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

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