[Measure Theory] Algebras and Monotone class theorem

Not all sets are measurable. Measure theory targets only sets that can assign measurements, not all sets. And, we will collect some measurable sets and this collection will be called $\sigma$-algebra.

Algebras and $\sigma$-algebras

Let $X$ be a set.

$\color{blue}{\mathbf{Definition.}}$ An algebra is a collection $\mathcal{A}$ of subsets of $X$ such that

1. $\emptyset \in \mathcal{A}$ and $X \in \mathcal{A}$;
2. If $A \in \mathcal{A}$, then $A^c \in \mathcal{A}$;
3. If $A_1, \cdots, A_n \in \mathcal{A}$, then $\bigcup_{i=1}^n A_i$ and $\bigcap_{i=1}^n A_i$ are in $\mathcal{A}$



$\mathcal{A}$ is a $\sigma$-algebra if in addition

4. whenever $A_1, A_2, \dots$ are in $\mathcal{A}$, then $\bigcup_{i=1}^\infty A_i$ and $\bigcap_{i=1}^\infty A_i$ are in $\mathcal{A}$

$\color{blue}{\mathbf{Definition.}}$ Measurable space
Let $X$ be a set and $\mathcal{A}$ be a $\sigma$-algebra on $X$. The tuple $(X, \mathcal{A})$ is called a measurable space. A set $A$ is called measurable, or $\mathcal{A}$ measurable if $A \in \mathcal{A}$.

$\mathbf{Remark.}$

1. A synonym for algebra is field.
2. A minimal set of postulates for $\mathcal{A}$ to be algebra is
   1. $X \in \mathcal{A}$;
   2. $A \in \mathcal{A}$ implies $A^c \in \mathcal{A}$;
   3. $A, B \in \mathcal{A}$ implies $A \cup B \in \mathcal{A}$;
3. Similarly, a minimal set of postulates for $\mathcal{A}$ to be $\sigma$-algebra is
   1. $X \in \mathcal{A}$;
   2. $A \in \mathcal{A}$ implies $A^c \in \mathcal{A}$;
   3. $A_1, A_2, \dots \in \mathcal{A}$ implies $\bigcup_{i=1}^\infty A_i \in \mathcal{A}$;

Here are some examples.

click to expand

$\color{green}{\mathbf{Example.}}$ Let $X = \mathbb{R}$, and let $\mathcal{A}$ be the collection of all subsets of $\mathbb{R}$. Then $\mathcal{A}$ is a $\sigma$-algebra.

$\color{green}{\mathbf{Example.}}$ Cocountable topology

Let $X = \mathbb{R}$, and let \(\begin{aligned} \mathcal{A} = \{A \subset \mathbb{R} \mid A \text{ is countable or } A^c \text{ is countable } \}. \end{aligned}\)

Then $\mathcal{A}$ is a $\sigma$-algebra.

$\color{green}{\mathbf{Example.}}$ Let $X = [0, 1]$ and let \(\mathcal{A} = \{\emptyset, X, [0, \frac{1}{2}], (\frac{1}{2}, 1] \}\). Then $\mathcal{A}$ is a $\sigma$-algebra.

$\color{green}{\mathbf{Example.}}$ Let $X = [0, 1]$ and let $B_1, \cdots, B_8$ be subsets of $X$ which are pairwise disjoint and whose union is all of $X$. Let $\mathcal{A}$ be the collection of all finite unions of the $B_i$’s as well as the empty sets. (Thus $\mathcal{A}$ consists of $2^8$ elements.) Then $\mathcal{A}$ is a $\sigma$-algebra.

One would like to assign a size to every subset of $X$, but in many settings, this is not possible. Instead, we consider a smaller collection of subsets of $X$. Then we can define another $\sigma$-algebra with this collection. If we have a collection $\mathcal{C}$ of subsets of $X$, define the intersection of all $\sigma$-algebras containing $\mathcal{C}$.

\[\begin{aligned} \sigma(\mathcal{C}) = \bigcap \{ \mathcal{A}_{\alpha} \mid \mathcal{A}_{\alpha} \text{ is a } \sigma \text{-algebra, } \mathcal{C} \subset A_{\alpha} \} \end{aligned}\]

Then, by the next lemma, $\sigma(\mathcal{C})$ is clearly $\sigma$-algebra. And it is referred to as $\sigma$-algebra generated by $\mathcal{C}$.

