• algebra
  • semi-algebra
  • algebra
  • $\sigma$-algebra
  • intersection of algebra
  • generate algebra from a class of sets
  • generate from semi-algebra
• additive and $\sigma$-additive
  • additive function
  • $\sigma$-additive function
  • additive but not $\sigma$-additive

In this section, we construct the set for measures, and we define the function on these sets.

In the previous section, we show that the property of measure can’t hold for the family of subsets on $\mathbb{R}$. Now we focus on the subsets with special properties.

algebra

semi-algebra

Do not be confused by the name algebra, you can just inage it like a set. On set $\Omega$, the semi-algebra $\mathcal{S}$, which is a subset, should satisfy:

1. $\Omega \in \mathcal{S}$
2. $\forall A, B\in \mathcal{S}$, $A\bigcap B \in \mathcal{S}$
3. $\forall A\in \mathcal{S}$, $\exists B_1, \cdots, B_n \in \mathcal{S}$, such that $A^c=\bigcup_{i=1}^nB_i$, where $B_1,\cdots, B_n$ are disjoint. or we say $A^c=\sum_{i=1}^nB_i$.

algebra

We extend semi-algebra to algebra by replacing the third property to

3. $\forall A \in \mathcal{S}$, $A^c\in \mathcal{S}$

$\sigma$-algebra

In the second property of semi-algebra and algebra, we assume FINITE union is contained in the semi-algebra or algebra. For $\sigma$-algebra, we assume infinite union is still contained in the $\sigma$-algebra.

2. $\forall A_j\in \mathcal{S}$, $j\geq 1$, $\bigcap_{j\geq 1}A_j \in \mathcal{S}$.

intersection of algebra

The intersection of algebra is algebra, same for $\sigma$-algebra.

generate algebra from a class of sets

On $\Omega$, we have the family of subsets $\mathcal{P}(\Omega)$, suppose $\mathcal{C}\subseteq \mathcal{P}(\Omega)$, we say $\mathcal{A}(\mathcal{C})$ is the algebra generated from the class $\mathcal{C}$ if

1. $\mathcal{C} \subseteq \mathcal{\mathcal{A}}$
2. $\forall \mathcal{B} \supseteq \mathcal{C} $, and $\mathcal{B}$ is algebra, we have $\mathcal{A}\subseteq \mathcal{B}$. Or to say, $\mathcal{C}$ is the smallest algebra.

Suppose $\mathcal{A_{\alpha}}$ is all the algebras containing $\mathcal{C}$, We claim that $\mathcal{A}=\bigcap_{\alpha}\mathcal{A_{\alpha}}=\mathcal{A}(\mathcal{C})$. Same results holds for $\sigma$-algebra.

generate from semi-algebra

If $\mathcal{C}$ is a special class, like semi-algebra, we have the following lemma.

Lemma Suppose semi-algebra $\mathcal{S}$ on $\Omega$, $\mathcal{A}(\mathcal{S})$ is the algebra generated by $\mathcal{S}$, then

Proof First we prove “$\Longleftarrow$”. The key point is the finite union of set is contained in algebra, i.e., $E,F\in \mathcal{S} \subset \mathcal{A}$, $E\bigcup F=(E^c\bigcap F^c)^c\in\mathcal{A}$.

Then we prove “$\Longrightarrow$”. Define $\mathcal{B}=\left\{\sum_{j=a}^n F_j,F_j\in\mathcal{S} \right\}$, we claim that $\mathcal{B}$ is an algebra and $\mathcal{B}\supset \mathcal{S}$, then by definition of $\mathcal{A}(\mathcal{S})$, we know $\mathcal{A}(\mathcal{S})\subseteq \mathcal{B}$, thus $A$ can be write in the characterization form.

Now we prove $\mathcal{B}$ is an algebra.

Proof

1. Known $\Omega\in \mathcal{S}$, $\Omega \in \mathcal{B}$.

2. Suppose $A=\sum_{j=1}^n E_j\in \mathcal{B}$ and $B=\sum_{k=1}^mF_k\in \mathcal{B}$,

   \[A\bigcap B=\left(\sum_{j=1}^n E_j\right)\bigcap \left(\sum_{k=1}^mF_k\right)\\ =\sum_{j=1}^n\sum_{k=1}^m\left( E_j\bigcap F_k \right)\]

   $E_j, F_k\in \mathcal{S}\Longrightarrow E_j\bigcap F_k\in \mathcal{S}$. Thus $A\bigcap B\in \mathcal{B}$.

