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Let $X^2$ be a real random variable. Is $X$ a random variable, too?

I figured out that $X$ is not a random variable.

I want to give a counter example.

So, I tried:

Let $A \subset \mathbb{R}$ be a non-measurable subset.

$X(\omega)=\mathbf{1}_A(\omega)-\mathbf{1}_{A^c}(\omega),\quad\omega\in\mathbb{R}$

So $X$ is not a random variable.

Is this way correct or is there another counter example for that?

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    $\begingroup$ The map $x\mapsto \sqrt x$ is measurable so $X = \sqrt{X^2}$ is a random variable (here we are considering the positive square root). $\endgroup$
    – Math1000
    Nov 30, 2019 at 18:55
  • $\begingroup$ Your example wouldn't work anyway as then $$X^2(\omega) = \mathsf 1_A(\omega) - 2\mathsf 1_A(\omega)\mathsf 1_{A^c}(\omega) + \mathsf 1_{A^c}(\omega) = \mathsf 1_A(\omega) + \mathsf 1_{A^c}\omega) = X(\omega),$$ since $\mathsf 1_A(\omega)\mathsf 1_{A^c}(\omega) = 0$. $\endgroup$
    – Math1000
    Nov 30, 2019 at 18:59

1 Answer 1

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Yes, your example is valid. For this case $X(\omega)$ is not a random variable since $X^{-1}(\{1\})=A\not\in \mathcal F$ and $X^2(\omega)\equiv 1$ is measurable.

All counterexamples here should have the same structure. The only difference can be that you do not need to take $\Omega=\mathbb R$.

Let $\Omega=\{-1,1\}$ with trivial $\sigma$-algebra $\mathcal F=\{\varnothing, \Omega\}$. Let $X(\omega)=\omega$. It is not measurable since for Borel set $\{1\}$ $$ X^{-1}(\{1\})=\{1\}\not\in \mathcal F. $$ And $X^2(\omega)=1$ is $\mathcal F$-measurable.

Note that it is just the same example as you provided: $X(\omega)=\mathbf{1}_{\{1\}}(\omega)-\mathbf{1}_{\{-1\}}(\omega),\quad\omega\in\Omega$.

So the right statement should be: if $X^2$ is a random variable than $X$ need not to be a random variable.

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