2. Classes of subsets and set function

In order to define the concept of measure, we need to make some preparations. In this section, we will introduce the concepts of semi-algebra, algebra, and $\sigma$-algebra, which are sets closed for some operations.

Definition 2.0.1.

Consider a set $\Omega$ and assume that $\mathcal{S}\subset \Im (\Omega )$, we say that $\mathcal{S}$ is a semi-algebra if

•	$\Omega \in \mathcal{S}$
•	$A,B\in \mathcal{S}$ $⟹$ $A\cap B\in \mathcal{S}$
•	$A\in \mathcal{S}$ $⟹$ $\exists E_1,...,E_n\in \mathcal{S}$ such that $A^c={\sum }_{j=1}^n{E}_j$

Remark 2.0.2. We writhe ${\sum }_{j=1}^n{E}_j$ to be the disjoint union of these sets.

Example 2.0.3. Let $\Omega =\mathbb{R}$ and $\mathcal{S}=\{\mathbb{R},\left\{(a,b]|a<b,a,b\in \mathbb{R}\right\},\left\{(-\infty ,b]|b\in \mathbb{R}\right\},$$\{(a,+\infty )|a\in$$\mathbb{R}\},\varnothing \}$, it is easy to check that $\mathcal{S}$ is a semi-algebra.

Remark 2.0.4. This example can be extend to ${\mathbb{R}}^n$, and it inspire us to define the concept of semi-algebra.

Definition 2.0.5. Given a set $\Omega$ and we say that $\mathcal{A}\subset \Im (\Omega )$ is an algebra if

•	$\Omega \in \mathcal{A}$
•	$A,B\in \mathcal{A}$ $⟹$ $A\cap B\in \mathcal{A}$
•	$A\in \mathcal{A}$ $⟹$ $A^c\in \mathcal{A}$

Observation $\mathcal{A}$ is an algebra $⟹$ $\mathcal{A}$ is a semi-algebra.

Definition 2.0.6. Given a set $\Omega$ and we say that $\mathcal{F}\subset \Im (\Omega )$ is a $\sigma$-algebra if

•	$\Omega \in \mathcal{A}$
•	${\left(A_j\right)}_{j⩾1}\in \mathcal{F}$ $⟹$ ${\bigcap }_{j⩾1}{A}_j\in \mathcal{F}$
•	$A\in \mathcal{F}$ $⟹$ $A^c\in \mathcal{F}$

Observation $\mathcal{F}$ is a $\sigma$-algebra $⟹$ $\mathcal{F}$ is an algebra.
Since algebra is closed to complements and finite intersection, it is obvious that the countable intersection of an algebra is also an algebra, and thus we can discuss the concept of an algebra generated by a set. Similar discussion can be carried out on the $\sigma$-algebra.

Claim 1

Consider a set $\Omega$ and $({\mathcal{A}}_{\alpha }{)}_{\alpha \in I}\subset \Im (\Omega )$ a family of algebra over $\Omega$ where $I$ is an index set. Then $\mathcal{A}={\bigcap }_{\alpha \in I}{\mathcal{A}}_{\alpha }$ is an algebra.

It is clear that $\Omega \in \mathcal{A}$ since $\Omega \in {\mathcal{A}}_{\alpha }$ for all $\alpha$. If $A,B\in \mathcal{A}$, then $A,B\in {\mathcal{A}}_{\alpha }$ for every $\alpha$, hence $A\cap B\in {\mathcal{A}}_{\alpha }$ for every $\alpha$ and therefore $A\cap B\in \mathcal{A}$. If $A\in \mathcal{A}$, then $A\in {\mathcal{A}}_{\alpha }$ for every $\alpha$ hence $A^c\in {\mathcal{A}}_{\alpha }$ for every $\alpha$, so $A^c\in \mathcal{A}$.

Claim 2

Given a set $\Omega$ and $({\mathcal{F}}_{\alpha }{)}_{\alpha \in I}\subset \Im (\Omega )$ a family of $\sigma$-algebra where $I$ is an index set. Then $\mathcal{F}={\bigcap }_{\alpha \in I}{F}_{\alpha }$ is a $\sigma$-algebra.

It is clear that $\Omega \in \mathcal{A}$, and $A\in \mathcal{F}$ implies that $A^c\in \mathcal{F}$ is clear too. The property $(A_j{)}_{j⩾1}\in \mathcal{F}⟹A^c\in \mathcal{F}$ is similar to that of claim 1.

Definition 2.0.7. Given a set $\Omega$ and $\mathcal{C}\subset \Im (\Omega )$, we say that an algebra $\mathcal{A}$ is generated by $\mathcal{C}$ if

•	$\mathcal{C}\subset \mathcal{A}$
•	$\{\begin{array}{l}\mathcal{B}\text{is an algebra over}\Omega \\ \mathcal{C}\subset \mathcal{B}\end{array}⟹\mathcal{A}\subset \mathcal{B}$

and we denote $\mathcal{A}$ by $\mathcal{A}(\mathcal{C})$.

