Sum of Discrete Random Variables

Theorem
Let $X$ and $Y$ be discrete random variables on the probability space $\left({\Omega, \Sigma, \Pr}\right)$. Let $U: \Omega \to \R$ and $V: \Omega \to \R$ be defined as:
 * $\forall \omega \in \Omega: U \left({\omega}\right) = X \left({\omega}\right) + Y \left({\omega}\right)$

Then $U$ is also a discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$.

Proof
To show that $U$ and $V$ are discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$, we need to show that:


 * $(1): \quad$ The image of $U$ and $V$ are countable subsets of $\R$;
 * $(2): \quad \forall x \in \R: \left\{{\omega \in \Omega: U \left({\omega}\right) = x}\right\} \in \Sigma$ and $\left\{{\omega \in \Omega: V \left({\omega}\right) = x}\right\} \in \Sigma$.

First we consider any $U_u = \left\{{\omega \in \Omega: U \left({\omega}\right) = u}\right\}$ such that $U_u \ne \varnothing$.

We have that $U_u = \left\{{\omega \in \Omega: X \left({\omega}\right) + Y \left({\omega}\right) = u}\right\}$.

Consider any $\omega \in U_u$.

Then:
 * $\omega \in X_x \cap Y_x$

where:
 * $X_x = \left\{{\omega \in \Omega: X \left({\omega}\right) = x}\right\}, Y_x = \left\{{\omega \in \Omega: Y \left({\omega}\right) = u - x}\right\}$

Because $X$ and $Y$ are discrete random variables, both $X_x \in \Sigma$ and $Y_x \in \Sigma$.

As $\left({\Omega, \Sigma, \Pr}\right)$ is a probability space, then $X_x \cap Y_x \in \Sigma$.

Now note that:
 * $\displaystyle U_u = \bigcup_{x \mathop \in \R} \left({X_x \cap Y_x}\right)$

That is, it is the union of all such intersections of sets whose discrete random variables add up to $u$.

As $X_x$ is a countable set it follows that $U_u$ is a countable union of countable sets.

From Countable Union of Countable Sets is Countable it follows that $X_x$ is a countable set.

And, by dint of $\left({\Omega, \Sigma, \Pr}\right)$ being a probability space, $U_u \in \Sigma$.

Thus $U$ is a discrete random variables on $\left({\Omega, \Sigma, \Pr}\right)$.