axiomatic sets 8 (subsets)

By axiom of extensionality, we defined two same sets.

Its condition is all elements of each sets is same.
If $a=\left\{ x,y  \right\}$  and $b=\left\{ x,y  \right\}$ , then $a=b$ .

If there are different elements in two sets,
or there are not some elements in each sets,
then two sets are not same.

Given two sets $a,b$ , if
\[ \forall x[x\in a\rightarrow x\in b] ,  \]
then the set $a$  is called the subset of the set $b$ .
We will write it $a\subset b$ .

Usually $a\subset b$  accepts the case of $a=b$ .

Hence, in addition if
\[ \exists y[y\notin a\wedge y\in b ] , \]
then $a$  is called the proper subset of $b$ .

The proper subset means the elements of the subset is always in the set ,
and there are some elements of the set not in the subset . Therefore, $a\neq b$ .

Please understand the difference of $\in$  and $\subset$ .

Although both are the binary relationship, $\in$  is the relation of a element and a set,
and $\subset$  is the relation of a set and a set.

Given two sets $a=\left\{ x,y  \right\},b=\left\{ x,y,z  \right\}$ .
As $a\subset b$ ($a$  is the proper subset of $b$ ) , $a\notin b$ .

As $x$ (and $y$) is in both $a$ and $b$ , $x\in a$ and $x\in b$ .
However, $x$ are not the subset of $a$  and $b$ ,
because the elements of $x$ are not in $a$ and $b$ .

If $a=\left\{ x,y  \right\},b=\left\{ x,y,\left\{ x,y  \right\}  \right\}$ , then $a\in b$  and $a\subset b$ .

Please note again the difference of $\left\{ x  \right\}$  and $\left\{ \left\{ x  \right\}  \right\}$ .


(This is not axiom of ZFC. We will use in axiom of power set. )

axiomatic sets 7 (union of sets)

By using axiom of pairing, we can define the union of sets.

\[ \forall x\exists z\forall w[w\in z\leftrightarrow \exists v[w\in v\wedge v\in x]]  \]

It is axiom of union.
The collection of all the elements $w$ of the elements $x,y$ of a set $a$ becomes a set.
We will write the set $\cup a$ .

If you know the naive set theory, you might feel the axiom of union strange.
Why the elements of the elements of a set are needed?
Because $a=\left\{ x,y  \right\}$  is needed.

Given a set (non ordered pair) $a=\left\{ x,y  \right\}$ and
$x=\left\{ x_1,x_2  \right\},y=\left\{ y_1,x_2  \right\}$ .

Axiom of union asserts all elements of the elements $x,y$ of a set $a$ is a set.
\[  \cup a=\left\{ x_1,x_2,y_1  \right\} . \]
We will define $\cup a=\cup\left\{ x,y  \right\}=x\cup y$ .

We will also write $\cup a=\left\{ x_1,x_2  \right\}\cup \left\{ y_1,x_2  \right\}$ .
In the same way,
\[ \left\{ x_1,\cdots ,x_n  \right\}=\left\{ x_1\right\}\cup \left\{ x_2,\cdots ,x_n  \right\}   \]
will be defined.

By Aixiom of union, we are able to use a set whose elements are three or more.

Using the symbol $x\cup y$ , the union of sets is stated by
\[ \forall w[w\in x\cup y\leftrightarrow w\in x\vee w\in y] .  \]

If a union of three sets $x_1,x_2,x_3 $ is needed, then put $a=\left\{ x_1,\left\{ x_2,x_3 \right\}  \right\}$  and
\[  \cup a=\cup \left\{ x_1, \cup\left\{ x_2,x_3  \right\}  \right\}=x_1\cup x_2\cup x_3 .  \]
Continuing this operation, the union of $x_1,\cdots x_n$ will be gotten by
\[ x_1\cup x_2\cup \cdots \cup x_n .   \]

