Confusion over marginalising out probability I have seen the probability equation below somewhere and I'm reasoning that it holds because the integral marginalises out $a$ leaving $p(c|s)$ 
$$ p(c|s) = \int p(a|s) \cdot p(c,a|s) \  da $$
where all the probabilities are PDFs.
I am wondering then whether the following is also true given that when you marginalise out $a$ then you are also left with $p(c|s)$ :
$$ p(c|s) = \int p(a) \  \ \  \cdot p(c,a|s) \  da $$
If this is also true, then I am left wondering how to (intuitively) reason that:
$$ \int p(a|s) \cdot p(c,a|s) \  da = \int p(a)  \cdot p(c,a|s) \  da $$
Note: I don't come from a statistics background so apologies if the answer is obvious.
Second Edit: I got the first equation from here which is the first equation in Section 2.3 but replacing $s'$ with $c$ and replacing $\pi(a|s)$ for $p(a|s)$.
 A: Your second expression seems to suggest (in conjunction with the first) that:
$$\int p(a|s) \cdot p((c,a)|s) \  da = \int p(a) \cdot p((c,a)|s) \  da $$
Now let $c=a$ (they are the same event). Then you get:
$$\int  p(a|s)^2 \  da = \int p(a) \cdot p(a|s) \  da $$
This seems to not be true in general. Let's say for example that $a$ is normal and conditional on $s$, it becomes normal with a different mean. It's easy to see from this answer: Integral of the square of the normal distribution. that the two sides will not be the same.

EDIT: Let's clarify some terminology.
Let's call the random variables $A$ and $S$. The $a$ is the event $A=a$ and $a|s$ is the event $A=a$ conditional on $S=s$ This is equivalent to the event $A=a \cap S=s$ in the space $S=s$ see here for more details:
No such thing as conditional events?. There is some abuse of notation here in terms of $a$ being interpreted as a scalar within the context of the integrals (otherwise, $da$ wouldn't make sense), but apart from that, everything is well-defined.
Building on this, another way to see why your second expression is wrong. You had:
$$ p(c|s) = \int p(a) \  \ \  \cdot p((c,a)|s) \  da $$
Let $a' = a|s$ since $a'$ and $a$ have the same domain (just their densities at those points in the domain are potentially different) so it doesn't matter which of them we integrate over we get:
$$ p(c|s) = \int p(a) \  \ \  \cdot p(c|s ,a') \  da' $$
You can see that this is only true if $a'=a$. Another point is that the event $A=a$ is not even in the same probability space as $C=c \& A=a | S=s$. So those two probabilities are apples to oranges and it's probably problematic to multiply them.
