This question comes from page 25, line 4, of Halmos' "Measure Theory", as the image below shows. This line says that $\bf C$ is a $\sigma$-ring which can be verified easily, but I just can not verify this easy statement because I can not make up the difference $[B_1\cup(C_1-A)]-[B_2\cup(C_2-A)]$ into a form of $B\cup(C-A)$. Could you please help me with this problem? Thanks!
NOTE: Here $\bf S(E)$ means the $\sigma$-ring generated by $\bf E$.
2. To show that $\bf C$ is a $\sigma$-ring, it is necessary to prove that it is closed under at most countable union and difference. The union part is really very easy and we prove the difference part. Suppose $B_1\cup(C_1-A), B_2\cup(C_2-A)$ are two arbitrary elements of $\bf C$ where $B_1,B_2\in{\bf S(E}\cap A)$ and $C_1,C_2\in\bf S(E)$, we need to show that their difference also assumes the form of $B\cup(C-A)$ and therefore belongs to $\bf C$. By taking at most countable union of sets in $\bf E$, the Theorem B in page 22 of this book can be extended to $\sigma$-ring case, that is, If $\bf E$ is any class of sets, then every set in $\bf S(E)$ can be covered by at most countable union of sets in $\bf E$. So $B_1\subseteq\bigcup\limits_{i = 1}^\infty(E_i\cap A)=(\bigcup\limits_{i = 1}^\infty E_i)\cap A$ where $E_i\in\bf E$, it means $B_1\subseteq A$, so $B_1$ is disjoint from $C_1-A$, and both $B_1,B_2$ is disjoint from both $C_1-A$ and $C_2-A$ by the same argument. It can be proved that if both sets $A_1$ and $A_2$ are disjoint from both sets $B_1$ and $B_2$, we have $(A_1\cup B_1)-(A_2\cup B_2)=(A_1-A_2)\cup(B_1-B_2)$, so $[B_1\cup(C_1-A)]-[B_2\cup(C_2-A)]=(B_1-B_2)\cup[(C_1-A)-(C_2-A)]$ $=(B_1-B_2)\cup[(C_1-C_2)-A]$. Because $B_1-B_2$ is in ${\bf S(E}\cap A)$ and $C_1-C_2$ is in $\bf S(E)$ by definition of $\sigma$-ring, the result above is of the desired form and the easy statement is verified not very easily.