For concrete example, the union of $a=\left\{ 0,1  \right\}$  is
\[ \cup a=0\cup 1=\left\{ \right\}\cup \left\{ \left\{ \right\} \right\} =\left\{ \left\{ \right\} \right\}=1 .  \]
This is not able to be gotten in the naive set theory.

axiomatic sets 6 (pairing)

As a set is a gathering of some matters, the elements of a set have no orders.
Given a set having elements $x,y,z$ , then
$\left\{ x,y,z \right\}=\left\{ x,z,y  \right\}=\left\{ y,x,z  \right\}=\left\{ y,z,x  \right\}=\left\{ z,x,y  \right\}=\left\{ z,y,x  \right\} $ .

We want to give an order for elements and
distinguish two sets having same elements of which the orders are not same.

At first we will define a set having only two elements.

\[ \forall x\forall y\exists a\forall w[w\in a\leftrightarrow w=x\vee w=y] \]

Namely, $a=\left\{ x,y  \right\}$  or $\left\{ y,x  \right\}$ . It is called axiom of pairing.

If we write $a=\left\{ x,y  \right\}$ , then it means $x\neq y$ , and
if $x=y$ , then we will write $a=\left\{ x  \right\}$ or $a=\left\{ y  \right\}$ .
Therefore, $\left\{ x,x  \right\}=\left\{ x  \right\}$ .

Given two sets $a=\left\{ x,y  \right\}$ and $b=\left\{ u,v  \right\}$ ,
if $a=b$ , then $x=u$  and $y=v$ , or $x=v$  and $y=u$ by axiom of extensionality.

That is to say $\left\{ x,y  \right\}=\left\{ u,v  \right\}$ and  $\left\{ x,y  \right\}=\left\{ v,u  \right\}$ .

However, if $a=b$ means only the case of $\left\{ x,y  \right\}=\left\{ u,v  \right\}$
and $\left\{ x,y  \right\}\neq \left\{ v,u  \right\}$ ,
then the only conditions of $x=u$ and $y=v$ will be required.

We will define an ordered pair for two elements by
\[ \left\{ \left\{ x  \right\}, \left\{ x,y  \right\} \right\}  \]
and write it $(x,y)$ or $\lt x,y \gt$ . Hence, definitely $\lt x,y \gt\neq \lt y,x \gt$ .

With concrete description, $\lt 0,1 \gt\neq\lt 1,0 \gt$ . Because
$\lt 0,1 \gt=\left\{ \left\{ 0  \right\}, \left\{ 0,1  \right\} \right\}$ ,
$\lt 1,0 \gt=\left\{ \left\{ 1  \right\}, \left\{ 1,0  \right\} \right\}$ ,
although $\left\{ 0,1  \right\} =\left\{ 1,0  \right\}$ .
(and you might know $\left\{ \left\{ 0  \right\}, \left\{ 0,1  \right\} \right\}=\left\{ 1, \left\{ 1,0  \right\} \right\}$ . )

By this definition, an order of two elements in a set will be decided and
two same sets having two different elements become equivalent,
 just only when the order of elements is equivalent.

If you want to set out three elements, you may put
\[  \lt a,b,c \gt=\lt a,\lt b,c \gt \gt=\left\{ \left\{ a  \right\}, \left\{ a,\left\{ \left\{ b  \right\}, \left\{ b,c  \right\} \right\}  \right\} \right\} \]
More elements if you need, then
\[ \lt a_1,a_2,a_3,\cdots,a_n \gt=\lt a_1,\lt a_2,a_3,\cdots,a_n \gt \gt   \]

Axiom of paring means that there is a set such that has only one or two elements.
Using the axiom, we are able to get an order of elements.
For example, an arity of functions will be noted by a set.

axiomatic sets 5 (extensionality)

Second axiom of ZFC is the extensionality of a set.
By this axiom, we are able to distinguish sets by means of members.

