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This question comes from the 4th line of the proof of Theorem E of Halmos' "Measure Theory", in page 25, which says that C is a sigma-ring. Because this website does not allow new users to link images, I rephrase it as follows: Suppose A is any subset of the whole space X, E is any collection of subsets of X, S(E) denotes the sigma-ring generated by E, $E\cap A$ means the collection formed by all intersections of elements from E with A. Then the collection of all sets of the form $B\cup(C-A)$ where B is from S($E\cap A$) and C is from S(E) is a sigma-ring.

I just can not prove this because I can not make up the difference $[B1\cup(C1-A)]-[B2\cup(C2-A)]$ into a form of $B\cup(C-A)$. Could you please help me prove this statement? Thanks!

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up vote 1 down vote accepted

How about choosing $B = B_1 - B_2 $ and $C = C_1 -C_2$, we have $ B \in S(E \cap A)$ by property of $ S(E \cap A) $, and $ C \in S(E) $, by property of $S(E)$.

The fact that this works can be seen by observing that both $B1 \cup (C1−A)$, and $B2\cup (C2−A)$ are in fact both disjoint unions. $B_1,B_2$ has everything to do with $A$ and $C_1 -A, C_2 -A$ has nothing to do with $A$ at all.

ps: I don't know why you use the and and or symbol here instead of intersection and union symbol.

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I got it. I managed to fill the gaps of your exposition and got it, Thank you very much. ps: I don't know how to input latex formula in this website so I use /\ and \/ to represent set intersection and union, but they are later editted to \vee and \wedge by Yemon Choi. – zzzhhh Apr 1 '10 at 8:41
you are welcome! – abcdxyz Apr 1 '10 at 10:21
I've re-edited to change $\wedge$ and $\vee$ to $\cap$ and $\cup$, respectively. – José Figueroa-O'Farrill Apr 1 '10 at 14:18

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