907: σ-Algebra of Set

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of $\sigma$-algebra of set

---

Topics

About: measure

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

---

Starting Context

• The reader knows a definition of set.

---

Target Context

• The reader will have a definition of $\sigma$-algebra of set.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$S$: $\in \{\text{ the sets }\}$
$\ast A$: $\subseteq PowS$
//

Conditions:
1) $S\in A$
$\wedge$
2) $\mathrm{\forall }a\in A(S\setminus a\in A)$
$\wedge$
3) $\mathrm{\forall }s:\mathbb{N}\to A({\cup }_{j\in \mathbb{N}}s(j)\in A)$
//

Each element of $A$ is called "measurable subset".

---

2: Note

In other words, $s$ is a sequence, $s(0),s(1),...$.

$s$ can be practically any finite sequence, because $s$ can map to the same $s(N)$ for each $N+1\le j$, and ${\cup }_{j\in \mathbb{N}}s(j)=s(1)\cup ...\cup s(N)$.

According to the conditions, these typical subsets are inevitably contained in $A$ among others: $\mathrm{\varnothing }=S\setminus S\in A$; ${\cap }_{j\in \mathbb{N}}s(j)=S\setminus {\cup }_{j\in \mathbb{N}}(S\setminus s(j))\in A$: see the proposition that for any set, any subset minus the union of any subsets is the intersection of the 1st subset minus the 2nd chunk of subsets.

The motivation for considering this concept is to define 'measure' on $S$: we want to measure some subsets of $S$, but we do not necessarily really need to measure all the subsets of $S$, so, we determine what subsets we need to measure, and they are exactly the elements of $A$, which is the reason why each element of $A$ is called "measurable subset".

Of course, $A$ is not automatically determined for $S$: it is a choice. $A=PowS$ or $A=\{S,\mathrm{\varnothing }\}$ is a possible choice.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>