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I have been reviewing some real analysis, here's is question 1.1 from Folland:

A family of sets $\mathcal{R} \in \mathcal{P}(X)$ is a $\sigma$-ring if it is closed under differences (i.e. if $E,F \in \mathbb{R}$ then $E \setminus F \in \mathcal{R}$) and countable unions.

Part (c) of the problem is:

If $\mathcal{R}$ is a $\sigma$-ring, then $\mathcal{M}= \{ E\subset X : E \in \mathcal{R} \text{ or } E^c \in \mathcal{R} \} $ is a $\sigma$- algebra.

If I take $X = \mathbb{R}$, $\mathcal{R} = \{\{0\}\}$, which is definitely a $\sigma$-ring, then $\{0\}, \mathbb{R}\setminus \{0\} \in \mathcal{M}$, but $\mathbb{R} = \{0\} \cup \mathbb{R}\setminus \{0\} \not\in \mathcal{R}$, nor $\emptyset \in \mathcal{R}$. What's wrong here?

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Your $\mathcal{R}$ is not a $\sigma$-ring because it is not closed under differences: $\{ 0 \} \setminus \{ 0 \} = \emptyset \notin \mathcal{R}$. –  Arthur Fischer Sep 20 '12 at 9:36
    
Can I ask for a hint here? I'm not so sure what to do when $A \in \mathcal{R}$, $B \not\in \mathcal{R}$, then how do I show $A \cup B \in \mathcal{M}$? –  user41894 Sep 20 '12 at 10:22

2 Answers 2

If $A\in\mathcal R,B\notin\mathcal R$, then $B^c\in\mathcal R$, and because $\mathcal R$ is closed under differences, $B^c\setminus A=A^c\cap B^c\in\mathcal R$. Hence $(A^c\cap B^c)^c=A\cup B\in\mathcal M$.

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Hint: To prove that $\mathcal{M}$ is closed under countable unions, note that the original $\sigma$-ring $\mathcal{R}$ is closed under nonempty countable intersections, and that $E \cup F^c = ( F \setminus E )^c$.

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