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Math Help - Ask a easy question

  1. #1
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    Ask a easy question

    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.
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  2. #2
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    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.
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