$\color{blue}{\mathbf{Definition.}}$ An $\sigma$-algebra generated by the collection $\mathcal{C}$ is defined by the smallest $\sigma$–algebra containing all the sets of $\mathcal{C}$. In other words, $$ \begin{aligned} \sigma(\mathcal{C}) = \bigcap \{ \mathcal{A}_{\alpha} \mid \mathcal{A}_{\alpha} \text{ is a } \sigma \text{-algebra, } \mathcal{C} \subset A_{\alpha} \} \end{aligned} $$

And this intersection of $\sigma$-algebra is also a $\sigma$-algebra, according to the following lemma:

$\color{green}{\mathbf{Lemma.}}$

If $\mathcal{A}_{\alpha}$ is a $\sigma$-algebra for each $\alpha$ in some non-empty index set $I$, then $\bigcap_{\alpha \in I} \mathcal{A}_{\alpha}$ is a $\sigma$-algebra.

In other words, any intersection of $\sigma$-algebra is also a $\sigma$-algebra.

$\mathbf{Remark.}$

1. Since there exists at least one $\mathcal{A}_{\alpha}$ e.g. power set of $X$, it never takes the intersection over an empty class of $\sigma$-algebra.
2. $\sigma$ is a monotone operation: if $\mathcal{C}_1 \subset \mathcal{C}_2$, then $\sigma(\mathcal{C}_1) \subset \sigma(\mathcal{C}_2)$.
3. $\sigma$ is an idempotent operation: $\sigma(\sigma(\mathcal{C})) = \sigma(\mathcal{C})$.

Like the previous lemma, there are some cases that we can ensure the preservation of $\sigma$-algebra. For example:

$\color{green}{\mathbf{Lemma.}}$

If $\mathcal{A}$ is a $\sigma$-algebra on $Y$ and $f$ is a mapping from $X$ into $Y$, then $$\mathcal{B} = \{ f^{-1}(A) \mid A \in \mathcal{A} \}$$ is a $\sigma$-algebra on $X$.

In other words, inverse image of $\sigma$-algebra is also a $\sigma$-algebra.

But, any union of $\sigma$-algebra doesn’t need to be $\sigma$-algebra.

$\color{#bf1100}{\mathbf{Remark.}}$

1. Union of $\sigma$-algebra doesn't need to be $\sigma$-algebra. For counterexample, consider $X = \{ 0, 1, 2 \}$ and $\mathcal{A}_1 = \{ \emptyset, X, \{ 0 \}, \{ 1, 2 \} \}$ and $\mathcal{A}_2 = \{ \emptyset, X, \{ 2 \}, \{ 0, 1 \} \}$.
2. Suppose $\mathcal{A}_1 \subset \mathcal{A}_2 \subset \cdots$ are $\sigma$-algebras consisting of subsets of $X$. Still, $\bigcup_{i=1} A_i$ is not necessarily a $\sigma$-algebra. For counterexample, consider $X = \mathbb{N}$ and $\mathcal{A}_n = \{ A \mid A \subseteq \{ 0, 1, \cdots, n \} \text{ or } A^c \subseteq \{ 0, 1, \cdots, n \}\}$.

The monotone class theorem

The monotone class theorem is a result that can be used to extend certain properties that hold for an algebra $\mathcal{C}$ to the $\sigma(\mathcal{C})$.

$\color{blue}{\mathbf{Definition.}}$ Monotone class
A monotone class is a collection of subsets $\mathcal{M}$ of $X$ such that

1. if $A_i \uparrow A$ and each $A_i \in \mathcal{A}$, then $A \in \mathcal{M}$;
2. if $A_i \downarrow A$ and each $A_i \in \mathcal{A}$, then $A \in \mathcal{M}$;

In other words, a monotone class is a collection (i.e. class) $\mathcal{M}$ of sets that is closed under countable monotone unions and also under countable monotone intersections.