Let ${\mathcal{A}}_{\alpha }$ be all algebra over $\Omega$ which contain $\mathcal{C}$. Then $\mathcal{A}={\bigcap }_{\alpha }{\mathcal{A}}_{\alpha }$ is an algebra generated by $\mathcal{C}$.

1. $\mathcal{C}\subset \mathcal{A}$ is clear since $\mathcal{C}\in {\mathcal{A}}_{\alpha }$ for all $\alpha$.
2. If $\mathcal{B}$ is any algebra which contain $\mathcal{C}$, then it must be ${\mathcal{A}}_{\alpha }$ for some $\alpha$ and hence $\mathcal{A}\subset \mathcal{B}$.

And $\mathcal{A}={\bigcap }_{\alpha }{\mathcal{A}}_{\alpha }$ is the smallest algebra over $\Omega$ which contain $\mathcal{C}$. Similarly, we can talk about a $\sigma$-algebra generated by $\mathcal{C}\subset \Im (\Omega )$.
The following Lemma shows that the elements of the algebra $\mathcal{A}(\mathcal{S})$ generated by the semi-algebra $\mathcal{S}$ can be decomposed into the elements of the semi-algebra $\mathcal{S}$.

Lemma 2.0.8.

Consider a set $\Omega$ and $\mathcal{S}\subset \Im (\Omega )$ a semi-algebra and $\mathcal{A}(\mathcal{S})$ an algebra generated by $\mathcal{S}$, then $$A\in \mathcal{A}(\mathcal{S})⟺\exists {\left\{E_j\in \mathcal{S}\right\}}_{1⩽j⩽n}\text{such that}A=\sum _{j=1}^n{E}_j$$

Proof.   ($⟸$) Assume that $\exists$ $E_j\in \mathcal{S}$ such that $A={\sum }_{j=1}^n{E}_j$, then $E_j\in \mathcal{A}(\mathcal{S})$ since $\mathcal{S}\subset \mathcal{A}(\mathcal{S})$. $$\begin{array}{rl}{A}^c& =(\sum _{j=1}^n{E}_j{)}^c\\ & =(\underset{j=1}{\overset{n}{⨆}}{E}_j{)}^c\\ & =\bigcap _{j=1}^n(E_j{)}^c\in \mathcal{A}(\mathcal{S})\end{array}$$Hence $A=(A^c{)}^c\in \mathcal{A}(\mathcal{S})$.

($⟹$) We construct $\mathcal{B}=\left\{{\sum }_{j=1}^n{F}_j|F_j\in \mathcal{S}\right\}$ and claim that $\mathcal{B}$ is an algebra.

1. $\Omega \in \mathcal{B}$ since $\Omega \in \mathcal{S}\subset \mathcal{B}$.

2. If $A,B\in \mathcal{B}$, then $A={\sum }_{j=1}^n{F}_j$ and $B={\sum }_{k=1}^m{G}_k$ where $F_j,G_k\in \mathcal{S}$ and $$\begin{array}{rl}A\bigcap B& =(\sum _{j=1}^n{F}_j)\bigcap (\sum _{k=1}^m{G}_k)\\ & =\sum _{j=1}^n\sum _{k=1}^m\underset{\text{contain in}\mathcal{S}}{\underbrace{\left(F_j\bigcap G_k\right)}}\in \mathcal{B}\end{array}$$

3. If $A\in \mathcal{B}$, then $A={\sum }_{j=1}^n{E}_j$ for $E_j\in \mathcal{S}$. Hence $A^c=({\sum }_{j=1}^n{E}_j{)}^c={\bigcap }_{j=1}^n(E_j{)}^c$ and

$$\begin{array}{rl}(E_1{)}^c& =\sum _{k_1=1}^{l_1}{F}_{1,k_1}{F}_{1,j}\in \mathcal{S}\\ & ⋮\\ (E_n{)}^c& =\sum _{k_n=1}^{l_n}{F}_{n,k_n}{F}_{n,k}\in \mathcal{S}\end{array}$$

since $E_j\in \mathcal{S}$ for all $j$. Hence $$\begin{array}{rl}{A}^c& =\bigcap _{j=1}^n(E_j{)}^c\\ & =\bigcap _{j=1}^n(\sum _{k_j=1}^{l_j}{F}_{j,k_j})\\ & =\sum _{k_1=1}^{l_1}\cdots \sum _{k_n=1}^{l_n}\underset{\text{contain in}\mathcal{S}}{\underbrace{\left(\bigcap _{1,k_1}^{n,k_n}{F}_{j,k_j}\right)}}\in \mathcal{B}\end{array}$$Hence $\mathcal{B}$ is an algebra. By the definition of $\mathcal{A}(\mathcal{S})$, it follows that $\mathcal{A}(\mathcal{S})\subset \mathcal{B}$ and it completed the proof.

$\square$

Then there are two important concepts, the so-called additive function and the $\sigma$-additive function.

