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The question is asking to prove that the family: $\{(−a, a) : a \in \mathbb{R}\}$ does not generate the Borel $\sigma$-algebra.

It is known that the family $\{(a,b) : a < b\}$ generates the Borel $\sigma$-algebra. I am confused about the restriction in the above problem preventing it from generating the Borel $\sigma$-algebra.

Thanks!

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Let $A$ be a set generated by the family $\{(-a,a), a\in \mathbb R\}$. If $x\in A$ then also $-x \in A$. Hence, this family can only generate sets that are symmetric with respect to the origin.

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Let $B_1=\{X:X\in B(R)~\&~-X=X\}$. Clearly, $B_1$ is $\sigma$ algebra of subsets of $R$ such that $\{(-a,a):a \in R\} \subseteq B_1$. Hence $\sigma(\{(-a,a):a \in R\}) \subseteq B_1$, but $B(R) \setminus B_1 \neq \emptyset$ (because $\{1\} \notin B_1$) which implies $B(R) \setminus \sigma(\{(-a,a):a \in R\}) \neq \emptyset$.

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