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.