# Generating a $\sigma$-algebra from a $\sigma$-ring

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?

-

## 2 Answers

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 $\bigcup \mathcal{A} \in \mathcal{M}$.

-

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

-