Here, for a collection of sets \(\{ A_n \}\) and $A$, $A_n \uparrow A$ stands for that $A_n$ is an increasing sequences of sets, i.e. $A_n \subseteq A_{n+1}$ and the countable union $\bigcup_{n \in \mathbb{N}} A_n = A$. In opposite way, we denote $A_n \downarrow A$ that $A_n$ is an decreasing sequences of sets, i.e. $A_{n+1} \subseteq A_{n}$ and the countable intersection $\bigcap_{n \in \mathbb{N}} A_n = A$.

$\mathbf{Remark.}$

1. It is obvious that $\emptyset \in \mathcal{M}$, by setting $A_i = \emptyset$ for all $i = 1, 2, \dots$.
2. The intersection of monotone classes is a monotone class
3. The intersection of all monotone classes containing a given collection of sets is the smallest monotone class containing that collection.

Now, the monotone class theorem is introduced:

$\color{red}{\mathbf{Thm.}}$ Monotone class theorem

Suppose $\mathcal{A}_0$ is an algebra, $\mathcal{A}$ is the smallest $\sigma$-algebra containing $\mathcal{A}_0$, and $\mathcal{M}$ is the smallest monotone class containing $\mathcal{A}_0$. Then $\mathcal{M} = \mathcal{A}$.

$\mathbf{Proof.}$

The idea of proof is that we extract some portion of $\mathcal{M}$, $\mathcal{N}$ where is closed under complements or countable intersection. And we show that $\mathcal{M} \subseteq \mathcal{N}$ for $\mathcal{M} = \mathcal{N}$, by proving that $\mathcal{N}$ is a monotone class containing $\mathcal{A}_0$.

(1) $\sigma$-algebra is clearly a monotone class, hence $\mathcal{M} \subseteq \mathcal{A}$.

(2) And we will show that $\mathcal{M}$ is a $\sigma$-algebra that contains $\mathcal{A}_0$. Then, $\mathcal{A} \subseteq \mathcal{M}$ obviously.

1. $A \in \mathcal{M}$ implies $A^c \in \mathcal{A}$

   
   Let $\mathcal{N}_1 = \{ A \in \mathcal{M} \mid \mathcal{A}^c \in \mathcal{M} \}$. We claim that $\mathcal{N}_1 = \mathcal{M}$. Since it is clear that $\mathcal{N}_1 \subseteq \mathcal{M}$, we have to show inverse direction. We claim that $\mathcal{N}_1$ is a monotone class containing $\mathcal{A}_0$.

   Let $A \in \mathcal{M}$. Without loss of generality, suppose $A_i \uparrow A$ and each $A_i \in \mathcal{N}_1$, then $A_i^c \in \mathcal{M}$ and $A_i^c \downarrow A^c$. Hence, $A^c \in \mathcal{M}$. Hence, $A \in \mathcal{N}_1$.

   Thus we conclude that $\mathcal{M}$ is closed under complement.

2. $A, B \in \mathcal{M}$ implies $A \cup B \in \mathcal{M}$

   
   Let $\mathcal{N}_2 = \{ A \in \mathcal{M} \mid \mathcal{A} \cap B \in A_0 \text{ for all } B \in \mathcal{M} \} \subseteq \mathcal{M}$.

   1. $\mathcal{N}_2$ is a monotone class

      Suppose each $A_i \in \mathcal{N}_2 \subseteq \mathcal{M}$ and $A_i \uparrow A$. Let any $B \in \mathcal{M}$ be given. Then
      $$ \begin{aligned} A \cup B = \bigcup_{i=1}^\infty (A_i \cup B) \in \mathcal{M} \end{aligned} $$ since $\mathcal{M}$ is monotone class and $(A_i \cup B) \uparrow (A \cup B)$. In analogous way, $\downarrow$ case can be proved. Thus, $A \in \mathcal{N}_2$ and $\mathcal{N}_2$ is monotone class.

   2. $\mathcal{N}_2$ contains $\mathcal{A}_0$.

      Let $A \in \mathcal{A}_0$. And let any $B \in \mathcal{M}$ be given. We have to show that $A \cup B \in \mathcal{M}$.
      To prove this, we will construct $$ \begin{aligned} \mathcal{N}_3 = \{ A \in \mathcal{M} \mid \mathcal{A} \cap B \in A_0 \text{ for all } B \in \mathcal{A}_0 \} \subseteq \mathcal{M} \end{aligned} $$ and show that $\mathcal{M} = \mathcal{N}_3$. Then it will imply that for any $A \in \mathcal{A}_0$ and $B \in \mathcal{M}$, $A \cup B \in \mathcal{M}$.

