I am reading Probability and Random Process and in chapter 4.2 where the author says something like that given a random variable $X$, $f(X)$ is also a valid random variable if $f$ is sufficiently smooth or regular by being continuous or monotonic. This explanation is a little bit abstract for me. Can anybody explain this further or maybe provide a counter example? Thanks a lot!

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    $\begingroup$ The real answer is that f must not be measurable in the Borel sense. $\endgroup$ Commented Mar 5, 2021 at 2:47
  • $\begingroup$ @MartínVacasVignolo But what does measurability have to do with smoothness? $\endgroup$
    – XXX
    Commented Mar 5, 2021 at 2:49
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    $\begingroup$ @Gracie continuous functions are Borel-Borel measurable, as are monotone functions. You can see this fails if you let $f$ be the indicator function of a nonmeasurable set on $[0,1]$ and $X$ is just the identity. $\endgroup$ Commented Mar 5, 2021 at 3:04

1 Answer 1


As defined by Grimmett and Stirzaker in sec. 2.1, a random variable $X$ on the sample space $\Omega$ with $\sigma$-algebra $\mathcal F$ is a real-valued function $X$ on $\Omega$ such that for every real $x$, $\{\omega \in \Omega : X(\omega) \le x \} \in \mathcal F$. Equivalently, for every Borel set $B \subseteq \mathbb R$, $\{\omega \in \Omega: X(\omega) \in B\} \in \mathcal F$.

Now $\{\omega \in \Omega: f(X(\omega)) \in B\} = \{\omega \in \Omega: X(\omega) \in f^{-1}(B)\}$. So in order for $f(X)$ to be a random variable (with the same $\mathcal F$), it suffices for $f^{-1}(B)$ to be Borel whenever $B$ is Borel. That is true if $f$ is a Borel measurable function. In particular it is true if $f$ is continuous.

On the other hand, for an example where $f(X)$ is not a random variable, you might take $\Omega = \mathbb R$ with Borel $\sigma$-algebra $\mathcal F$, and $X(\omega)=\omega$, and $f$ some function that is not Borel measurable.


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