# If $|X|$ is $\mathcal{F}$-measurable. Is $X$ $\mathcal{F}$-measurable, too?

My thoughts: Yes, it is measurable.

$|X|:\Omega\to\mathbb{R}$ is a random Variable which is $\mathcal{F}$-measurable. Then $\left\{ \omega:|X\left(\omega\right)|\leq x\right\} \in\mathcal{F}$ for all $x\in\mathbb{R}$. $\Rightarrow$ $\left\{ \omega:-x\leq X\left(\omega\right)\leq x\right\} \in\mathcal{F}$ for all $x\in \mathbb{R}$.

So what I have to show now, is that every element of the Borel $\sigma$-Algebra is constructable by these sets.

Because of the symmetry I seem not to be able to do that. Any hints? But I am pretty sure, that it is possible. Thanks for any advice.

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You have show that $X^{-1}[-x,x]$ is measurable for each $x$, but since the intervals $[-x,x]$ are symmetric with respect to $0$, the $\sigma$-algebra generated by such sets will consist of sets which are symmetric with respect to $0$. – Davide Giraudo May 2 '12 at 10:59
@DavideGiraudo Thank you very much for pointing that out! – Aufwind May 2 '12 at 11:02

If $\mathcal F\neq 2^{\Omega}$, we have a counter-example: take $A$ a non-measurable set, and put $f(x)=\begin{cases}1&\mbox{ if }x\in A\\\ -1&\mbox{ if }x\notin A \end{cases}$. Then $|f|=1$ and is measurable, but $f$ is not since $\{f>0\}=A$ is not measurable.