monotone class theorem, proof I am having difficulty with this proof:


It is the three sentences I have colored that is very difficult. Could someone please explain why they are true?
red line: I understand "Since A is an algebra", but why the conclusion?
green line: "Why does it follow that M(A)=M"?
blue text: I dont understand the conclusion?
 A: Red Line: Take any $A_0 \in \mathcal A$. Since $\mathcal A$ is an algebra, it must be closed under finite intersection and complementation. Thus, $A_0 \cap A \in \mathcal A \subseteq \mathcal M$ and $A_0 $\ $A = A_0 \cap A^c \in \mathcal A \subseteq \mathcal M$. Then by definition of $\mathcal M(A)$, $A_0 \in \mathcal M(A)$ and since $A_0$ was arbitrary, $\mathcal A \subseteq \mathcal M(A)$. 
Green Line: The monotone class generated by $\mathcal A$, which we call $\mathcal M$, is the smallest monotone class containing $\mathcal A$, meaning no other monotone class containing $\mathcal A$ is properly contained inside $\mathcal M$. It was shown above that $\mathcal M(M)$ is a a monotone class for any $M \in \mathcal M$. In particular, since any $A \in \mathcal A \subseteq \mathcal M$, we know that $\mathcal M(A)$ is a monotone class. The Red Line shows that $\mathcal A \subseteq \mathcal M(A)$. In words, it is a monotone class containing the algebra $\mathcal A$. Since $\mathcal M$ is the smallest monotone class containing $\mathcal A$, it must be contained in any other monotone class containing $\mathcal A$. In particular, $\mathcal M \subseteq \mathcal M(A)$. Certainly, since $\mathcal M(A)$ consists only of sets from $\mathcal M$ by definition, we get the opposite containment. So $\mathcal M(A) = \mathcal M$. 
Blue Line: A few words previous to the blue line says that $\mathcal A \subseteq \mathcal M(M)$ for all $M \in \mathcal M$. Again, in words this means that it is a monotone class that contains the algebra $\mathcal A$. Since $\mathcal M$ is the minimal/smallest monotone class containing $\mathcal A$, we must have $\mathcal M \subseteq \mathcal M(M)$. The other inclusion is trivial since $\mathcal M(M)$ is made up of sets from $\mathcal M$. Thus $\mathcal M(M) = \mathcal M$ for all $M \in \mathcal M$.
