1.2. Classes of Sets#

Definition 1.5

A collection of subsets of \(\Omega\), \(\mathcal{S}\), is a semi-algebra if it satisfies the following properties:

  • ➀ \(\Omega \in \mathcal{S}\)

  • ➁ \(A, B \in \mathcal{S} \implies A \cap B \in \mathcal{S}\)

  • ➂ The complement of every set in \(\mathcal{S}\) can be written as a finite disjoint union of sets in \(\mathcal{S}\). Formally, \(\forall A \in \mathcal{S}, \; \exists E_1, \ldots, E_n \in \mathcal{S}\) such that \(A^\complement = \biguplus_{i=1}^n E_i\).

Example 1.1

Let \(\Omega = \R\). Suppose \(\mathcal{S}\) is a collection of subsets in \(\R\) consisting of intervals of the following forms:

  • ➀ \((a, b]\)

  • ➁ \((-\infty, b]\)

  • ➂ \((a, \infty)\)

  • ➃ \((-\infty, \infty) = \R\)

One may check that \(\mathcal{S}\) is indeed a semi-algebra by definition. And it is by studying this type of collection of sets that the definition of semi-algebra arises.

Definition 1.6

A collection of subsets of \(\Omega\), \(\mathcal{A}\), is called an algebra if

  • ➀ \(\Omega \in \mathcal{A}\)

  • ➁ \(A, B \in \mathcal{A} \implies A \cap B \in \mathcal{A}\)

  • ➂ \(A \in \mathcal{A} \implies A^\complement \in \mathcal{A}\)

Proposition 1.1

Let \(\{\mathcal{A}_\alpha\}_{\alpha \in \Lambda}\) be a collection of algebras in \(\Omega\) where \(\Lambda\) is any index set, then

\[\begin{align*}\bigcap_{\alpha \in \Lambda}\mathcal{A}_{\alpha}\end{align*}\]

is also an algebra.

Proof. TODO

Definition 1.7

A collection of subsets of \(\Omega\), \(\mathcal{F}\), is called a \(\sigma\)-algebra if

  • ➀ \(\Omega \in \mathcal{F}\)

  • ➁ \(\mathcal{F}\) is closed under countable intersections, i.e,. \(A_n \in \mathcal{F} \; \forall n \in \N^\ast \implies \bigcap_{n=1}^\infty A_n \in \mathcal{F}\)

  • ➂ \(A \in \mathcal{F} \implies A^\complement \in \mathcal{F}\)

Of course, a \(\sigma\)-algebra is also closed under finite intersections (and unions) since we can always write

\[\begin{align*}\bigcap_{n=1}^m A_n = \bigcap_{n=1}^m A_n \cap\bigcap_{n=m+1}^\infty\Omega = \bigcap_{n=1}^\infty A_n \quad\text{where $A_n = \Omega$ for $n \geq m+1$}\end{align*}\]

Therefore, the \(\sigma\)-algebra is of course an algebra.

Analogous to Proposition 1.1, any intersections of \(\sigma\)-algebras is also itself a \(\sigma\)-algebra.

Proposition 1.2

Let \(\{\mathcal{F}_\alpha\}_{\alpha \in \Lambda}\) be a collection of \(\sigma\)-algebras in \(\Omega\) where \(\Lambda\) is any index set, then

\[\begin{align*}\bigcap_{\alpha \in \Lambda}\mathcal{F}_{\alpha}\end{align*}\]

is also a \(\sigma\)-algebra.

Having proved Proposition 1.1 and Proposition 1.2, we may introduce the algebra and \(\sigma\)-algebra generated by a class of subsets in \(\Omega\).

Definition 1.8

Let \(\mathcal{C}\) be a class of subsets in \(\Omega\). The algebra generated by \(\mathcal{C}\), is defined by

\[\begin{align*}\mathcal{A}(\mathcal{C}) := \bigcap_{ \text{$\mathcal{A}$ is an algebra, }\mathcal{A} \supseteq \mathcal{C} }\mathcal{A}\end{align*}\]

That is, it is the intersection of all algebras containing \(\mathcal{C}\). Similarly, the \(\sigma\)-algebra generated by \(\mathcal{C}\) is defined by

\[\begin{align*}\sigma(\mathcal{C}) := \bigcap_{ \text{$\mathcal{F}$ is a $\sigma$-algebra, }\mathcal{F} \supseteq \mathcal{C} }\mathcal{F}\end{align*}\]

Note that the set

\[\begin{align*}\set{\mathcal{A} \subset \mathcal{P}(\Omega)}{\text{$\mathcal{A}$ is an algebra, }\mathcal{A} \supseteq \mathcal{C}}\end{align*}\]

is clearly nonempty since the power set, \(\mathcal{P}(\Omega)\), itself is one of its element. And then by Proposition 1.1, \(\mathcal{A}(\mathcal{C})\) is indeed well-defined, and so is \(\sigma(\mathcal{C})\).

One important property for \(\mathcal{A}(\mathcal{C})\)(resp. \(\sigma(\mathcal{C})\)) is that it is the smallest algebra (resp. \(\sigma\)-algebra) containing \(\mathcal{C}\), as stated in the following proposition.

Proposition 1.3

Let \(\mathcal{C} \subseteq \mathcal{P}(\Omega)\). We have the following:

  • ➀ If \(\mathcal{A}\) is an algebra containing \(\mathcal{C}\), then \(\mathcal{A} \supseteq \mathcal{A}(\mathcal{C})\).

