Problem
In any basic probability course, we learn that the Bernoulli distribution is a discrete probability distribution. However, when using measure theory, it seems that it is actually not.
Context
I could not find any measure-theoretic description of a Bernoulli random variable and distribution, so I had to come up with one, but this makes the Bernoulli distribution not discrete.
Let $([0, 1], \mathcal{B}([0,1]), \text{Leb})$ be a probability space, where $\mathcal{B}([0, 1])$ denotes the Borel sigma-algebra on $[0, 1]$ and $\text{Leb}$ the Lebesgue measure on it. Given $p\in[0, 1]$ define the Bernoulli random variable as $X_p:[0, 1]\to\{0, 1\}$
$$
X_p(\omega) =
\begin{cases}
1 & \omega \leq p \\
0 & \omega > p
\end{cases}\qquad \forall\, \omega\in[0, 1].
$$
The distribution of this random variable is $\mu_p = \text{Leb} \circ X^{-1}_p$ is a probability measure on $(\{0, 1\}, \mathcal{P}(\{0, 1\}))$, where $\mathcal{P}(\{0, 1\})$ is the power set on $\{0, 1\}$ and the distribution can be written as
$$
\mu_p(\mathsf{A}) = \sum_{x\in\{0,1\}} p^x(1-p)^{1-x} \delta_x(\mathsf{A}) \qquad \forall\, \mathsf{A}\in\mathcal{P}(\{0, 1\}).
$$
Clearly, this distribution is dominated by the counting measure $\mu_p \ll d\text{count}$
$$
d\text{count}(\mathsf{A}) =
\begin{cases}
|\mathsf{A}| & \mathsf{A} \text{ is finite.} \\
\infty & \mathsf{A} \text{ is infinite.}
\end{cases}
$$
However, $X_p$ is not a discrete random variable, since a random variable is discrete if its pre-image is countable. In this case, $X^{-1}(\{0, 1\}) = [0, 1]$ and this is uncountable. Am I missing something?
Notice that one could argue that the base probability space I should use, should not be $([0, 1], \mathcal{P}([0, 1]), \text{Leb})$ but rather any countable/finite measurable space with a normalized counting measure. However, this would make it impossible to have Bernoulli random variables for $p\in[0,1]$ irrational.
Best Answer
A random variable $X$ is discrete if it takes values in a discrete set (so a finite set or a set with the discrete topology).
Discreteness is about the image, not the pre-image.