Can non-random variables be expressed as random variables with probability 1? Suppose that $a = 5$. In this case, $a$ is not random since it is always equal to $5$. Can $a$ instead be interpreted as a random variable $X$ such that
\begin{equation}
p(X=x) =
\begin{cases}
1 &\text{if} \quad x = 5 \\
0 &\text{otherwise}
\end{cases}
\end{equation}
If so, would $a$ be interpreted as a discrete random variable, as indicated by the probability mass function above, or as a continuous random variable?
 A: Given a probability space $(\Omega, \mathscr F, P)$, a random variable is a Borel function $X : \Omega\to\mathbb R$. That's all there is to it.
For $a\in\mathbb R$ let $X_a : \Omega\to\mathbb R$ be defined by $X_a(\omega) = a$ for all $\omega\in\Omega$. Consider a Borel set $B \subseteq \mathbb R$. Then
$$
X_a^{-1}(B) = \{\omega\in \Omega : X_a(\omega) \in B\} = \begin{cases}\Omega & a \in B \\ \emptyset & a \notin B\end{cases}.
$$
No matter how coarse the $\sigma$-algebra $\mathscr F$ is, it will always be the case that $\{\emptyset, \Omega\}\subseteq \mathscr F$ so this means that $X_a$ is Borel.
The distribution for $X_a$ is $P\circ X_a^{-1}$ and
$$
(P\circ X_a^{-1})(B) = \mathbf 1_{a\in B}
$$
i.e. any set containing $a$ has a measure of $1$ and any set not containing $a$ has a measure of zero. Effectively this is a point mass of $1$ on $a$, and the corresponding pmf is just $P(X=x) = \mathbf 1_{x = a}$.
This distribution is not absolutely continuous w.r.t. the Lebesgue measure so it does not have a pdf in the usual sense. The Dirac delta is a way to treat it as if it does, but really integrating with respect to a Dirac delta is just using a different measure.
