Sigma Algebras and Borel Sets

Every member of $\mathfrak{S}$ contains $\mathrm{\varnothing }$; if $A$ belongs to every member, then so does $X\setminus A$; and if every $A_n$ belongs to every member, then so does ${\bigcup }_n{A}_n$. Hence $\bigcap \mathfrak{S}$ is a sigma-algebra on $X$.

The power set $\mathcal{P}(X)$ is a sigma-algebra on $X$ containing $\mathcal{E}$, so $\mathcal{P}(X)\in {\mathrm{\Sigma }}_X(\mathcal{E})$ and the defining intersection for ${\sigma }_X(\mathcal{E})$ is taken over a nonempty family.

By step 1.1, ${\sigma }_X(\mathcal{E})$ is a sigma-algebra. Every set in $\mathcal{E}$ belongs to every member of ${\mathrm{\Sigma }}_X(\mathcal{E})$, so $\mathcal{E}\subseteq {\sigma }_X(\mathcal{E})$; and the defining intersection is contained in every sigma-algebra containing $\mathcal{E}$. Thus it is the unique smallest such sigma-algebra.

Forward direction: assume, in this step and in every later step that cites it, that $\mathcal{D}$ is a lambda-system. Then $X\in \mathcal{D}$ by [L1], which is clause 1. For $A\in \mathcal{D}$ we have $A\subseteq X$ and $X\in \mathcal{D}$, so $X\setminus A\in \mathcal{D}$ by the relative-difference clause of [L1]; this is clause 2.

Reverse direction, whose hypothesis is independent of the forward branch: assume, in this step and in every later step that cites it, clauses 1, 2 and 3. Then $X\in \mathcal{D}$, which is the first lambda-system clause of [L1], and $\mathrm{\varnothing }=X\setminus X\in \mathcal{D}$ by clauses 1 and 2.

Return to the forward direction, so that the hypothesis in force is again the one of step 1.1, namely that $\mathcal{D}$ is a lambda-system. Let $A_0,A_1,\cdots \in \mathcal{D}$ be pairwise disjoint and put $B_n:={\bigcup }_{k\le n}{A}_k$. We show $B_n\in \mathcal{D}$ by induction on $n$. For $n=0$, $B_0=A_0\in \mathcal{D}$. Suppose $B_n\in \mathcal{D}$. Disjointness gives $A_{n+1}\subseteq X\setminus B_n$, and $X\setminus B_n\in \mathcal{D}$ by step 1.1, so $(X\setminus B_n)\setminus A_{n+1}\in \mathcal{D}$ by [L1]. Its complement in $X$ is $X\setminus ((X\setminus B_n)\setminus A_{n+1})=B_n\cup A_{n+1}=B_{n+1}$, which lies in $\mathcal{D}$ by step 1.1.

Still under the clauses 1, 2 and 3 assumed in step 1.2, let $A,B\in \mathcal{D}$ with $A\subseteq B$. Then $X\setminus B\in \mathcal{D}$ by clause 2, and $A\cap (X\setminus B)=\mathrm{\varnothing }$ because $A\subseteq B$. The sequence $A,X\setminus B,\mathrm{\varnothing },\mathrm{\varnothing },\dots$ is therefore a pairwise disjoint sequence in $\mathcal{D}$ by step 1.2, so clause 3 gives $A\cup (X\setminus B)\in \mathcal{D}$, and clause 2 then gives $X\setminus (A\cup (X\setminus B))=B\setminus A\in \mathcal{D}$, which is the relative-difference clause of [L1].

Still in the forward direction, the sets $B_n$ of step 2.1 increase and satisfy ${\bigcup }_n{B}_n={\bigcup }_n{A}_n$, so ${\bigcup }_n{A}_n\in \mathcal{D}$ by the increasing-union clause of [L1]. This is clause 3, and with step 1.1 it proves the forward direction.

