Sigma algebra generated by an absolute value random variable I need to find out what the sigma algebra generated by $Y$ looks like for $$Y: [0,1] \ni (\omega) \to 1- |2\omega -1| \in \mathbb{R}.$$ 
The graph of $Y$ is symmetric with respect to $\omega = \frac{1}{2}$. 
I know the sigma algebra won't be all borel sets on $[0,1]$, because of the symmetry.
$Y^{-1}([a,b]) = [\frac{2-b}{2},\frac{2-a}{2} ] \cup [\frac{a}{2}, \frac{b}{2}]$
This is as far as I got. Could this help me determine what $\sigma(Y)$ is?
EDIT:
If we change the domain of Y so that $$Y: [0,2] \ni (\omega) \to 1- |2\omega -1| \in \mathbb{R}.$$ 
will then $$\sigma(Y) = \{A \subset [0,1] : A \ \text{is a borel set and} \ \forall x \in A: \frac{1}{2} -x \in A \} \cup \mathcal{B}[1,2]?$$ (point in $[1,2]$ have no common values with those in $[0,1]$)
 A: It's the $\sigma$-algebra $$\mathcal{F}=\left\{A\subseteq [0,1]:A\text{ is a borel set }\wedge\left(\forall x\in A,\ {1\over 2}-x\in A\right)\right\}$$
The easy part is showing that $\sigma(Y)\subseteq\mathcal{F}$.
As for the harder part, let $B\in\mathcal{F}$ a borel subset symmetrical with respect to ${1\over2}$. You can show that $Y(B)$ is a borel subset and that $Y^{-1}(Y(B))=B$.
Hint: $Y(B)=Y\left(B\cap\left[{1\over2},1\right]\right)$ and $Y|_{\left[{1\over2},1\right]}$ is the map $(x\mapsto 2-2x)$
Added:
The statement in the second part is false, substantially because the domain is not symmetrical with respect to ${1\over 2}$.
In that case, applying (again) the definition quoted in the other answer:
$$\sigma(Y)=\mathcal{F}':=\left\{A\subseteq [0,2]:A\text{ is a borel set }\wedge\forall x\in A\left({1\over2}-x\in A\vee{1\over2}-x\notin [0,2]\right)\right\}=\\=\left\{A\subseteq [0,2]:A\text{ is a borel set }\wedge\left(\forall x\in A\cap[0,1],\ {1\over2}-x\in A\right)\right\}$$
A: If $\mathcal{B}_{\left[0,1\right]}$ denotes the Borel $\sigma$-algebra
on $\left[0,1\right]$ then $$\sigma\left(Y\right)=\left\{ Y^{-1}\left(B\right)\mid B\in\mathcal{B}_{\left[0,1\right]}\right\}$$
Have a look here
