Given a probability space $(\Omega, F, P)$, I know that for each event $A \in F$, we can define an indicator random variable $I_A$, then $A$ corresponds to $I_A=1$. I am wondering does there exist a "universal" random variable $X$, such that $\forall A \in F, A$ can be represented as $a \leq X \leq b$ for some $a$ and $b \in \mathbb{R}.$

Moreover, given a probability space, does the set of all random variables have any structure? If the "universal" random variable exists, is it unique and what is its role in this set?

  • 1
    $\begingroup$ The set of all random variables is the set of all $\mathfrak F$-measurable functions on $\Omega$ and it has lots of natural, useful structures. The question of the existence of a "universal" variable is quickly and simply addressed by considering two different measurable sets and their complements: soon you will see it's impossible for certain combinations of those sets to correspond to intervals of real numbers. That, unfortunately, doesn't leave one much to write about! $\endgroup$
    – whuber
    Mar 9, 2021 at 19:04

1 Answer 1


The result is simple but it's surprisingly hard to demonstrate rigorously: when such a "universal random variable" $X$ exists, its image has at most two elements, which implies the sigma algebra $\mathfrak F$ has at most four events.

Let's begin by establishing definitions and notation.

I understand that an event "$A$ can be represented as $a\le X \le b$" means $a=a(A)$ and $b=b(A)$ are numbers determined by $A$ and $X$ and that $A$ is the set of all outcomes for which the value of $X$ lies between $a$ and $b,$

$$A = X^{-1}([a,b]) = \{\omega\in\Omega\mid a \le X(\omega)\le b\}.$$

To avoid discussing infinities, compose $X$ with the (measurable, strictly increasing) function $x\to 1/(1 + \exp(-x))$ to assure the image of $X$ is in the interval $[0,1].$ This does not change the universal representation property of $X.$

Now we can carry out the analysis.

Suppose the image $X(\Omega)$ of $X$ contains more than two values. Pick three of them and call them $x \lt y \lt z.$

Because $\{y\}$ is measurable, $B = X^{-1}(y)$ is a (nonempty) event. The complement $B^\prime = \Omega \setminus B$ also is an event. But

$$X(\omega) = y \in [x,z] \subseteq [a(B^\prime), b(B^\prime)]$$

implies, by virtue of the representation property of $X,$ that $\omega\in B^\prime;$ that is, $\omega\notin B.$ This contradiction of the assumption $\omega\in B$ implies the original supposition is false: it is not possible for the image of $X$ to contain more than two values, QED.


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