Does the collection of all $0$-$1$ random variables form a set?

A rather trivial example but consider the subset of all random variables such that:

1. The sample space is an ordinal
2. The event space is just $\{0\}$

This collection has a random variable for each ordinal, and hence is not a set.

The Shiranai answer explains that we can make any nonempty set $\Omega$ a sample space, and the "set" of all nonempty sets $\Omega$ is too large to be a set.

However, there is a sense in which there are "just as many" Bernoulli random variables as there are general random variables: We can define two random variables $X$ and $Y$ to be in the same equivalence class if they have the same cumulative distribution function (CDF), that is, if $P[X\leq x] = P[Y\leq x]$ for all $x \in \mathbb{R}$. Now define:

• $\mathcal{B}$ is the set of all equivalence classes of Bernoulli random variables.

• $\mathcal{V}$ is the set of all equivalence classes of random variables.

Let $|A|$ denote the cardinality of a set $A$.

Claim:

$|\mathcal{B}|=|\mathcal{V}| =|\mathbb{R}|$, where $\mathbb{R}$ denotes the set of real numbers.

Proof:

Each Bernoulli random variable $X$ has a distribution that is characterized by a probability $p \in [0,1]$ (where $p=P[X=1]$). Thus, $|\mathcal{B}|=|[0,1]|=|\mathbb{R}|$. It is clear that $$ \mathcal{B} \subseteq \mathcal{V}$$ and so $$|\mathcal{B}|\leq |\mathcal{V}|$$ By definition, $|\mathcal{V}|$ is equal to the cardinality of all CDF functions. A general random variable $X$ has a CDF $F_X:\mathbb{R}\rightarrow\mathbb{R}$ that is continuous from the right. Thus, the CDF $F_X:\mathbb{R}\rightarrow\mathbb{R}$ is fully determined by specifying its value $F_X(q)$ for each rational number $q \in \mathbb{Q}$. Indeed, for any $x \in \mathbb{R}$, we can obtain $F_X(x)$ as $$ F_X(x) = \lim_{i\rightarrow\infty} F_X(q_i)$$ where $\{q_1, q_2, q_3, ...\}$ is a sequence of rational numbers that approaches $x$ from the right. So $|\mathcal{V}|\leq |\mathbb{R}^{\mathbb{Q}}| = |\mathbb{R}|$. In particular: $$ |\mathbb{R}|=|\mathcal{B}|\leq |\mathcal{V}| \leq |\mathbb{R}^{\mathbb{Q}}| = |\mathbb{R}|$$ $\Box$