Introduction;

온도계는 어떻게 온도를 측정할까? 측정을 한다는 것은 매우 실질적이지만, 깊게 생각하면 할수록 추상적인 개념이 등장한다. 이번 시리즈에서는 기본적인 측도를 위한 추상적 개념을 정의하고, 이것이 일상생활의 측도와 어떠한 연관이 있는지, 그리고 어떻게 응용이 되는지에 대해 알아보고자 한다.

Definition 1.1: Semi-Algebra;

Let $X$ be any nonempty set. Then we define semi-algebra $C$ : a collection of subset of $X: C \subseteq \mathcal{P}(X)$ such that:

(i) $\emptyset, X \in C$ ;

(ii) $\{ A, B \in C \} \Rightarrow \{ A \cap B \in C \}$ ;

(iii) For $\{i, \ j, \ n \in \mathbb{N}\} \wedge \{ 1 \leq i, \ j \leq n \}:$ $\{ A \in C \} \Rightarrow \{ A^C = X \setminus A = \bigcup_{k = 1}^{n}C_{n} \}$ where $\{C_i, C_j \in C \} \wedge \{C_i \ \cap \ C_j = \emptyset \}$ ;

Example 1.1;

Let $X$ be any nonempty set and $C = \mathcal{P}(X)$ ; Then $C$ is a semi-algebra;

Proof)

(i) $\emptyset, X \in C = \mathcal{P}(X)$ ;

(ii) Claim: $\{ A, B \in C \} \Rightarrow \{ A \cap B \in C \}$ ;

Proof of (ii):
If $A, B \in C$,
then $\{A \cap B \subseteq X\}\Rightarrow \{ A \cap B \in \mathcal{P}(X) \}$
$\iff (A \cap B) \in C \; \blacksquare$

(iii) For $\{i, \ j, \ n \in \mathbb{N}\} \wedge \{ 1 \leq i, \ j \leq n \}:$ $\{ A \in C \} \Rightarrow \{ A^C = X \setminus A = \bigcup_{i=1}^{n}C_{i} \}$ where $\{C_i \in C \} \wedge \{C_i \ \cap \ C_j = \emptyset \}$ ;

Proof of (iii):
If $A \in \mathcal{P}(X) = C$ then $X \setminus A \in \mathcal{P}(X) \; \blacksquare$

Definition 1.2: Algebra;

Let $X$ be a nonempty set. Then we define algebra $\mathcal{F}$ : a collection of subset of $X: \mathcal{F} \subseteq \mathcal{P}(X)$ such that:

(i) $\emptyset, X \in \mathcal{F}$ ;

(ii) $\{ A, B \in \mathcal{F} \} \Rightarrow \{ A \cap B \in \mathcal{F} \}$ ;

(iii) $\{ A \in \mathcal{F} \} \Rightarrow \{ A^C = X \setminus A \in \mathcal{F} \}$ ;

Example 1.2;

Every algebra is semi algebra;

Proof)

Let $X$ be a nonempty set and $\mathcal{F} \subseteq \mathcal{P}(X)$ be an algebra;
Claim: $\mathcal{F}$ is an semi-algebra;

(i) $\emptyset, X \in \mathcal{F}$ ;

(ii) $\{ A, B \in \mathcal{F} \} \Rightarrow \{ A \cap B \in \mathcal{F} \}$ ;

(iii)
Given: $\{ A \in \mathcal{F} \} \Rightarrow \{ A^C = X \setminus A \in \mathcal{F} \}$ ;
Let $n = 1: A^C = \bigcup_{i=1}^{n}C_{i} \; where \; C_i = A^C \in \mathcal{F}$ ;

By (i), (ii) and (i) algebra is semi algebra $\blacksquare$

Example 1.3;

Let $X$ be a nonempty set and let $\mathcal{G} \subseteq P(X)$ with the following properties:

(i) $\emptyset, X \in \mathcal{G}$ ;

(ii) $\{ A, B \in \mathcal{G} \} \Rightarrow \{ A \cup B \in \mathcal{G} \}$ (Union);

(iii) $\{ A \in \mathcal{G} \} \Rightarrow \{ A^C = X \setminus A \in \mathcal{G} \}$ ;

Then following set $\mathcal{G}$ is an algebra;

Proof)

Claim: $\{ Complement \} \wedge \{ Intersection \} \iff \{ Complement \} \wedge \{ Union \}$ ;

($\Rightarrow$ part)

Let $A, B \in \mathcal{G}$

$\Rightarrow A^C, B^C \in \mathcal{G}$ (Complement)

$\Rightarrow (A^C \cap B^C) \in \mathcal{G}$ (Intersection)

$\Rightarrow (A \cup B)^C \in \mathcal{G}$

$\Rightarrow (A \cup B) \in \mathcal{G}$ (Complement);

($\Leftarrow$ part)

$A, B \in \mathcal{G}$

$\Rightarrow A^C, B^C \in \mathcal{G}$ (Complement)

$\Rightarrow (A^C \cup B^C) \in \mathcal{G}$ (Union)

$\Rightarrow (A \cap B)^C \in \mathcal{G}$

$\Rightarrow A \cap B \in \mathcal{G}$ (Complement);

Hence $\mathcal{G}$ is an algebra $\blacksquare$