$\sigma$-Fields

If $\mathcal{F}$ is both a field and a monotone class, then $\mathcal{F}$ is a $\sigma$-field.

A field is a monotone class if and only if it is a $\sigma$-field.

$\mathcal{F}$ satisfies (S1), (S2), and closure under amc intersections iff $\mathcal{F}$ is a $\sigma$-field.

Every field that is closed under countable disjoint unions is a $\sigma$-field.

If $\mathcal{F}$ satisfies (S1), closure under differences, and closure under countable unions or closure under countable intersections, then $\mathcal{F}$ is a $\sigma$-field.

A finite union of $\sigma$-fields on $\Omega$ is not necessarily a field on $\Omega$.

If a finite union of $\sigma$-fields on $\Omega$ is a field, then it is a $\sigma$-field as well.

Given $\sigma$-fields $\mathcal{F}_{1} \subsetneq \mathcal{F}_{2} \subsetneq \cdots$ on $\Omega$, it is not necessarily the case that $\bigcup_{n=1}^{\infty } \mathcal{F}_{n}$ is a $\sigma$-field.

An amc union of sets of the first category is of the first category.

Let $\mathcal{F}=\left\{ A \subseteq \mathbb {R}: A \text{ or } A^{c} \text{ is a set of the first category}\right\}$. Then $\mathcal{F}$ is a $\sigma$-field of subsets of $\mathbb {R}$.

Every $\sigma$-ring is closed under finite unions and amc intersections.

$\mathcal{F}$ is a $\sigma$-field iff $\mathcal{F}$ is a $\sigma$-ring with $\Omega \in \mathcal{F}$.

State and prove an existence and uniqueness result regarding the [minimal] $\sigma$-ring generated by a collection $A$ of subsets of $\Omega$.

For any collection $\mathcal{C}$ of sets, let $\mathcal{C}^{*}$ denote the collection of all countable unions of differences of $\mathcal{C}$-sets. Define $\mathcal{A}_{0}=\mathcal{A}$, and for any ordinal $\alpha \succ 0$, define $\mathcal{A}_{\alpha }=\left(\bigcup_{0 \preceq \beta \prec \alpha } \mathcal{A}_{\beta }\right)^{*}$. Then $0 \preceq \alpha \prec \beta$ implies $\mathcal{A} \subseteq \mathcal{A}_{\alpha } \subseteq \mathcal{A}_{\beta } \subseteq S(\mathcal{A})$.

Letting $\omega$ denote the first uncountable ordinal, show that $S(\mathcal{A})=\bigcup_{0 \preceq \alpha \prec \omega } \mathcal{A}_{\alpha }$.

If $\operatorname {card}(\mathcal{A}) \leq \mathfrak {c}$, then $\operatorname {card}(S(\mathcal{A})) \leq \mathfrak {c}$.

If $\Omega$ is finite, then any $\sigma$-field on $\Omega$ is finite.

For the remainder, assume that $\Omega$ is infinite and that there is a countably infinite $\sigma$-field $\mathcal{F}=\left\{ A_{1}, A_{2}, \ldots \right\}$ on $\Omega$. For each $\omega \in \Omega$, let $B_{\omega }=\bigcap \left\{ A_{n}: \omega \in A_{n}\right\}$. For any distinct $\omega , \omega^{\prime } \in \Omega$, either $B_{\omega }=B_{\omega^{\prime }}$ or $B_{\omega } \cap B_{\omega^{\prime }}=\varnothing$.

From (b), there exists a disjoint collection $\left\{ C_{1}, C_{2}, \ldots \right\}$ of nonempty $\mathcal{F}$-sets with $\bigcup_{n=1}^{\infty } C_{n}=\Omega$.

For each sequence $e=\left(e_{1}, e_{2}, \ldots \right)$ of 0's and 1's, let $D_{e}=\bigcup \left\{ C_{n}: e_{n}=1\right\}$. Then $e \neq e^{\prime }$ forces $D_{e} \neq D_{e^{\prime }}$, and a contradiction ensues. Hence, there cannot exist a countably infinite $\sigma$-field on our infinite $\Omega$.

If $\mathcal{A} \subseteq \mathcal{A}^{\prime } \subseteq \sigma (\mathcal{A})$, then $\sigma \left(\mathcal{A}^{\prime }\right)=\sigma (\mathcal{A})$.

For any collection $\varnothing \neq \mathcal{A} \subseteq 2^{\Omega }, \pi (\mathcal{A}) \subseteq \lambda (\mathcal{A}) \subseteq \sigma (\mathcal{A})$.

