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Thread: 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 $\displaystyle \bf C$ is a $\displaystyle \sigma$-ring which can be verified easily, but I just can not verify this easy statement because I can not make up the difference $\displaystyle [B_1\cup(C_1-A)]-[B_2\cup(C_2-A)]$ into a form of $\displaystyle B\cup(C-A)$. Could you please help me with this problem? Thanks!

    NOTE: Here $\displaystyle \bf S(E)$ means the $\displaystyle \sigma$-ring generated by $\displaystyle \bf E$.
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  2. #2
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    To show that $\displaystyle \bf C$ is a $\displaystyle \sigma$-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 $\displaystyle B_1\cup(C_1-A), B_2\cup(C_2-A)$ are two arbitrary elements of $\displaystyle \bf C$ where $\displaystyle B_1,B_2\in{\bf S(E}\cap A)$ and $\displaystyle C_1,C_2\in\bf S(E)$, we need to show that their difference also assumes the form of $\displaystyle B\cup(C-A)$ and therefore belongs to $\displaystyle \bf C$. By taking at most countable union of sets in $\displaystyle \bf E$, the Theorem B in page 22 of this book can be extended to $\displaystyle \sigma$-ring case, that is, If $\displaystyle \bf E$ is any class of sets, then every set in $\displaystyle \bf S(E)$ can be covered by at most countable union of sets in $\displaystyle \bf E$. So $\displaystyle B_1\subseteq\bigcup\limits_{i = 1}^\infty(E_i\cap A)=(\bigcup\limits_{i = 1}^\infty E_i)\cap A$ where $\displaystyle E_i\in\bf E$, it means $\displaystyle B_1\subseteq A$, so $\displaystyle B_1$ is disjoint from $\displaystyle C_1-A$, and both $\displaystyle B_1,B_2$ is disjoint from both $\displaystyle C_1-A$ and $\displaystyle C_2-A$ by the same argument. It can be proved that if both sets $\displaystyle A_1$ and $\displaystyle A_2$ are disjoint from both sets $\displaystyle B_1$ and $\displaystyle B_2$, we have $\displaystyle (A_1\cup B_1)-(A_2\cup B_2)=(A_1-A_2)\cup(B_1-B_2)$, so $\displaystyle [B_1\cup(C_1-A)]-[B_2\cup(C_2-A)]=(B_1-B_2)\cup[(C_1-A)-(C_2-A)]$$\displaystyle =(B_1-B_2)\cup[(C_1-C_2)-A]$. Because $\displaystyle B_1-B_2$ is in $\displaystyle {\bf S(E}\cap A)$ and $\displaystyle C_1-C_2$ is in $\displaystyle \bf S(E)$ by definition of $\displaystyle \sigma$-ring, the result above is of the desired form and the easy statement is verified not very easily.
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