# Why are random variables measurable?

Why do we require random variables to be measurable?

The only reason I can think of is being able to identify which event happened knowing that the value of random variable (and therefore we can also calculate the probability of random variable assuming that particular value)?

This is what the definition of random variable basically says where it requires the random variable function to be measurable. Do the reasons I gave above make sense?

• This is about the loosest condition required to even make sense of a definition of expected value etc. Nov 9, 2015 at 16:55
• @HagenvonEitzen But is the reason I gave correct? Nov 9, 2015 at 17:08

## 1 Answer

The reason you give in parentheses is the correct one - we need to be able to compute probabilities with random variables, and the only way to do this is to find the pre-image of sets and compute probabilities with them. This requires that the pre-image of (measurable) sets be measurable, which is the definition of a measurable function.

To be more precise, suppose $(\Omega_1,\mathcal{F}_1,P)$ is your probability space, $(\Omega_2,\mathcal{F}_2)$ is your "target" space (this is usually taken to be $\mathbb{R}$ with its Borel sets). Then, suppose we want $X:\Omega_1\rightarrow \Omega_2$ to be a random variable. This means that we want to be able to measure the probability of events like $\{X\in A\}$ where $A\in \mathcal{F}_2$. Well, the way we do this is to find $X^{-1}(A)$, and try to compute $P(X^{-1}(A))$. We can only do this if $X^{-1}(A)\in \mathcal{F}_1$, in other words, if $X$ is a measurable function.

• I'm still not sure why this is defined this way. Intuitively, I would have defined it the other way - for every $A \in \mathcal{F}_1$, $X(A) \in \mathcal{F}_2$. Why doesn't this ensure that the probabilities of all events are defined? And does measurability imply this? Dec 15, 2021 at 11:30