• Continuous set function
• $\sigma$-additive and continuity on algebra
• Extension of set function from semi-algebra to algebra

In this section, we talk about the continuity of set functions.

Continuous set function

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

1. is continuous from below at $E\in \mathcal{C}$ if $\forall E_n\in \mathcal{C}$ where $n \geq 1$, $E_n\uparrow E$, i.e., $E_n\subseteq E_{n+1}$ and $\bigcup_{n\geq 1}E_n=E$,

   \(\mu(E_n)\to \mu(E)\) ​

2. is continuous from above at $E\in \mathcal{C}$ if $\forall E_n\in \mathcal{C}$ where $n \geq 1$, $E_n\downarrow E$ ($E_n\supseteq E_{n+1}$ and $\bigcap_{n\geq 1}E_n=E$), and $\exists n_0$, $\mu(E_{n_0})<\infty$,

   \[\mu(E_n)\to \mu(E)\]

Note that $\exists n_0$, $\mu(E_{n_0})<\infty$. For example $E_n=[n, \infty)$, $\mu(E_n)=\infty$, $\mu(\emptyset)=0$, while $E_n\to \emptyset$.

$\sigma$-additive and continuity on algebra

Now we consider the relation between $\sigma$-additive and continuous function on algebra.

Suppose $\mathcal{A}$ an algebra on $\Omega$, function $\mu: \mathcal{A}\to \mathbb{R}_+\bigcup \{+\infty\}$ is additive.

1. If $\mu$ is $\sigma$-additive $\Longrightarrow$ $\forall E\in \mathcal{A}$, $\mu$ is continuous at $E$
2. If $\mu$ is continuous from below $\Longrightarrow\, \mu$ is $\sigma$-additive
3. If $\mu$ is continuous from above at $\emptyset$ and $\mu$ is finite $\Longrightarrow$ $\mu$ is $\sigma$-additive.

Proof

1. (a) First we prove continuous from below.

   Suppose $E_n\uparrow E$,

   

   define $F_1=E_1$, $F_k=E_k\backslash E_{k-1}$, then $\bigcup_{n\geq 1} E_n=\bigcup_{k\geq 1}F_k$ and $F_k$ are disjoint. Remember $\sigma$ means infinite disjoint, thus

   \[\mu(E)=\sum_{k\geq 1}\mu(F_k)\\ =\lim_{n\to \infty}\sum_{k=1}^n\mu(F_k)\\ =\lim_{n\to \infty}\mu(\sum_{k=1}^nF_k)\\ =\lim_{n\to \infty}\mu(E_n)\]

   is continuous.

   (b) Now we prove $\mu$ is continuous from above. Suppose $E_n\downarrow E$.

   

   define $G_1=E_{n_0}\backslash E_{n_0+1}$, $G_k=E_{n_0}\backslash E_{n_0+k}$, then $G_k\uparrow E_{n_0}\backslash E$, known that $\mu$ is continuous from below, we have

   \[\mu(G_k)\uparrow \mu(E_{n_0}\backslash E)\]

   Thus

   \[\mu(E_{n_0}\backslash E)=\lim_k \mu(E_{n_0}\backslash E_{n_0+k})\\ \overset{additive}{\Longrightarrow} \mu(E_{n_0})-\mu(E)=\lim_k\mu(E_{n_0})-\mu(E_{n_0+k})\\ \Longrightarrow \mu(E)=\lim_k \mu(E_{n_0+k})\]

2. Suppose $E=\bigcup_{i\geq 1}E_i$ where $E, E_i\in \mathcal{A}$, and $E_i$ are disjoint. Denote $F_n=\sum_{i=1}^n E_i$, which is monotone and converges to $E$. Known $\mu$ is continuous from below, we have

   \[F_n\uparrow E\\ \overset{cont. f. b.}{\Longrightarrow}\lim_n\mu(F_n)=\lim_n\mu(\sum_{k=1}^nE_k)= \mu(E)\\ \overset{addtive}{\Longrightarrow} \sum_{k\geq 1}\mu(E_k)=\mu(E)\]

3. In the notation of 2, denote $G_n=E-F_n$, then $G_n$ is monotone decreasing converges to $\emptyset$, thus $\mu(G_n)\downarrow 0$.

   \[\mu(E)=\sum_{i=1}^n \mu(E_i)+\mu(G_n)\\ \overset{G_n\downarrow 0}{\Longrightarrow} \mu(E)=\sum_{i\geq 1}\mu(E_i)\]

   $\blacksquare$

Extension of set function from semi-algebra to algebra

We now show that if we extend a semi-algebra, on which there is a additive function, to algebra, then the extension of set function is unique.

Suppose that $\mathcal{S}$ is a semi-algebra, $\mu:\mathcal{S}\to \mathbb{R_+}\bigcup\{+\infty\}$ is additive, then there exist a function $\nu:\mathcal{A}(\mathcal{S})\to \mathbb{R_+}\bigcup\{+\infty\}$ satisfying

1. additive
2. $\nu(A)=\mu(A)$, $\forall A \in \mathcal{S}$
3. If there exist $\mu_1$ and $\mu_2$, which are set functions on the algebra, and $\mu_1(A)=\mu_2(A)$, $\forall A\in \mathcal{S}$, then $\mu_1(E)=\mu_2(E)$, $\forall E \in \mathcal{A}(\mathcal{S})$.

If $\mu$ is $\sigma$-additive, then the extension is also $\sigma$-additive.