Let $C=\{(-a, a): a \in \mathbb{R}\}$ and $F=\sigma(C)$.
Prove that $F=\mathcal{B}(\mathbb{R})\cap\{A\subseteq\mathbb{R}: A=-A\}$.
I don't have problems in proving $F\subseteq \mathcal{B}(\mathbb{R})\cap\{A\subseteq\mathbb{R}: A=-A\}$, I'd like a confirm for the other inclusion.
Let $F^+=\{A \cap [0,+\infty): A \in F\}$, then $F^+=\mathcal{B}[0,+\infty)$:
$F\subseteq \mathcal{B}(\mathbb{R}) $ and $[0,+\infty) \in \mathcal{B}(\mathbb{R})$ imply $F^+\subseteq \mathcal{B}[0,+\infty)$;
$\forall a,b\in[0,+\infty)$ we have $(a,b)=[0,b) \cap \cup_n[a+n^{-1},+\infty) \in F^+$, but $\mathcal{B}[0,+\infty) $ is generated by those intervals so $F^+\supseteq \mathcal{B}[0,+\infty)$; hence the equality.
Now take $B\in \mathcal{B}(\mathbb{R})\cap\{A\subseteq\mathbb{R}: A=-A\}$; we have $B\cap[0,+\infty)\in \mathcal{B}[0,+\infty)=F^+$. So there exists $A\in F$ such that $A\cap[0,+\infty)=B\cap[0,+\infty) $ which means that A and B have the same positive elements, and since they are equals to their opposites they also have the same negative elements. Therefore $B=A\in F$, from which $$F\supseteq \mathcal{B}(\mathbb{R})\cap\{A\subseteq\mathbb{R}: A=-A\}$$