If X is a random variable then so is |X|

Question

Resnick claims that the converse is not necessarily true, but before I can find a counterexample on $\mathbb{R},B(\mathbb{R})$ I need to prove the initial claim.

My understanding is that a formal definition of a random variable is a mapping from a given measurable space onto $\mathbb{R},B(\mathbb{R})$, so intuitively I am guessing I need to show that the inverse image under |X| is a subset of the inverse image under X. Is this the right idea, and if so, how do I show this formally?

If it is a correct intuition, would it then be sufficient to show the two inverse images are not equal, or should I attempt to enumerate a specific example?

Any help would be greatly appreciated.

Answer 1

Your intuition is almost right: $|X|^{-1}[A]=X^{-1}[A]\cup X^{-1}[-A]$, where $-A=\{-a:\ a\in A\}$.

$X$ may not be measurable, even though $|X|$ is. Take a modified characteristic function of any non-measurable subset $A$ of your space: $X(\omega)=1$ if $\omega\in A$ and $X(\omega)=-1$ otherwise. Then $|X|$ is measurable (it is a constant function), but $X$ is not.