
& = 0\cdot\Pr(C=0)+1\cdot\Pr(C=1) \\[10pt]
& = \mathbb E(C).
\end{align}
$$
Now let's recall how $\mathbb E(C\mid A)$ is defined when $A$ and $C$ are random variables in general. It is a function $g(A)$ of $A$ – thus a random variable – whose conditional expected value $\mathbb E(g(A)\mid A\in\mathcal A)$ given any event of the form $[A\in\mathcal A]$ is the same as the expected value $\mathbb E(C\mid A\in\mathcal A)$ of $C$ given that event.
Now apply this to $\Pr(E\mid A)$ when $A$ is a random variable.  It is a function of $A$ – thus a random variable – whose expected value given any event of the form $[A\in\mathcal A]$ is the conditional probability $\Pr(E\mid A\in\mathcal A)$.  Hence
$$
\mathbb E(\Pr(E\mid A)) = \Pr(E). \tag 1
$$
That can be expressed as a Riemann–Stieltjes integral
$$
\int_{\mathbb R} \Pr(E\mid A=x)\,dF_A(x). \tag 2
$$
Applying this to the case $E=A+B$, we get
$$
\Pr(A+B \le s) = \int_{\mathbb{R}} \Pr(A+B \le s\mid A=x ) \, dF_A(x).
$$
Postscript responding to comments:  It appears that the question in the comments below may be how to get from $(1)$ to $(2)$ above.  The random variable $\Pr(E\mid A)$ is a function of $A$.  Call it $g(A)$.  Its expected value is
$$
\mathbb E(\Pr(E\mid A)) = \mathbb E(g(A)) = \int_{\mathbb R} g(x)\,dF_A(x).
$$
But $g(x) = \Pr(E\mid A=x)$, so this integral is
$$
\int_{\mathbb R} \Pr(E\mid A=x)\,dF_A(x).
$$