It is stated by following formula;
\[ \forall x\forall y[\forall z[z\in x\leftrightarrow z\in y]\leftrightarrow x=y]   \]

Using a familiar expression,
\[ x=y\leftrightarrow \forall z[z\in x\leftrightarrow x\in y]  \]

This axiom means three things.
one is a set (except for empty set) has elements,
two is the difference of two sets is only dependent on each elements, and
three is a set can be extended by adding distinct elements.

If $x\neq y$ are two sets, then there are some elements (or a element)
that belong to one but not the other.
For example, if
$x=\left\{ x;1\leq x\leq \sqrt{2}, x=\mbox{ real number}  \right\}$ , and
$y=\left\{ y;1\leq y\lt \sqrt{2}, y=\mbox{ real number}  \right\}$ , then $x\neq y$ .
If
$x=\left\{ x;1\leq x\leq \sqrt{2}, x=\mbox{ rational number} \right\}$ , and
$y=\left\{ y;1\leq y\lt \sqrt{2}, y=\mbox{ rational number} \right\}$ , then $x= y$ .

The axiom of extensionality requires that a set will be only identified by the elements.
When we deal with a set, we have to observe the elements of a set carefully.

By contraries if the elements are not defined or are defined ambiguously,
 the set can not exist.
(However, you will see the gathering of some matters not a set,
 although the elements are defined with no doubt. )

As the elements create a set, empty set is always one in any spaces.
There do not exist two different empty sets.

It can be possible that empty set becomes a element.
\[ \left\{ \phi \right\} =\left\{ \left\{ \right\}\right\}  \]

We will define this set $\left\{\phi\right\}$ as $1$ ;
\[ \left\{\phi\right\}=1   \]
because in the theory $1$ also has to be a set.

By this definition, you must understand that  $\phi$  is not equivalent to $\left\{\phi\right\} $ .

In a addition, we will always be able to use the number $0$ and $1$ in any spaces.

axiomatic sets 4 (empty set)

First axiom of ZFC is the existence of empty set.

It is stated by following logical formula;
\[ \exists x\forall y[y\notin x]  \]

Using A familiar expression,
\[ \phi=\left\{ \right\}=\left\{x ; \forall y\notin x\right\},  \]
or,
\[  \forall y\notin \phi \]

This axiom means three things. One is a set(e.g. empty set) always exists in any spaces,
two is there is a set which has no elements,
and three is any sets has empty set as a element

Namely, it is always true that if $x\in\phi$ , then any $y$ has the member $x$ .

As the number of elements of empty set is zero, $0$ can be used instead of $\phi$ .
Please note that $0$  is a set in the theory.

This axiom is proved by another axioms.
However, it is one of ZFC-axioms.

axiomatic sets 3 (ZFC "a class")

In "naive set thory" (initiated by G. Cantor and R. Dedekind in the 1870s),
a set has been given the following conditions.

A set is a group of some matters which are usually called elements or members and satisfies ;
(1)It is possible to determine whether an element (in the space) should be included in the set or not,
(2)It is possible to determine whether two elements in the set are identical or not.

However, such weak conditions of a set are fraught with some paradoxes and contradictions.

Zermelo-Fraenkel set theory (referred to as the ZFC including axiom of Choice ) begins
with putting some axioms in order to avoid such problems.
Let us have a look at these axioms.

Before beginning with the axioms, we have to define  "a class".
In the naive set theory, any gathering of some matters is a set
whereas in the ZFC, it would be merely be a class rather than a set.
A class consists of sets and proper classes.
Although a proper class is a gathering of some matters, it is not a set.

Only a gathering of elements which is adapted to ZFC-axioms is deemed as a set.
A proper class is a gathering of elements, but it is not a set.

Please do remember that it is the great difference between the ZFC and naive set theory.

axiomatic sets (ZFC) 2

Axiomatic set theory was in large part found in 1908-1922 by E. Zermelo and A. Fraenkel. 

