Sigma Algebras and Borel Sets — Examples

The family $\{\mathrm{\varnothing },X\}$ satisfies the axioms in [L1]; when $X=\mathrm{\varnothing }$, it is the one-member family $\{\mathrm{\varnothing }\}$.

The power set $\mathcal{P}(X)$ satisfies the axioms in [L1].

Every sigma-algebra contains $\mathrm{\varnothing }$ and hence $X$, and every one is a subfamily of $\mathcal{P}(X)$. Thus the two verified families are the extremes, coinciding when $X=\mathrm{\varnothing }$.

The empty set is at most countable. Complementation exchanges the two alternatives in the definition of $\mathcal{C}$, so $\mathcal{C}$ contains $\mathrm{\varnothing }$ and is complement-closed.

Let $(A_n)$ lie in $\mathcal{C}$. If every $A_n$ is at most countable, [L1] makes ${\bigcup }_n{A}_n$ at most countable. If some $A_j$ is cocountable, then $X\setminus {\bigcup }_n{A}_n\subseteq X\setminus A_j$ is at most countable by [L3]. In either case the union lies in $\mathcal{C}$.

Steps 1.1 and 1.2 verify all axioms in [L4], so $\mathcal{C}$ is a sigma-algebra on $X$.

Applying [L1] to the finite index set $k$ gives the displayed sigma-algebra and a bijection from $\mathcal{P}(k)$ to it.

The finite power set $\mathcal{P}(k)$ has $2^k$ members. For $k=0$, necessarily $X=\mathrm{\varnothing }$ and there is one union; for $k=1$, the two unions are $\mathrm{\varnothing }$ and $X$.

If $A$ is $F_{\sigma }$, [L1] writes it as a countable union of closed sets. Closed sets are complements of the open generators in [L2], so they and their countable union are Borel.

If $A$ is $G_{\delta }$, [L1] writes it as a countable intersection of open, hence Borel, sets; [L3] makes the intersection Borel.

By [L1] and [L3], ${\mathbb{Q}}_{\mathbb{R}}$ is a countable union of closed Borel sets and is therefore Borel and $F_{\sigma }$.

Density of the complement in [L2] prevents ${\mathbb{Q}}_{\mathbb{R}}$ from containing a nonempty open interval, so it is not open; density of ${\mathbb{Q}}_{\mathbb{R}}$ and its being a proper subset prevent it from being closed.

The final claim, that ${\mathbb{Q}}_{\mathbb{R}}$ is not $G_{\delta }$, is exactly the third conclusion of [L1].

The family is nonempty, and $(-\mathrm{\infty },a]\cap (-\mathrm{\infty },b]=(-\mathrm{\infty },\min \{a,b\}]$, so it is a pi-system by [L1].

Complementation exchanges $(-\mathrm{\infty },a]$ with $(a,\mathrm{\infty })$. A generated sigma-algebra is complement-closed, so [L2] implies that the closed left rays generate $\mathcal{B}(\mathbb{R})$.

The description in [L1] makes $C$ a specified subset of $\mathbb{R}$, equipped here with the subspace topology.

The subspace theorem [L2] applies to this inclusion $C\subseteq \mathbb{R}$.

Therefore [L2] gives $\mathcal{B}(C)=\{B\cap C:B\in \mathcal{B}(\mathbb{R})\}$, as claimed.

The family $\mathcal{D}$ contains $X$ and is closed under complements. Apart from $\mathrm{\varnothing }\subseteq X$, its proper containments are $\mathrm{\varnothing }\subseteq A\subseteq X$ for a two-element set $A$, and every corresponding difference remains in $\mathcal{D}$; every increasing sequence in the finite family stabilizes. Thus $\mathcal{D}$ is a lambda-system by [L1].

Both $\{1,2\}$ and $\{1,3\}$ lie in $\mathcal{D}$, but their intersection $\{1\}$ does not. Hence $\mathcal{D}$ is not a pi-system by [L2], refuting the statement.

Every increasing or decreasing sequence in the finite family ${\mathcal{M}}_n$ stabilizes, so its union or intersection belongs to ${\mathcal{M}}_n$. Thus each ${\mathcal{M}}_n$ is a monotone class, and ${\mathcal{M}}_n\subseteq {\mathcal{M}}_{n+1}$.

