Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here is exercise 1.1(c) from Folland's Real Analysis:

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.

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

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?

share|improve this question

2 Answers 2

First note that your proposed $\sigma$-ring is not a $\sigma$-ring: $\{ 0 \} \in \mathcal{R}$, but $\varnothing = \{ 0 \} \setminus \{ 0 \} \notin \mathcal{R}$.


Note that since $\bigcap_{n=1} E_n = E_1 \setminus \bigcup_{n=2} ( E_1 \setminus E_n )$ it follows that $\mathcal{R}$ is closed under intersections of nonempty countable subfamilies.

We will also make use of the following set identity: $$F \cup E = ( E^\text{c} \setminus F )^\text{c}.$$

Note, first, that $\mathcal{M}$ is clearly closed under complements, so it suffices to show that it is closed under countable unions. If $\mathcal{A}$ is a countable subfamily of $\mathcal{M}$, let $\mathcal{A}^+ := \mathcal{A} \cap \mathcal{R}$, and $\mathcal{A}^- := \mathcal{A} \setminus \mathcal{R}$. As $\mathcal{A}^+$ is a countable subfamily of $\mathcal{R}$, then $A := \bigcup \mathcal{A}^+ \in \mathcal{R}$. If $\mathcal{A}^- = \varnothing$, there is nothing else to do, so we assume that $\mathcal{A}^- \neq \varnothing$. Using the above set identity (and de Morgan's laws) it follows that $$\begin{multline} {\textstyle \bigcup} \mathcal{A} = {\textstyle \bigcup} \mathcal{A}^+ \cup {\textstyle \bigcup} \mathcal{A}^- = A \cup {\textstyle \bigcup_{E \in \mathcal{A}^-}} E = %\\ = {\textstyle \bigcup_{E \in \mathcal{A}^-}} ( A \cup E ) % = {\textstyle \bigcup_{E \in \mathcal{A}^-}} ( E^\text{c} \setminus A )^\text{c} = ( {\textstyle \bigcap_{E \in \mathcal{A}^-}} ( E^\text{c} \setminus A ) )^\text{c}. \end{multline}$$ Since $E^\text{c} \in \mathcal{R}$ for each $E \in \mathcal{A}^-$, the closure properties of $\mathcal{R}$ imply that $\bigcap_{E \in \mathcal{A}^-} ( E^\text{c} \setminus A ) \in \mathcal{R}$, and so $\bigcup \mathcal{A} \in \mathcal{M}$.

share|improve this answer

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 $(A^c\cap B^c)^c=A\cup B\in\mathcal M$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.