Does the collection of all $0$-$1$ random variables form a set? Where $(\Omega,\mathcal{F},\mu)$ be a probability space, a random variable is a function $X:\Omega\to \mathbb{R}$.
Let us make this simpler... a $0$-$1$ random variable is a function $p:\Omega\to \{0,1\}$.
And now harder...

Consider the collection of ALL 0-1 random variables (over all probability spaces). Does this form a set or is there some strange set theory going on when we consider such a huge collection?

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

*

*The sample space is an ordinal

*The event space is just $\{0\}$
This collection has a random variable for each ordinal, and hence is not a set.
A: 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:
$$ |\mathbb{R}|=|\mathcal{B}|\leq |\mathcal{V}| \leq |\mathbb{R}^{\mathbb{Q}}| = |\mathbb{R}|$$ $\Box$