Still under the clauses assumed in step 1.2, let $A_0\subseteq A_1\subseteq \cdots$ lie in $\mathcal{D}$. Put $C_0:=A_0$ and $C_{n+1}:=A_{n+1}\setminus A_n$; each $C_{n+1}$ lies in $\mathcal{D}$ by step 2.2, the $C_n$ are pairwise disjoint, and ${\bigcup }_n{C}_n={\bigcup }_n{A}_n$. Clause 3 then gives ${\bigcup }_n{A}_n\in \mathcal{D}$, which is the increasing-union clause of [L1]. With steps 1.2 and 2.2 this proves the reverse direction. This proves the stated claim.

A nonempty intersection of lambda-systems on $X$ contains $X$. If $A\subseteq B$ lie in every member, then $B\setminus A$ lies in every member; and if $(A_n)$ is increasing and lies in every member, then ${\bigcup }_n{A}_n$ lies in every member. Thus the intersection is a lambda-system.

The power set $\mathcal{P}(X)$ is a lambda-system containing $\mathcal{E}$, so the family intersected in the definition of ${\lambda }_X(\mathcal{E})$ is nonempty.

Steps 1.1 and 1.2 make ${\lambda }_X(\mathcal{E})$ a lambda-system. Every generator lies in every lambda-system being intersected, while an intersection is contained in each of its factors, so $\mathcal{E}\subseteq {\lambda }_X(\mathcal{E})$ and ${\lambda }_X(\mathcal{E})$ is minimal.

A nonempty intersection of monotone classes is closed under increasing countable unions and decreasing countable intersections, because each operation is performed in every class being intersected.

The power set $\mathcal{P}(X)$ is a monotone class containing $\mathcal{E}$, so the family intersected in the definition of $m_X(\mathcal{E})$ is nonempty.

By steps 1.1 and 1.2, $m_X(\mathcal{E})$ is a monotone class. Every generator belongs to every class in the intersection, and the intersection is contained in each such class; hence it contains $\mathcal{E}$ and is minimal.

De Morgan's identity gives ${\bigcap }_n{A}_n=X\setminus {\bigcup }_n(X\setminus A_n)\in \mathcal{A}$ by [L1]. Finite and empty intersections are included by repeating terms and by $\bigcap \mathrm{\varnothing }=X=X\setminus \mathrm{\varnothing }$.

If $A,B\in \mathcal{A}$, then $A\setminus B=A\cap (X\setminus B)\in \mathcal{A}$ by step 1.1, and $A\mathrm{△}B=(A\setminus B)\cup (B\setminus A)\in \mathcal{A}$.

Each tail intersection and tail union of $(A_n)$ belongs to $\mathcal{A}$ by step 1.1 and [L1]. Applying countable union and intersection closure once more to the formulas in [L2] puts both ${\mathrm{lim\hspace{0.17em}inf}}_n{A}_n$ and ${\mathrm{lim\hspace{0.17em}sup}}_n{A}_n$ in $\mathcal{A}$.

By [L1], $x\in {\mathrm{lim\hspace{0.17em}inf}}_n{A}_n$ exactly when $x$ belongs to one tail intersection, which is exactly the existence of $N$ such that $x\in A_k$ for every $k\ge N$. This proves both directions of claim 1, including $N=0$.

By [L1], $x\in {\mathrm{lim\hspace{0.17em}sup}}_n{A}_n$ exactly when $x$ belongs to every tail union, which is exactly: for every $N$ there is $k\ge N$ with $x\in A_k$. This is equivalent to membership in infinitely many terms, since a finite set of successful indices has an index larger than all its members.

Eventual membership from step 1.1 implies the repeated-membership condition of step 1.2 by taking $k\ge \max \{N,N_0\}$ for each requested $N_0$. Hence ${\mathrm{lim\hspace{0.17em}inf}}_n{A}_n\subseteq {\mathrm{lim\hspace{0.17em}sup}}_n{A}_n$.

If $\mathcal{E}\subseteq \mathcal{F}$, then the sigma-algebra ${\sigma }_X(\mathcal{F})$ contains $\mathcal{E}$, so minimality in [L1] gives ${\sigma }_X(\mathcal{E})\subseteq {\sigma }_X(\mathcal{F})$.