      
      

      We claim that $\mathcal{N}_3$ is a monotone class containing $\mathcal{A}_0$. First, it is clear that $\mathcal{N}_3$ contains $\mathcal{A}_0$ since $\mathcal{A}_0$ is an algebra. Second, it is a monotone class: let each $E_1, E_2, \dots \in \mathcal{N}_3$ and $E_i \uparrow E$. For any $F \in \mathcal{A}_0$, $E \cup F = \bigcup_{i=1}^\infty (E_i \cup F) \in \mathcal{M}$ since $\mathcal{M}$ is a monotone class and $(E_i \cup F) \uparrow (E \cup F)$. Thus, $E \in \mathcal{N}_3$.

3. $\mathcal{M}$ is closed under countable intersections

   
   Let $A_1, A_2, \dots \in \mathcal{M}$. Then, define $B_n = \bigcup_{i=1}^n A_i$ for each $n$. Then, as $\mathcal{M}$ is closed under finite union, $B_n \in \mathcal{M}$ for each $n$.

   
   

   And obviously, $B_n \uparrow \bigcup_{i=1}^\infty A_i$. Since $\mathcal{M}$ is monotone class, $\bigcap_{i=1}^\infty A_i \in \mathcal{M}$.

\[\tag*{$\blacksquare$}\]

Hence, this theorem connects monotone classes and $\sigma$-algebras. Note that it is necessary to $\mathcal{A}_0$ be an algebra in the proof in order to endow a monotone class containing $\mathcal{A}_0$ with closedness under complement and finite union/intersection. If not, this theorem does not hold; for a set \(X = \{ 0, 1, 2 \}\) and a monotone class \(\mathcal{M} = \{ \emptyset, \{0\}, X \}\) consisting of subsets of $X$, but it is not a $\sigma$-algebra.

We close the section by proving the interesting fact regarding $\sigma$-algebra:

$\color{red}{\mathbf{Thm.}}$ Every $\sigma$-algebra is either finite or uncountably infinite.

$\mathbf{Proof.}$

Let $\mathcal{A}$ be a countable infinite $\sigma$-algebra containing a set $X$. ($X$ must be infinite.) Thus, we can denote \(\mathcal{A} = \{ A_n : \; n \in \mathbb{N} \}\).

Define $f: X \to \mathcal{A}$ by the smallest measurable set containing $x \in X$:

\[\begin{aligned} f(x) = \bigcap_{x \in A_n} A_n \end{aligned}\]

Then, for any $x, y \in X$ with $x \neq y$, $f(x)$ and $f(y)$ are disjoint. Suppose not. Then $f(x) = f(x) \cap f(y)$ by definition of $f$. Similarly, we have $f(y) = f(x) \cap f(y)$. Then, we show that $\mathcal{A} = { \emptyset, X }$ in this case. It is because for any $A \mathcal{A}$, it is obvious that $A \subseteq X$ and for any $x \in X$, we can choose $y \in A$ if $A \neq \emptyset$. But, $x \in f(x) = f(y) \subset A$, hence $A = X$.

Therefore, there exists a partition of $\mathcal{A}$, $\mathcal{B}$, denoted by

\[\begin{aligned} \mathcal{B} = \{ f(x) \; | \; x \in X \}. \end{aligned}\]

And for any $N \subset \mathbb{N}$, $\bigcup_{n \in N} B_n \in \mathcal{A}$ as each $B_n$ is measurable. Thus,

\[\begin{aligned} 2^{| \mathbb{N} |} \leq | \mathcal{A} |. \end{aligned}\] \[\tag*{$\blacksquare$}\]

Borel $\sigma$-algebra

$\color{blue}{\mathbf{Definition.}}$ Borel $\sigma$-algebra
Let $\mathcal{G}$ be the collection of open subsets of a metric space $X$, we call $\sigma(\mathcal{G})$ the Borel $\sigma$-algebra on $X$, and it is often denoted $\mathcal{B}$. Then elements of $\mathcal{B}$ are called Borel sets and are said to be Borel measurable.

