A function on a probability space that is not a random variable

Question

I am suffering right now studying measure space, and I have a question. First of all, let $(\Omega, \mathcal{F}, P)$ be a given probability space. As I understood, to define a distribution function $F$, it is important that $\{X\in A\}\in \mathcal{F}$ for each Borel $A\in \mathbb{R}$, because since $X$ is measurable, $X^{-1}(A)=\{X\in A\}=\{\omega: X(\omega)<x\}$ is an event. Is it correct? Also, are there any examples of a function on the space which is not a random variable?

Answer 1

Consider $\Omega = \{a,b,c\}$. $\mathcal F = \big\{\emptyset, \{a,b\}, \{c\}, \{a,b,c\}\big\}$. Verify that $\mathcal F$ satisfies the requirements for being a $\sigma$-algebra. Note that $P(\cdot)$ assigns probabilities only to members of $\mathcal F$. In particular, $\{a\} \notin \mathcal F$ and hence has no probability assigned to it by $P(\cdot)$.

Now, consider the function $X$ that maps $a\to 0, b\to 1, c\to 2$. Notice that the Borel set $\{X\leq 0\}$ is such that $\{\omega \colon X \leq 0\} = \{a\} \notin \mathcal F$, and so the function $X\colon \Omega \to \mathbb R$ does not satisfy all the requirements for a function to be deemed a random variable.

And don't say that $X$ would indeed be a random variable if only I had defined $\mathcal F$ to be the power set of $\Omega = \{a,b,c\}$ instead of the sub-algebra that I chose. While your allegation is correct, it is irrelevant. The definition of $(\Omega, \mathcal F, P)$ is as given, and the function $X$ is not a random variable on this particular probability space.