The family ${\sigma }_X(\mathcal{E})$ is already a sigma-algebra. It is therefore the smallest sigma-algebra containing itself, and [L1] gives ${\sigma }_X({\sigma }_X(\mathcal{E}))={\sigma }_X(\mathcal{E})$.

From $\mathcal{E}\subseteq {\sigma }_X(\mathcal{F})$, monotonicity and idempotence in [L1] give ${\sigma }_X(\mathcal{E})\subseteq {\sigma }_X({\sigma }_X(\mathcal{F}))={\sigma }_X(\mathcal{F})$.

Interchanging $\mathcal{E}$ and $\mathcal{F}$ gives ${\sigma }_X(\mathcal{F})\subseteq {\sigma }_X(\mathcal{E})$; together with step 1.1 this proves equality.

For $n\in \mathbb{N}$ put $B_n:=A_n\setminus {\bigcup }_{k<n}{A}_k$. The union with $k<n$ is finite and is empty when $n=0$, so [L1] gives $B_n\in \mathcal{A}$. The sets $B_n$ are pairwise disjoint and ${\bigcup }_n{B}_n={\bigcup }_n{A}_n$.

The assumed disjoint-union closure applied to $(B_n)$ gives ${\bigcup }_n{A}_n\in \mathcal{A}$. Thus $\mathcal{A}$ has the countable-union axiom in [L2] and is a sigma-algebra.

Put $B_n:={\bigcup }_{k\le n}{A}_k$. By [L1], each $B_n$ lies in $\mathcal{A}$, and $B_n\subseteq B_{n+1}$.

The hypothesis gives ${\bigcup }_n{B}_n\in \mathcal{A}$, and ${\bigcup }_n{B}_n={\bigcup }_n{A}_n$. Hence $\mathcal{A}$ is closed under arbitrary countable unions and is a sigma-algebra by [L2].

Since $X\in \mathcal{D}$, [L1] gives $X\setminus A\in \mathcal{D}$ for every $A\in \mathcal{D}$.

For $A,B\in \mathcal{D}$, step 1.1 and intersection closure give $A\cup B=X\setminus ((X\setminus A)\cap (X\setminus B))\in \mathcal{D}$. Thus every finite union of members belongs to $\mathcal{D}$.

For a sequence $(A_n)$ in $\mathcal{D}$, the partial unions $B_n:={\bigcup }_{k\le n}{A}_k$ lie in $\mathcal{D}$ by step 2.1 and increase, so [L1] gives ${\bigcup }_n{A}_n={\bigcup }_n{B}_n\in \mathcal{D}$. Together with steps 1.1 and 2.1, this is the sigma-algebra criterion [L2].

One has $X\in \mathcal{D}$ and $A\cap X=A\in \mathcal{D}$, so $X\in {\mathcal{D}}_A$.

If $B,C\in {\mathcal{D}}_A$ and $B\subseteq C$, then $C\setminus B\in \mathcal{D}$ and $A\cap (C\setminus B)=(A\cap C)\setminus (A\cap B)\in \mathcal{D}$ by [L1]; hence $C\setminus B\in {\mathcal{D}}_A$.

If $(B_n)$ is increasing in ${\mathcal{D}}_A$, then ${\bigcup }_n{B}_n\in \mathcal{D}$ and $A\cap {\bigcup }_n{B}_n={\bigcup }_n(A\cap B_n)\in \mathcal{D}$ by [L1]. Therefore ${\bigcup }_n{B}_n\in {\mathcal{D}}_A$, and all lambda-system axioms hold.

Fix $A\in \mathcal{P}$. By [L3], ${\mathcal{D}}_A$ is a lambda-system. If $B\in \mathcal{P}$, then $A\cap B\in \mathcal{P}\subseteq \mathcal{D}$ by [L1] and [L2], so $\mathcal{P}\subseteq {\mathcal{D}}_A$. Minimality in [L2] yields $\mathcal{D}\subseteq {\mathcal{D}}_A$.

Now fix $B\in \mathcal{D}$. Symmetry of intersection and step 1.1 show $A\cap B\in \mathcal{D}$ for every $A\in \mathcal{P}$, so $\mathcal{P}\subseteq {\mathcal{D}}_B$. By [L3] and [L2], ${\mathcal{D}}_B$ is a lambda-system containing $\mathcal{P}$ and therefore contains $\mathcal{D}$.