If the nonempty collection $\mathcal{A}$ is finite, then $\sigma (\mathcal{A})=f(\mathcal{A})$.

For arbitrary collection $\mathcal{A}$, we have $\sigma (\mathcal{A})=\sigma (f(\mathcal{A}))$.

For arbitrary collection $\mathcal{A}$, we have $f(\sigma (\mathcal{A}))=\sigma (f(\mathcal{A}))$.

Let $\mathcal{C}$ denote the class of sets of the stated form. Then $\Omega \in \mathcal{C}$ and $\mathcal{C}$ is closed under intersection.

Show that $\mathcal{C}$ is closed under complementation by first showing that

where if $n_{i}<2$, $\bigcup_{j=2}^{n_{i}}\left(A_{i j}^{c} \cap \bigcap_{k=1}^{j-1} A_{i k}\right)$ is defined as $\varnothing$, $i=1, \ldots , m$.

For $i=1, \ldots , m$, $\bigcup_{j=2}^{n_{i}}\left(A_{i j}^{c} \cap \bigcap_{k=1}^{j-1} A_{i k}\right) \in \mathcal{C}$. Also, by closure under intersection, we have $\left(\bigcup_{i=1}^{m} \bigcap_{j=1}^{n_{i}} A_{i j}\right)^{c} \in \mathcal{C}$.

$\mathcal{C}$ is a field that contains $\mathcal{A}$, and thus $f(\mathcal{A}) \subseteq \mathcal{C}$.

Every $\mathcal{C}$-set must be in any field containing $\mathcal{A}$, hence $\mathcal{C} \subseteq f(\mathcal{A})$.

$f(\mathcal{A})=\left\{ A \subseteq \Omega : A \text{ is finite or } A^{c} \text{ is finite}\right\}$.

When $\Omega$ is amc, $\sigma (\mathcal{A})=2^{\Omega }$.

For general $\Omega$, $\sigma (\mathcal{A})=\left\{ A \subseteq \Omega : A \text{ or } A^{c} \text{ is amc}\right\}$.

Suppose that $\mathcal{A}$ is such that $\sigma (\mathcal{A})=2^{\Omega }$. For each distinct pair $\omega , \omega^{\prime } \in \Omega$ there exists $A \in \mathcal{A}$ with $\omega \in A$ but $\omega^{\prime } \notin A$.

Let $\Omega$ be countable, and suppose that for each distinct pair $\omega , \omega^{\prime } \in \Omega$, there exists a set $A \in \mathcal{A}$ such that $\omega \in A$ but $\omega^{\prime } \notin A$. Then $\sigma (\mathcal{A})=2^{\Omega }$.

$\sigma (\mathcal{A}) \cap B$ is a $\sigma$-field on $B$.

Next, define $\sigma_{B}(\mathcal{A} \cap B)$ to be the minimal $\sigma$-field over $B$ generated by the class $\mathcal{A} \cap B$. Then $\sigma_{B}(\mathcal{A} \cap B)=\sigma (\mathcal{A}) \cap B$.

To prove the claim, it is sufficient to show that $\sigma (\mathcal{F}) \subseteq m(\mathcal{F})$.

If $m(\mathcal{F})$ is a field, then $\sigma (\mathcal{F}) \subseteq m(\mathcal{F})$.

$\Omega \in m(\mathcal{F})$.

Let $\mathcal{G}=\left\{ A \subseteq \Omega : A^{c} \in m(\mathcal{F})\right\} . \mathcal{G}$ is a monotone class on $\Omega$ and $m(\mathcal{F}) \subseteq \mathcal{G}$.

$m(\mathcal{F})$ is indeed closed under complements.

Let \mathscr {H}_{1}=\left\{ A \subseteq \Omega : A \cup B \in m(\mathcal{F}) for all B \in \mathcal{F}\right\}. Then $\mathscr {H}_{1}$ is a monotone class such that $\mathcal{F} \subseteq \mathscr {H}_{1}$ and $m(\mathcal{F}) \subseteq \mathscr {H}_{1}$.

Let \mathscr {E}_{2}=\left\{ B \subseteq \Omega : A \cup B \in m(\mathcal{F}) for all A \in m(\mathcal{F})\right\}. Then $\mathscr {E}_{2}$ is a monotone class such that $\mathcal{F} \subseteq \mathscr {H}_{2}$, and $m(\mathcal{F}) \subseteq \mathscr {H}_{2}$.

$m(\mathcal{F})$ is closed under finie unions, and hence is a field.