3. $\mathcal{B}\ni A=\sum_{j=1}^nE_j $, where $E_j\in \mathcal{S}$. $A^c=\bigcap_{j=1}^nE_j^c$, where $E_j^c=\sum_{k_j=1}^{l_j}F_{j,k_j}$ and $F_{j,k_j}\in \mathcal{S}$. Thus \(A^c=\bigcap_{j=1}^n\left(\sum_{k_j=1}^{l_k}F_{j,k_j}\right)\\ =\sum_{k_1=1}^{l_1}\cdots\sum_{k_n=1}^{l_n}\left(\bigcap_{j=1}^{n}F_{j,k_j}\right)\\\) where the finite intersection $\bigcap_{j=1}^{n}F_{j,k_j}\in \mathcal{S}$. Thus $A^c\in \mathcal{B}$. $\blacksquare$

additive and $\sigma$-additive

additive function

Suppose $\mathcal{C}\subseteq \mathcal{P}(\Omega)$, $\emptyset\in \mathcal{C}$, function $\mu: \mathcal{C}\to \mathbb{R}_+\bigcup \{+\infty\}$ is additive if:

1. $\mu(\emptyset)=0$
2. If $E_1,\cdots,E_n\in \mathcal{C}$ are disjoint $E=\sum_{j=1}^nE_j\in \mathcal{C}$, then $\mu(E)=\sum_{j=1}^n\mu(E_j)$.

There are two observations:

1. if $\exists A\in \mathcal{C}$, $\mu(A)<\infty$, then the property 1: $\mu(\emptyset)=0$ is redundant.

   Proof

   \(A=A\bigcup \emptyset\Longrightarrow \mu(A)=\mu(A)+\mu(\emptyset)\\ \overset{\mu(A)<\infty}{\Longrightarrow}\mu(\emptyset)=0\) $\blacksquare$

2. if $E\subseteq F$, $F\backslash E\in \mathcal{C}$, then $\mu(E)\leq\mu(F)$.

   Proof

   If $\mu(E)<\infty$, according to property 2, we have $\mu(F)=\mu(E)+\mu(F\backslash E)\geq \mu(E)$.

   If $\mu(E)=\infty$, $\mu(F)=\mu(E)+\mu(F\backslash E)=\infty$. $\blacksquare$

An discrete measure example for additive function is assign each $X_j\in \Omega$ a measure $P_j\geq 0$. Then For $A\in \mathcal{C}\subseteq \mathcal{P}(\Omega)$, $\mu(A)=\sum_jP_j\mathbb{1}\{X_j\in A\}$.

$\sigma$-additive function

For additive function defined above, force the $E_j$ to be disjoint and INFINITE, we can get $\sigma$-additive function, i.e. replace the second property with

2. If $E_j\in \mathcal{C}$, $E_j\bigcap E_k=\emptyset$ for $j\ne k$, $E=\sum_{j\geq 1}E_j\in \mathcal{C}$, then $\mu(E)=\sum_{j\geq 1}\mu(E_j)$.

You may gradually understand the meaning for $\sigma$. Remember the sigma additive property for measure we define in the previous section and the $\sigma$-algebra we defined above. $\sigma$ means infinite disjoint.

additive but not $\sigma$-additive

Below is an example for a function additive but not $\sigma$-additive.

Suppose $\Omega=(0,1)$, $\mathcal{C}=\{(a,b], 0\leq a < b < 1 \}$,

$(a,b]=\sum_{j=1}^n(a_j,b_j]$, thus $\mu$ is additive.

$(0,\frac{1}{2}]=\sum_{j\geq 1}(x_{j+1},x_j]$, where $x_1=\frac{1}{2}$, $x_j\downarrow0$, $x_{j+1}>x_j>0$, then $\mu((0,\frac{1}{2}])=+\infty$ while $\mu(\sum_{j\geq 1}(x_{j+1},x_j])\to \frac{1}{2}$. Thus it’s not $\sigma$-additive.