  • ➁ If \(\mathcal{F}\) is a \(\sigma\)-algebra containing \(\mathcal{C}\), then \(\mathcal{F} \supseteq \sigma(\mathcal{C})\).

As we see in Example 1.1, it is easy to describe the elements in a semi-algebra directly in terms of subsets of \(\Omega\). However, it is difficult in general to describe the elements in an algebra, and it is even more difficult to do the same for the \(\sigma\)-algebra.

It is because there is no explicit form of describing a \(\sigma\)-algebra in terms of subsets in \(\Omega\) that makes the extension of a measure (or any properties) from the semi-algebra to \(\sigma\)-algebra extremely difficult. As we shall see in many proofs, we will exploit the property of a \(\sigma\)-algebra in an inductive way rather than a constructive one.

Fortunately, we do have an explicit form for the algebra generated by a semi-algebra \(\mathcal{S}\), \(\mathcal{A}(\mathcal{S})\), in terms of sets in \(\mathcal{S}\). It turns out that it is simply the collection of all finite disjoint unions of sets in \(\mathcal{S}\).

Theorem 1.2

Let \(\mathcal{S}\) be a semi-algebra on \(\Omega\). Then

(1.5)#\[\begin{align} A \in\mathcal{A}(\mathcal{S}) \iff\exists E_1, \ldots, E_n \in\mathcal{S}, \; A = \biguplus_{i=1}^n E_i \end{align}\]

Note

Note that the integer \(n\) in (1.5) is not fixed. That is the choice of \(n\) may vary for different set \(A\).

Proof. (Proof of \(\impliedby\)) The reverse direction of (1.5) is easy to prove. Suppose \(A = \biguplus_{i=1}^n E_i\) where each \(E_i \in \mathcal{S}\). Then clearly \(A \in \mathcal{A}(\mathcal{S})\) since the algebra is closed under finite unions.

(Proof of \(\implies\)) The proof of the forward direction is more interesting.

Note

Note that given \(A \in \mathcal{A}(\mathcal{S})\), it is tempting and yet difficult to find a representation of \(A\) as the right-hand side of (1.5). Hence, we shall prove this in an indirect way. Instead of finding a specific representation of \(A\), we shall look at all representations described by (1.5). That is, we consider the collection, say \(\mathcal{C}\), of all finite disjoint unions of sets in \(\mathcal{S}\). And then we show that \(\mathcal{A}(\mathcal{S}) \subseteq \mathcal{C}\).

Define

\[\begin{align*}\mathcal{C} = \set{E \subseteq \Omega}{ \exists E_1, \ldots, E_n \in \mathcal{S}, \; E = \biguplus_{i=1}^n E_i }\end{align*}\]

The goal is to show that this collection \(\mathcal{C}\) is actually an algebra containing \(\mathcal{S}\).

It is clear that \(\mathcal{C} \supseteq \mathcal{S}\) by the definition of \(\mathcal{C}\). What remains to show is that \(\mathcal{C}\) is an algebra. We need to check whether it satisfies each of the three properties one after another.

First, note that \(\Omega \in \mathcal{C}\) since \(\Omega \in \mathcal{S}\) and we have seen that \(\mathcal{C} \supseteq \mathcal{S}\).

Now, we show \(\mathcal{C}\) is closed under intersections. Let \(E, F \in \mathcal{C}\). By definition, \(E\) and \(F\) can be written as

\[\begin{align*} E = \biguplus_{i=1}^n E_i \quad\text{and}\quad F = \biguplus_{j=1}^m F_j \end{align*}\]

where \(E_i, F_j \in \mathcal{S}\). It then follows that

\[\begin{align*} E \cap F = \biguplus_{i=1}^n \brk{ E_i \cap \biguplus_{j=1}^m F_j } = \biguplus_{i=1}^n \biguplus_{j=1}^m \brk{E_i \cap F_j}\end{align*}\]

Note that \(E_i \cap F_j \in \mathcal{S}\). Hence, we have \(E \cap F \in \mathcal{C}\) since the above equation says \(E \cap F\) is a finite disjoint union of sets, \((E_i \cap F_j)\)’s, in \(\mathcal{S}\).

We also need to show \(\mathcal{C}\) is closed under complements. Let \(E \in \mathcal{C}\). The set \(E\) can be written as

\[\begin{align*} E = \biguplus_{i=1}^n E_i \end{align*}\]

where \(E_i \in \mathcal{S}\). Then, the complement of \(E\),

\[\begin{align*} E^\complement = \bigcap_{i=1}^n E^\complement_i \end{align*}\]

By the property of semi-algebra, each \(E^\complement_i\) can be written as

\[\begin{align*} E^\complement_i = \biguplus_{j=1}^{m_i} F_{i,j}\end{align*}\]

where \(F_{i,j} \in \mathcal{S}\). Therefore, \(E^\complement_i \in \mathcal{C}\) by the definition of \(\mathcal{C}\). And then \(E^\complement \in \mathcal{C}\) since we have already shown that \(\mathcal{C}\) is closed under finite intersections.

In summary, we have proved so far that \(\mathcal{C}\) is indeed an algebra containing \(\mathcal{S}\). It then follows that \(\mathcal{A}(\mathcal{S}) \subseteq \mathcal{C}\) since \(\mathcal{A}(\mathcal{S})\) is the smallest algebra containing \(\mathcal{S}\). Therefore, for each \(A \in \mathcal{A}(\mathcal{S})\), \(A\) can always be written as a finite disjoint union of sets in \(\mathcal{S}\) by the definition of \(\mathcal{C}\).