¿If a Borel $\sigma$-algebra is generated by a collection of subsets of algebra, then the Borel $\sigma$-algebra is generated by the algebra?

Question

Let $\left(M,\sigma\left(\tau\right)\right)$ a measure space with $\sigma\left(\tau\right)$ a Borel $\sigma$-algebra where $\tau$ is a topology in $M$. Suppose there is a algebra $\Gamma$ in $M$ containing a collection $\mathcal{B}$. If $\sigma\left(\tau\right)$ is generated by $\mathcal{B}$, then $\sigma\left(\tau\right)$ is generated by $\Gamma$.

Answer 1

This is not necessarily true, because nothing excludes the algebra $\Gamma$ from being “too large.” To see a counterexample, suppose that $M=\mathbb R$, let $\Gamma=2^{\mathbb R}$, the set of all subsets of $\mathbb R$, and $\mathcal B=\tau$, where $\tau$ is the usual topology on $\mathbb R$. Clearly, $\Gamma$ is an algebra, $\mathcal B\subseteq\Gamma$, and $\sigma(\mathcal B)=\sigma(\tau)$. But $\sigma(\tau)\neq 2^{\mathbb R}=\Gamma=\sigma(\Gamma)$.