Definition 2.0.9. Let $\mathcal{C}\subset \Im (\Omega )$ and $\varnothing \in \mathcal{C}$, we define a function $$\mu :\mathcal{C}⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$$and say that $\mu$ is additive if

•	$\mu (\varnothing )=0$
•	If $E_1,E_2,...,E_n\in \mathcal{C}$ and $E={\sum }_{j=1}^n{E}_j\in \mathcal{C}$, then $\mu (E)={\sum }_{j=1}^n\mu (E_j)$

Observation Assume that there exists $A\in \mathcal{C}$ such that $\mu (A)<+\infty$. We can write $A=A\cup \varnothing$, then $\mu (A)=\mu (A)+\mu (\varnothing )$ $⟹$ $\mu (\varnothing )=0$.

Observation Assume that $E\subset F$, then

•	If $\mu (E)=+\infty$, then $\mu (F)=\mu (E)+\mu (F\backslash E)$ by the additivity of $\mu$, so $\mu (F)=+\infty$
•	If $\mu (E)<+\infty$, then $\mu (F)-\mu (E)=\mu (F\backslash E)⩾0$, so $\mu (E)⩽\mu (F)$

It shows that $\mu$ is an increasing function with respect to the relation "$\subset$" of sets.
The following example is called a discrete measure, but when we say so, we do not acquiesce in defining the concept of measure. This is like after defining a regular submanifold, we need to prove that it is a manifold; after defining a linear subspace, we need to verify that it is a linear space. Please pay attention to this subtle difference.

Example 2.0.10. (discrete measure)

Given a set $\Omega$ and assume that we have a family of points ${\left\{X_j\in \Omega \right\}}_{j⩾1}$ in $\Omega$ and a sequence ${\left\{P_j\in {\mathbb{R}}_{+}\right\}}_{j⩾1}$. For $\mathcal{C}\subset \Im (\Omega )$, we define $$\begin{array}{rl}\mu :\mathcal{C}& ⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}\\ A& \mapsto \sum _{\alpha ⩾1}{P}_{\alpha }\cdot 1(X_j\in A)\end{array}$$where $1\left(X_j\right)=\{\begin{array}{l}1,j=\alpha \\ 0,j\ne \alpha \end{array}$. It is clear that $\mu (\varnothing )=0$. If $E_1,E_2,...,E_n\in \mathcal{C}$ such that $E={\sum }_{j=1}^n{E}_j\in \mathcal{C}$, then $\mu (E)={\sum }_{j=1}^n\mu (E_j)$ is clear too. Consequently, $\mu$ is additive.

Definition 2.0.11.

Let $\mathcal{C}\subset \Im (\Omega )$ and $\varnothing \in \mathcal{C}$, $\mu$ is a function $$\mu :\mathcal{C}⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$$We say that $\mu$ is $\sigma$-additive if

•	$\mu (\varnothing )=0$
•	If $E_j\in \mathcal{C}$ and $E_j\bigcap E_k=\varnothing$ with $j\ne k$ and $E={\sum }_{j⩾1}{E}_j\in \mathcal{C}$, then $\mu (E)=$${\sum }_{j⩾1}\mu (E_j)$

Observation $\mu$ is $\sigma$-additive $⟹$ $\mu$ is additive. The following example tells us $\mu$ is $\sigma$-additive $\nLeftarrow$ $\mu$ is additive.

Example 2.0.12. Let $\Omega =(0,1)$ and $\mathcal{C}=\left\{(a,b]|0⩽a<b<1\right\}$, we define $$\begin{array}{rl}\mu :\mathcal{C}& ⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}\\ (a,b]& \mapsto \{\begin{array}{l}+\infty ,a=0\\ b-a,a\ne 0\end{array}\end{array}$$If $(a,b]=\varnothing$, it implies that $a=b$, thus $\mu (\varnothing )=0$. If $(a,b]={\sum }_{j=1}^n(a_j,b_j]$, then there are two cases

•	If $a_k=0$ for some $k$, then $\mu (a_k,b_k]=+\infty$, it implies that $\mu (a,b]=+\infty$ since $(a_k,b_k]\subset (a,b]$
•	If $a_j\ne 0$ for all $j$, then

$$\begin{array}{rl}\mu (a,b]& =b-a\\ & =(b_1-a_1)+\cdots +(b_j-a_j)+\cdots +(b_n-a_n)\\ & =\mu (a_1,b_1]+\cdots +\mu (a_j,b_j]+\cdots +\mu (a_n,b_n]\\ & =\sum _{j=1}^n\mu (a_j,b_j]\end{array}$$

Thus $\mu$ is additive.

Claim 3

$\mu$ is not $\sigma$-additive.

Take $(0,\frac{1}{2}]={\sum }_{j=1}(x_{j+1},x_j]$ where the decreasing sequence ${\left\{x_j\right\}}_{j⩾1}$$⟶0$, moreover, we set $x_1=\frac{1}{2}$. $\mu (0,\frac{1}{2}]=+\infty$ but ${\sum }_{j=1}\mu (x_{j+1},x_j]=$$\frac{1}{2}$.