Why is the Bernoulli distribution not discrete?

Question

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.

Answer 1

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.

Answer 2

I don't think that definition of discrete random variable is a good one (countable preimage). Countable image is more like it, although still not quite right. Note that, since we allow an enormous variety of general measure spaces to model the "same" random variable (one with the same distribution) - in particular we could also model a Bernoulli variable with a finite measure space - such a definition is not even well-defined.

A measure (in total generality) is said to be discrete if it is a countable weighted sum of Dirac measures. A random variable is discrete if its induced distribution measure is discrete. This is clearly true here, since $\mu=(1-p)\delta(0)+p\delta(1)$.