Consider on $\Bbb{R}$ the family $\Sigma $ of all Borel sets which are symmetric w.r.t. the origin. Show that $\Sigma $ is a $\sigma $-algebra. 

  Consider on $\Bbb{R}$ the family $\Sigma $ of all Borel sets which are symmetric w.r.t. the origin. Show that $\Sigma $ is a $\sigma $-algebra.

Context: Preparing for my exam
Effort:


*

*To show that $\Bbb{R}\in \Sigma $, note that $\Bbb{R}$ is a Borel set that is symmetric w.r.t. to the origin.

*To show that $A\in \Sigma \Rightarrow A^c\in \Sigma $, assume $A\in \Sigma $. Then $A=B\cup -B$ for some $B\in\mathcal{B}([0,\infty ))$. Then $A^c=B^c\cap (-B)^c$. How do I show that this is writable as $B' \cup -B'$, where $B'\in\mathcal{B}([0,\infty))$ ?

*Showing that $\Sigma$ is stable under countable unions seems clear to me.

 A: Another way to say "$A$ is symmetric wrt the origin" is "$\forall x \in A, -x \in A$". Now if $A$ is symmetric wrt the origin and $x \in A^c$, then $-x \in A^c$; for if we had $-x \not\in A^c$, then $-x \in A \implies x \in A$, contradiction. Therefore $A^c$ is symmetric wrt to the origin (and it's Borel, being the complement of a Borel set).
If you wanted to go in your intended direction, then you can simply take $B' = A^c \cap [0, +\infty)$; it's of course Borel, but to prove $A^c = B' \cup -B'$ you would essentially have to repeat the reasoning of my first paragraph.
A: To show that $A\in \Sigma \Rightarrow A^c\in \Sigma $, it suffices to show that
$$
\forall x\in A:-x\in A \Longrightarrow  \forall y\in A^c:-y\in A^c,
$$
which is equivalent with showing that 
$$
\forall x\in A:-x\in A \Longrightarrow \forall y\not\in A : -y\not\in A,
$$
which is equivalent with showing that
$$
\exists y\not\in A:-y\in A \Longrightarrow \exists x\in A:-x\not\in A.
$$
This last statement hold if we set $x:=-y.$ 
