What exactly is $\cap$-stable here? From my lecture notes:

Theorem: Let $(\Omega, \mathcal A, P)$ be a probability space, $A \in \mathcal A, \mathcal M := \{ M_1, \ldots, M_n \} \subset \mathcal A$. The following statements are equivalent:
a) for all $E \in \sigma(\mathcal M): A$ and $E$ are independent;
b) for all $J\subset \{ 1, \ldots, n \}$: $A$ and $\cap_{j\in J} M_j$ are independent.
Proof: One direction is trivial. For the other direction note that $$\mathcal D_A:=\{ E \in \mathcal A \ | \   A \text{ and }  E  \text{ are independent} \}$$ is a Dynkin system. It contains all sets of type $\cap_{j\in J} M_j$. And since it is a $\cap$-stable system, $\delta(\mathcal M)=\sigma(\mathcal M)$. Thus, $\sigma(\mathcal M)\subset D_A$.

I don't understand the proof. What exactly is $\cap$-stable here? If $\mathcal M$ were $\cap$-stable, we could conclude that $\delta(\mathcal M)=\sigma(\mathcal M)$. But how does the assumption imply that $\mathcal M$ is $\cap$-stable?
If the proof is indeed false, is there any fix that uses $\mathcal D_A$?
 A: *

*A class $\mathcal{C}$ of sets is $\cap$-stable if it is closed under $\cap$. It means that, if $E,F\in\mathcal{C}$ then $E\cap F\in\mathcal{C}$.

*In the proof of the theorem it is used that the class $\mathcal{B}$ of all sets of type $\cap_{j\in J} M_j$, where $J\subset \{ 1, \ldots, n \}$, is $\cap$-stable. 
It is easy to prove such statement. 


*The final steps of the proof need some fixing/clarifcation. Here is, in detail, how they should be: 


$\mathcal{D}_A$ is a Dynkin system and $\mathcal{B}\subseteq \mathcal{D}_A$ so $\delta(\mathcal{B})\subseteq \mathcal{D}_A$, but since $\mathcal{B}$ is $\cap$-stable, we have $\delta(\mathcal{B})=\sigma(\mathcal{B})$, so we have that $\sigma(\mathcal{B})\subseteq \mathcal{D}_A$. Now, note that $\sigma(\mathcal{B})=\sigma(\mathcal{M})$. So we can conclude that $\sigma(\mathcal{M})\subseteq \mathcal{D}_A$.
A: Here is some detail you might find useful
Consider
$$\mathcal D_A:=\{ E \in \mathcal A \ | \   A \text{ and }  E  \text{ are independent} \}$$
Let $E_1,E_2 \in \mathcal D_A$ therefore 
$$P(A \cap E_1) = P(A)P(E_1)\\
P(A \cap E_2) = P(A)P(E_2) $$
We would like to see that $$P(A \cap E_1 \cap E_2) = P(A)P(E_1)P(E_2) $$
This follows for any pair of sets $E_1,E_2 \in \cap  M_j$?
$$E_1 =  A_{j_1}^1\cap A_{j_2}^1\ldots \cap A_{j_k}^1 \\
 E_2 =  A_{i_1}^2\cap A_{i_2}^2\ldots \cap A_{i_{k'}}^2$$
Consider 
$\{r_1, \ldots r_{k''}\} = \{j_1,\ldots, j_k\} \cap \{i_1,\ldots i_{k'} \}$
$\{J_1, \ldots J_{s'}\} = \{j_1,\ldots, j_k\} \setminus \{i_1,\ldots i_{k'} \}$
$\{I_1, \ldots I_{s'}\} =  \{i_1,\ldots i_{k'} \}  \setminus \{j_1,\ldots, j_k\}$
rewrite to obtain
$$E_1 =  A_{r_1}^1\cap\ldots \cap A_{r_k''}^1\cap A^1_{J_1}\ldots \cap A_{J_s}^1 \\
 E_2 =  A_{r_1}^2\cap \ldots A_{r_{k''}}^2\cap A^2_{I_1}\ldots \cap A_{I_{s'}}^2$$
Therefore 
$$E_1\cap E_2  =  \big(A_{r_1}^1\cap A_{r_1}^2\cap\big)\ldots \cap \big(A_{r_k''}^1\cap A_{r_{k''}}^2\big)\cap A^1_{J_1}\ldots \cap A_{J_s}^1 \cap A^2_{I_1}\ldots \cap A_{I_{s'}}^2 $$
So in order to assure that $E_1\cap E_2$ you must check that each $\big(A_{r_i}^1\cap A_{r_i}^2\cap\big) \in M_{r_i}$ which  will follow if $M_{r_i}$ is $\cap$- stable (in case $M_{r_i}$ is a sigma algebra that follows).