$\color{#bf1100}{\mathbf{Proposition.}}$ If $X = \mathbb{R}$, then the Borel $\sigma$-algebra on $X$, $\mathcal{B} = \sigma (\mathcal{G})$ is generated by each of the following collection of sets:

1. $\mathcal{C}_1 = \{ (a, b) \mid a, b \in \mathbb{R} \}$;
2. $\mathcal{C}_2 = \{ [a, b] \mid a, b \in \mathbb{R} \}$;
3. $\mathcal{C}_3 = \{ (a, b] \mid a, b \in \mathbb{R} \}$;
4. $\mathcal{C}_4 = \{ (a, \infty] \mid a \in \mathbb{R} \}$;

$\mathbf{Proof.}$

(1) Let $\mathcal{G}$ be the collection of open sets. By definition, $\sigma (\mathcal{G})$ is the Borel $\sigma$-algebra. Since every element of $\mathcal{C}_1$ is open, hence $\mathcal{C}_1 \subset \mathcal{G}$, so that $\sigma(\mathcal{C}_1) \subset \sigma(\mathcal{G})$.

Conversely, if $G$ is open, then it is the countable union of open intervals. Note that very finite open interval is in $\mathcal{C}_1$.

In case of unbounded open interval, since $(a, \infty) = \bigcup_{n=1}^\infty (a, a+n)$, hence $(a, \infty) \in \sigma(\mathcal{C}_1)$ if $a \in \mathbb{R}$. Similarly, $(- \infty, a) \in \sigma(\mathcal{C}_1)$ if $a \in \mathbb{R}$.

Hence, if $G$ is open, then $G \in \sigma(\mathcal{C}_1)$ since it is combination of countable unions and intersections of elements of $\sigma(\mathcal{C}_1)$, which shows $\sigma(\mathcal{G}) \subseteq \sigma(\mathcal{C}_1)$.

(2) If \([a, b] \in \mathcal{C}_2\), then $[a, b] = \bigcap_{n=1}^\infty (a - \frac{1}{n}, b + \frac{1}{n}) \in \sigma(\mathcal{G})$. Hence $\mathcal{C}_2 \subset \sigma(\mathcal{G})$, thus $\sigma(\mathcal{C}_2) \subseteq \sigma(\mathcal{G})$.

If $(a, b) \in \mathcal{C}_1$, choose $n_0 \geq 2 / (b - a)$. Then

\[\begin{aligned} (a, b) = \bigcup_{n=n_0}^\infty \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] \in \sigma(\mathcal{C}_2) \end{aligned}\]

Thus, $\mathcal{C}_1 \subset \sigma (\mathcal{C}_2)$, hence $\mathcal{B} = \sigma(\mathcal{C}_1) \subset \sigma (\mathcal{C}_2)$.

(3) If \((a, b] \in \mathcal{C}_3\), then $(a, b] = \bigcap_{n=1}^\infty (a, b + \frac{1}{n}) \in \sigma(\mathcal{G})$. Hence $\sigma(\mathcal{C}_3) \subseteq \sigma(\mathcal{G})$.

If \((a, b) \in \mathcal{C}_1\), choose $n_0 \geq 2 / (b - a)$. Then

\[\begin{aligned} (a, b) = \bigcup_{n=n_0}^\infty \left( a, b - \frac{1}{n} \right] \in \sigma(\mathcal{C}_3) \end{aligned}\]

Thus, $\mathcal{B} = \sigma(\mathcal{C}_1) \subset \sigma (\mathcal{C}_3)$.

(4) If \((a, \infty) \in \mathcal{C}_4\), then $(a, \infty) = \bigcup_{n=1}^\infty (a, b + n) \in \sigma(\mathcal{G})$. Hence $\sigma(\mathcal{C}_4) \subseteq \sigma(\mathcal{G})$.

If $(a, b] \in \mathcal{C}_3$, then $(a, b] = (b, \infty) - (a, \infty) \in \sigma(\mathcal{C}_4)$. Thus, $\mathcal{B} = \sigma(\mathcal{C}_3) \subset \sigma (\mathcal{C}_4)$.

\[\tag*{$\blacksquare$}\]

Reference

[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Wikipedia, Measurable space