Unifying the treatment of discrete and continuous random variable I have been working on the reconciliation of the treatment of discrete and continuous random variable in a measure theoretic sense. But I found myself blocked on fundamental results.
We know that If $X$ and $Y$ are respectively discrete and continuous we have the marginals :
\begin{align}
P(A) = \sum_i P(A,X_i)\\
p(A) = \int_{\Omega_Y} p(A,t)dt\\
\end{align} 
I am trying to find the equation that generalize these results and by derivation find those actual results.
Since the discrete case involve measures and the continuous case involve densities I don't know where to start.
I could use the density of the discrete variable using Dirac but then again I am stuck because to derive the continuous case we need the Radon-Nikodym theorem which is not available because $X$ is not absolutely continuous to the lesbegue measure.
1) Do I need to consider a special measure for the discrete case ?
2) Do I miss any layer of abstraction here that would allow me to derive these results ?
Thank you for any pointer !
EDIT: 
Part of my question is also how to define the marginal of a joint distribution if we don't know if the second random variable is discrete or continuous. If it is discrete we can define it in term of sum and if it is continuous we can define it in term of an integral of the joint density.
But in the general case how do we define it ( not knowing if discrete or continuous ).
Above notation :
$P(A) = P(a\in A, x \in \Omega_X)$ so $P$ is a probability distribution
$P(a \in A) = \int_A p(x)dx$ so $p$ is a density
 A: Your notation is still a bit imprecise, but the message is pretty much clear. Let $\Bbb P$ be a product distribution, that is some probability measure on the product space $\Omega\times \bar\Omega$. The marginals of $\Bbb P$ on each factor spaces are given (be definition) as
$$
  \mathsf P(A):=\Bbb P(A\times \bar\Omega)\qquad \bar{\mathsf P}(B) := \Bbb P(\Omega\times B)
$$
for all measurable $A$ and $B$. Now, is the factor spaces are nice measurable spaces (e.g. $\Bbb R^n$ with Borel $\sigma$-algebra) then there always exist conditional distributions such that
$$
  \Bbb P(A\times B) = \int_\Omega\kappa(B|\omega)\mathsf P(\mathrm d\omega) = \int_{\bar\Omega}\bar\kappa(A|\bar\omega)\bar{\mathsf P}(\mathrm d\bar\omega) \tag{1}
$$
for all measurable $A$ and $B$, and $(1)$ determines $\Bbb P$ uniquely since to construct the product measure we have to define it (in a consistent way) just on measurable rectangles. 
Consider now the two special cases that you have. 


*

*The fact that the first variable is discrete means that $\mathsf P = \sum_i\mathsf P(\omega_i)\delta_{\omega_i}$ for some countable collection of points $\omega_i\in \Omega$. In that case the conditional distribution does always exist, so the formula $(1)$ boils down to 
$$
  \bar{\mathsf P}(B) = \int_\Omega\kappa(B|\omega)\mathsf P(\mathrm d\omega) = \sum_{i}\mathsf P(\omega_i)\kappa(B|\omega_i) = \sum_{i}\mathsf P(\omega_i)\kappa(B|\omega_i) = \sum_i \Bbb P(\{\omega_i\}\times B).
$$

*For the continuous random variable, we have $\bar\Omega = \Bbb R$ and $\bar{\mathsf P} \ll \lambda$ where $\lambda$ is the Lebesgue measure, so let $\bar p$ be its density. From $(1)$ we obtain:
$$
  \mathsf P(A) = \int_{\Bbb R}\bar\kappa(A|\bar\omega)\bar{\mathsf P}(\mathrm d\bar\omega) = \int_{\Bbb R}\bar\kappa(A|\bar\omega)\bar p(\bar\omega)\lambda(\mathrm d\bar\omega)
$$
which recovers your second formula. So, that's why I was wondering what does "$p(A,t)$" mean in your case: the joint density (w.r.t Lebesgue measure) may not exist: for sure it does not exist if the first variable has discrete distribution. Not to say that density does not have  set-valued arguments. So in fact we are talking about conditional distribution multiplied by the density of the second variable. If both variables are continuous and the joint density $q$ does exist, then your "$p(A,t)$" is $\int_A q(s,t)\lambda(\mathrm ds)$.
