Let $G_n = \{U \subset \mathbb R^n \mid U \text{ open }\}$.
Why does it hold true that $\sigma(G_n) = m(G_n) = \delta(G_n)$, where $\sigma$ is the Borel $\sigma$-algebra, $m$ the smallest monotone class created by the open sets and $\delta$ the smallest Dynkin-system produced by the open sets?
Best Answer
$1)$ $G_n$ is a family of sets that is closed under finite intersections,so by a known theorem we have that $\sigma(G_n)=\delta(G_n)$
Also note that every closed set in $\Bbb{R}^n$ can be expresed as a countable intersection decreasing open of open sets.
Indeed if $F$ is open then $F=\bigcap_n\{x \in \Bbb{R}^n:d(x,O)<\frac{1}{n}\}$
$m(G_n)$ contains all open and closed sets.
So $m(G_n)$ contains the algebra generated by the open sets.
$2)$ Note that every Dynkin class is a monotone class.
$3)$ Also note that $m(G) \subseteq \delta(G)$, whre $G$ is any family of subsets of $\Bbb{R}^n$
From $1),2),3)$ you have the conclusion.