The union ${\bigcup }_n{\mathcal{M}}_n$ is the family of finite subsets of $\mathbb{N}$. The increasing sequence $0\subseteq 1\subseteq 2\subseteq \cdots$ lies in this union but has union $\mathbb{N}$, which is not finite. Hence the union is not a monotone class.

Let ${\mathcal{A}}_n$ be the family of unions of blocks of ${\mathcal{P}}_n$. Complements and countable unions correspond to complements and unions of block-index sets, so each ${\mathcal{A}}_n$ is a sigma-algebra; splitting $T_n$ into $\{n\}$ and $T_{n+1}$ gives ${\mathcal{A}}_n\subseteq {\mathcal{A}}_{n+1}$.

A member of some ${\mathcal{A}}_n$ is finite if it omits $T_n$ and cofinite if it contains $T_n$. Conversely every finite or cofinite subset belongs to some ${\mathcal{A}}_n$. Hence ${\bigcup }_n{\mathcal{A}}_n$ is the finite-cofinite algebra.

Every singleton $\{2j\}$ lies in the union, but their countable union is the set of even naturals, which is neither finite nor cofinite and so is absent by step 2.1. This violates [L1].

The families $\mathcal{A}:=\{\mathrm{\varnothing },X,A,X\setminus A\}$ and $\mathcal{B}:=\{\mathrm{\varnothing },X,B,X\setminus B\}$ each satisfy the sigma-algebra axioms.

Their union contains $A$ and $B$ but not $A\cap B=\{1\}$. It therefore fails the closure in [L1] and is not a sigma-algebra.

Every increasing or decreasing sequence in the finite chain $\mathrm{\varnothing }\subseteq \{1\}\subseteq X$ stabilizes, so its union or intersection belongs to $\mathcal{M}$. Thus $\mathcal{M}$ is a monotone class by [L1].

The complement $X\setminus \{1\}=\{2\}$ is absent from $\mathcal{M}$, so [L2] shows that $\mathcal{M}$ is not an algebra.

Suppose, for contradiction, that every subset of $\mathbb{R}$ is Borel, so $\mathcal{P}(\mathbb{R})=\mathcal{B}(\mathbb{R})$.

The Cantor set satisfies $C\subseteq \mathbb{R}$, and [L4] with [L5] gives $\mathrm{\mid }C\mathrm{\mid }=\mathrm{\mid }\{0,1{\}}^{\mathbb{N}}\mathrm{\mid }=2^{{\mathrm{\aleph }}_0}=\mathrm{\mid }\mathcal{P}(\mathbb{N})\mathrm{\mid }=\mathfrak{c}$.

Every subset of $C$ is a subset of $\mathbb{R}$, so $\mathcal{P}(C)\subseteq \mathcal{P}(\mathbb{R})$. Step 1.1 makes that inclusion land in $\mathcal{B}(\mathbb{R})$, whose cardinality is $\mathfrak{c}$ by [L1]; hence $\mathrm{\mid }\mathcal{P}(C)\mathrm{\mid }\le \mathfrak{c}=\mathrm{\mid }C\mathrm{\mid }$ by step 1.2.

By [L2], $\mathrm{\mid }C\mathrm{\mid }<\mathrm{\mid }\mathcal{P}(C)\mathrm{\mid }$, contradicting step 2.1. Therefore some subset of $\mathbb{R}$ is not Borel. The cardinality argument selects no particular subset, but [L3] shows that the omitted subset cannot be empty.

Suppose, for contradiction, that a countably infinite sigma-algebra exists.

This contradicts [L1], so the asserted object does not exist.

Suppose, for contradiction, that every Borel set belongs to the displayed fixed finite-stage class.

That class would then equal the Borel sigma-algebra. Since the Borel sigma-algebra is already closed under complements, countable unions, and countable intersections, every further stage would add no set, so the hierarchy would stabilize at that countable stage.

The stabilization in step 2.1 contradicts [L1]. Hence the asserted finite description does not contain every Borel subset of $\mathbb{R}$.