Thus for arbitrary $A,B\in \mathcal{D}$ one has $A\in {\mathcal{D}}_B$, which means $A\cap B\in \mathcal{D}$.

By [L1], [L2], and [L3], ${\lambda }_X(\mathcal{P})$ is a sigma-algebra containing $\mathcal{P}$. Hence [L4] gives ${\sigma }_X(\mathcal{P})\subseteq {\lambda }_X(\mathcal{P})$.

The sigma-algebra ${\sigma }_X(\mathcal{P})$ is a lambda-system: it contains $X$, is closed under relative differences because it is closed under complements and intersections, and is closed under increasing countable unions. Since it contains $\mathcal{P}$, [L3] gives ${\lambda }_X(\mathcal{P})\subseteq {\sigma }_X(\mathcal{P})$.

Steps 1.1 and 1.2 prove equality. Minimality in [L3] also gives ${\lambda }_X(\mathcal{P})\subseteq \mathcal{D}$, so ${\sigma }_X(\mathcal{P})\subseteq \mathcal{D}$.

Put $\mathcal{C}:=\{E\in \mathcal{M}:X\setminus E\in \mathcal{M}\}$. If $(E_n)$ increases in $\mathcal{C}$, then $X\setminus E_n$ decreases, so monotone closure of $\mathcal{M}$ places both ${\bigcup }_n{E}_n$ and its complement ${\bigcap }_n(X\setminus E_n)$ in $\mathcal{M}$. The decreasing case is the same with union and intersection interchanged. Hence $\mathcal{C}$ is a monotone class.

If $E\in \mathcal{A}$, then $X\setminus E\in \mathcal{A}\subseteq \mathcal{M}$ by [L1] and [L2], so $\mathcal{A}\subseteq \mathcal{C}$. Minimality in [L2] gives $\mathcal{M}\subseteq \mathcal{C}$, which is the asserted complement closure.

Let ${\mathcal{C}}_B:=\{E\in \mathcal{M}:E\cap B\in \mathcal{M}\}$. Intersection with $B$ commutes with increasing unions and decreasing intersections, so the two monotone closure axioms for $\mathcal{M}$ show that ${\mathcal{C}}_B$ is a monotone class.

If $E\in \mathcal{A}$, then $E\cap B\in \mathcal{A}\subseteq \mathcal{M}$ by [L1] and [L2]. Thus $\mathcal{A}\subseteq {\mathcal{C}}_B$, and minimality gives $\mathcal{M}\subseteq {\mathcal{C}}_B$.

Fix $B\in \mathcal{M}$ and put ${\mathcal{C}}_B:=\{E\in \mathcal{M}:E\cap B\in \mathcal{M}\}$. As before, intersection with $B$ commutes with increasing unions and decreasing intersections, so ${\mathcal{C}}_B$ is a monotone class. By symmetry of intersection and [L2], every $E\in \mathcal{A}$ lies in ${\mathcal{C}}_B$.

Minimality in [L1] gives $\mathcal{M}\subseteq {\mathcal{C}}_B$. Since $B$ was arbitrary, $E\cap B\in \mathcal{M}$ for all $E,B\in \mathcal{M}$.

By [L2], $\mathcal{M}$ contains $X$ and is closed under complements and binary intersections, hence under finite unions; it is therefore an algebra.

Every sigma-algebra is closed under increasing unions and, by De Morgan's law, decreasing intersections. Thus ${\sigma }_X(\mathcal{A})$ is a monotone class containing $\mathcal{A}$, and minimality in [L1] gives $\mathcal{M}\subseteq {\sigma }_X(\mathcal{A})$.

Since $\mathcal{M}$ is a monotone class by [L1], it is closed under increasing countable unions. Step 1.1 and [L3] make it a sigma-algebra containing $\mathcal{A}$, so [L4] gives ${\sigma }_X(\mathcal{A})\subseteq \mathcal{M}$.

The inclusions of steps 2.1 and 1.2 prove the equality.

