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 $$F\supseteq \mathcal{B}(\mathbb{R})\cap\{A\subseteq\mathbb{R}: A=-A\}$$

  • $\begingroup$ With $-A$ I mean $\{-a: a \in A \}$. For example, $(-a, a)= -(-a, a)$. $\endgroup$ – user May 10 at 14:15
  • 1
    $\begingroup$ This seems correct to me. EDIT: This proof is actually very elegant $\endgroup$ – Maximilian Janisch May 16 at 19:24
  • $\begingroup$ I am much too tired right now to check the details but this is exactly the right idea. $\endgroup$ – DanielWainfleet May 18 at 4:57
  • $\begingroup$ I wonder if it would be possible to use that $\Bbb R$ and $(0,a)$ are homeomorphic (e.g. by $g(x)= a\cdot e^x/(1+e^x)$) $\endgroup$ – Mirko May 23 at 2:38

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 $A\subseteq B\cup(-B)$.)


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