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Let sample space be [0,1]. let $\tau$ be the collections of all the open intervals on [0,1]. Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing $\tau$. Is the statement any $\sigma$-algebra including $\tau$ on [0,1] contains the Borel $\sigma$ true? If so can anyone show me the proof?

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Yes, this is true.

Let $F$ be the smallest set from the family of sets $\mathcal F$ which contains $A$. Then $F\subseteq G$ whenever $G\in \mathcal F$ contains $A$. This is in fact the definition of the smallest set in $\mathcal F$ containing $A$. For example, it applies for Borel $\sigma$-algebras.

Little disclaimer: I'm not sure I shall use "sets" or "classes" here, but hope it's fine.

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The intersection of an arbitrary collection of $\sigma$-algebras is a $\sigma$-algebra.

The Borel $\sigma$-algebra $\cal B$ generated by $\tau$ is the smallest $\sigma$-algebra containing $\tau$, that is the intersection of all $\sigma$-algebras containing $\tau$. Hence if $\cal A$ is a $\sigma$-algebras containing $\tau$, then ${\cal B} \subset {\cal A}$.

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