The family $\mathcal{A}$ contains $X$; if $A\subseteq B$ lie in $\mathcal{A}$, then $B\setminus A=B\cap (X\setminus A)$ lies in $\mathcal{A}$; and every increasing countable union lies in $\mathcal{A}$. Hence the axioms in [L2] hold.

Increasing unions lie in $\mathcal{A}$ by [L1]. If $(A_n)$ decreases in $\mathcal{A}$, then ${\bigcap }_n{A}_n=X\setminus {\bigcup }_n(X\setminus A_n)$ lies in $\mathcal{A}$, so the axioms in [L3] hold.

Since $\mathrm{\varnothing }=\mathrm{\varnothing }\cap Y$, the empty set lies in the trace. If $A\cap Y$ lies in the trace, then $Y\setminus (A\cap Y)=(X\setminus A)\cap Y$ lies in it.

For a sequence of traced sets, ${\bigcup }_n(A_n\cap Y)=({\bigcup }_n{A}_n)\cap Y$ lies in the trace. Together with step 1.1 these are exactly the sigma-algebra axioms on $Y$.

By [L1], ${\sigma }_X(\mathcal{E}){\mathrm{\mid }}_Y$ is a sigma-algebra on $Y$, and it contains $\mathcal{E}{\mathrm{\mid }}_Y$. Hence [L2] gives ${\sigma }_Y(\mathcal{E}{\mathrm{\mid }}_Y)\subseteq {\sigma }_X(\mathcal{E}){\mathrm{\mid }}_Y$.

Let $\mathcal{G}:=\{A\subseteq X:A\cap Y\in {\sigma }_Y(\mathcal{E}{\mathrm{\mid }}_Y)\}$. The identities $(X\setminus A)\cap Y=Y\setminus (A\cap Y)$ and $({\bigcup }_n{A}_n)\cap Y={\bigcup }_n(A_n\cap Y)$ show that $\mathcal{G}$ is a sigma-algebra on $X$; it contains $\mathcal{E}$.

By [L2], ${\sigma }_X(\mathcal{E})\subseteq \mathcal{G}$, so tracing gives ${\sigma }_X(\mathcal{E}){\mathrm{\mid }}_Y\subseteq {\sigma }_Y(\mathcal{E}{\mathrm{\mid }}_Y)$. Together with step 1.1 this proves equality.

Let ${\mathcal{I}}_U$ be the family of rational open intervals contained in $U$. By [L1] with $n=1$, this is a subfamily of a countable family, so [L2] makes it at most countable.

Every member of ${\mathcal{I}}_U$ lies in $U$. Conversely, the basis clause of [L1] puts each $x\in U$ in some $I\in {\mathcal{I}}_U$. Thus $U=\bigcup {\mathcal{I}}_U$; when $U$ is empty both sides are empty.

Open and closed sets generate the same sigma-algebra by complementation. Open intervals are open, while [L2] expresses every open set using rational open intervals; hence the open intervals and the rational open intervals each generate the sigma-algebra in [L1].

For $a<b$, rational density [L3] gives $(a,b]=\bigcap \{(a,q):q\in \mathbb{Q},q>b\}$ and $(a,b)=\bigcup \{(a,q]:q\in \mathbb{Q},a<q<b\}$. It also gives $(a,\mathrm{\infty })=\bigcup \{(a,q]:q\in \mathbb{Q},q>a\}$ and $(a,\mathrm{\infty })=\bigcup \{(q,\mathrm{\infty }):q\in \mathbb{Q},q>a\}$. Finally $(-\mathrm{\infty },b]=\mathbb{R}\setminus (b,\mathrm{\infty })$ and $(a,b)=(a,\mathrm{\infty })\cap {\bigcup }_{q<b,q\in \mathbb{Q}}(-\mathrm{\infty },q]$. By [L5] all displayed rational-indexed unions and intersections are countable, so these identities give both generator inclusions for the half-open, real-ray, and rational-ray families.

Applying [L4] to the inclusions in steps 1.1 and 2.1 shows that every displayed family generates $\mathcal{B}(\mathbb{R})$.