Exercise about sub-$\sigma$-algebra of $\mathcal{B}(\mathbb{R})$ 
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 $$F\supseteq \mathcal{B}(\mathbb{R})\cap\{A\subseteq\mathbb{R}: A=-A\}$$
 A: I go to prove: $\mathcal{B}(\mathbb{R}) \cap \{ A\subseteq \mathbb{R} \mid A=-A \} \subseteq \mathcal{F}$.
Let $\mathcal{B}=\mathcal{B}(\mathbb{R})$, $\mathcal{B}_{1}=\mathcal{B}([0,\infty))$,
$\mathcal{G}=\{A\subseteq\mathbb{R}\mid A=-A\}$, $\mathcal{C}=\{(-a,a)\mid a>0\}$,
and $\mathcal{F}=\sigma(\mathcal{C})$ on $\mathbb{R}$. Note that
$\mathcal{B}$ and $\mathcal{G}$ are $\sigma$-algebras on $\mathbb{R}$
while $\mathcal{B}_{1}$ is an $\sigma$-algebra on $[0,\infty)$.
Define $f:\mathbb{R}\rightarrow[0,\infty)$ by $f(x)=|x|$. Let $\mathcal{D}=\{[0,a)\mid a>0\}$.
It is well-known that $\mathcal{D}$ is a generator for $\mathcal{\mathcal{B}}_{1}$
(in the sense that $\sigma(\mathcal{D})=\mathcal{B}_{1}$ on $[0,\infty)$).
Clearly, for each $A\in\mathcal{D}$, $f^{-1}(A)\in\mathcal{B}\cap\mathcal{G}$
because $f^{-1}\left([0,a)\right)=(-a,a)$. Therefore $f$ is $(\mathcal{B}\cap\mathcal{G})/\mathcal{B}_{1}$-measurable.
Moreover, we have,
\begin{eqnarray*}
f^{-1}(\mathcal{B}_{1}) & = & f^{-1}(\sigma(\mathcal{D}))\\
 & = & \sigma(f^{-1}(\mathcal{D}))\\
 & = & \sigma(\mathcal{C})\\
 & = & \mathcal{F}.
\end{eqnarray*}
Given $A\in\mathcal{B}\cap\mathcal{G}$, we define $B=[0,\infty)\cap A$.
Note that $A\in\mathcal{B}\Rightarrow B\in\mathcal{B}_{1}$. Therefore
$f^{-1}(B)\in\mathcal{F}$. However, $f^{-1}(B)=B\cup(-B)=A$. This
shows that $\mathcal{B}\cap\mathcal{G}\subseteq\mathcal{F}$. (It
is elementary and routine to check that $B\cup(-B)=A$ by observing
that $A=-A$. For, $B\subseteq A$ by definition. Let $x\in-B$, then
$-x\in B\subseteq A$, and hence $x\in-A=A$. Therefore $-B\subseteq A$ and hence
$B\cup(-B)\subseteq A$. To show the reverse inclusion, let $x\in A$.
If $x\geq0$, then $x\in B$ by the very definition. Suppose that
$x<0$, then $-x>0$. Since $A=-A$, we have $-x\in A$. Therefore
$-x\in B$ and hence $x\in-B$. This shows that $A\subseteq B\cup(-B)$.)
