Let be $\mathcal{M}$ a family of sets that is $\cap$-stable and $\delta(\mathcal{M})$ denotes the Dynkin-system generated by $\mathcal{M}$.
We define $$ \mathcal{G}_1:=\{A\in\delta(\mathcal{M})\mid A\cap B \in\delta(\mathcal{M})\text{ for all } B\in\mathcal{M}\} $$ and see that $\mathcal{M}\in\mathcal{G}_1$ and that $\mathcal{G}_1$ is a Dynkin-system.
Next, we define $$ \mathcal{G}_2:=\{A\in\delta(\mathcal{M})\mid A\cap B \in\delta(\mathcal{M})\text{ for all } B\in\delta(\mathcal{M})\}. $$
Now, our professor stated that we can easily see that $\mathcal{M}\subseteq \mathcal{G}_2$. However, I don't see it. Moreover, I doubt that it's true but I couldn't produce a counter example. Maybe someone else can explain $\mathcal{M}\subseteq \mathcal{G}_2$ or show that this subset property is false.
Fix $A\in \mathcal M$. $\{B \in \delta (\mathcal M): A\cap B \in \delta (\mathcal M) \}$ is a Dynkin-system which contains $\mathcal M$. Hence, it contains $\delta (\mathcal M)$. This means that every $B$ in $\delta (\mathcal M)$ belongs to $\{B \in \delta (\mathcal M): A\cap B \in \delta (\mathcal M) \}$. In other words, $B\in \delta (\mathcal M)$ implies $A\cap B \in \delta (\mathcal M)$, so $A \in \mathcal G_2$. This proves that $\mathcal{M}\subseteq \mathcal{G}_2$.