Halmos Measure Theory Sec.6 Theorem B The following is the exact statement of the theorem. $\pmb M(\pmb R)$ stands for the monotone class generated by $\pmb R$ and $\pmb S(\pmb R)$ the generated sigma ring. I do not understand why $\pmb M$ is a ring. Halmos constructed the class $\pmb K(F)$, but what is the point of having such a class? I can understand everything before and including showing $\pmb K(F)$ is a monotone class. (Cannot comprehend at all what Halmos is doing after that.)
EDIT: Picture taken from google books.


 A: "$M \subset K(E)$. The validity of this relation for every $E$ in $M$ is equivalent to the assertion that $M$ is a ring."
Proof: Suppose $E$ and $F$ are in $M$. Then, by the subset relation, $E$ and $F$ are in $K(F)$. By the definition of $K$, $E - F$, $F-E$, and $E \cup F$ are in $M$. As $E$ and $F$ were chosen arbitrarily, we see that $M$ is closed under relative complementation and finite union. Therefore, $M$ is a ring.
Does that help?
Note that this sort of argument is very common in measure theory. The idea is the following. We want to show, for example, that every set in some class of sets has some property $P$. Usually the class of sets in question has some structure: for example, it may be a monotone class $M(R)$ generated by a ring $R$. To show that all $E \in M(R)$ have property $P$, we define $K$ to be the class of sets in $M(R)$ with property $P$. Now, if we can show that $R \subset K$ (all $E \in R$ have property $P$) and that $K$ is monotone, then by the minimality of $M(R)$ we can conclude that $M(R) \subset K$. And then by the definition of $K$ we can conclude that every member of $M(R)$ has property $P$.