It is much different from the set theory which we study in beginner class.
We will start from;

(1)undefined terms
(2)axioms
(3)notations and symbols which are restricted

For example, if a set $A$  and a set $B$ are sets, then the couple $\left\{ A,B \right\}$ is a set. This is "axiom of pairing" . We will formally express;
\[  \forall x\forall y\exists z\forall w[w\in z \leftrightarrow w=x\vee w=y]  \]

When a set $A$ and a set $B$  are sets, we have definitely believed $A \cup B$  is a set. However, the axiomatic system requires the condition.

Why was the axiomatic set theory needed?

The classical set theory, we know as "the naive set theory", is very intuitive and easy to understand, but it had been fraught with some contradictions and problems which are not solved.

axiomatic sets (ZFC)

In axiomatic set theory, all objects is a set.
Although it is very natural, it is not easy to understand.

For example, $\mathbb{N}$  is "natural number" and must be a set in the theory.

The definition is as follow;

You must know a empty set $\phi$ .

$\exists y \forall x\neg [x\in y]$ .

$y$ means $\phi$ .

We represent $\phi=0$ and $\left\{ \phi \right\}=1$ .

Next, $2=1+\left\{ 1  \right\}=\left\{\phi, \left\{ \phi \right\}  \right\}$ ,

$\cdots \cdots \cdots$ , and

$n+1=1+\left\{ n  \right\}$ .

In addition, axiom of infinity is defined;

$\exists x[0\in x\wedge \forall y(y\in x\rightarrow (y\cup \left\{y \right\})\in x)]$ .

How do you feel this.

measures 14

We defined an inner measure $m^i$ for an arbitrary set $C$;
\[  m^i(C)=|J|-m^o(J\cap C^c) . \]
, where $m^o$ was an outer measure, and $J$ was a measurable set which covered $C$ fully.

An outer measure will be also defined
\[  m^o(C)=\inf \left\{\sum m(J_i)| C\subset\cup J_i  \right\} .  \]

You may think an inner measure defined by
\[  m^{ii}(C)=\sup\left\{ \sum m(I_i)| \cup I_i\subset C   \right\} .   \]

However, $m^i$  can handle more sets than $m^{ii}$ .

For example, given $\Omega=[0,1]$ and
\[  f(x)=\left\{ \begin{array}{ll}
x=1 & x\in\mathbb{Q} \\
x=0 & x\notin\mathbb{Q}
\end{array}    \right. \]

What does the measure of $f$ ?

measures 13

Supposed that a set $C$ , and a set $J$  which fully covers $C$ .
$C\subset J$

The set $C$  is arbitrary.   Its shape may be complicated, or not .

However $J$  is measurable.
For example, $J$  is a big square, and $m(J)$ is height multiplied by width.
$m(C)\leq m(J)$

If a set is measurable, the value of the measure is equal to it of the outer measure. 
$m(J)=m^o(J)=|J|$

Then, an inner measure $m^i$  will be defined by the outer measure,
\[  m^i(C)=|J|-m^o(J\cap C^c) . \] 
, where $C^c$  is the complementary set of $C$ .

We must accept that there exist some measurable sets $J$ which fully cover an aribtrary set $C$ .
It would not be strong opposition.

measures 12

You may think it is strange that an outer measure becomes a measure.

An outer measure is a measure of a certain set to cover its area fully.

This means to allow measurement of a given set by approximating it from the outside of its perimeter.

Namely, an outer measure contains all elements of the set and some elements outside of the set.

However, an inner measure contains just only the elements in the set although it is not able to cover fully.

$\cup I^k\subset C\subset \cup I^l$ .

The theorem of Darboux needs that an outer measure will be consistent with an inner measure.

$m(\cup I^k)\leq m(C)\leq m(\cup I^l)$  .

How do we understand an inner measure?

measures 11

If you carefully watch the proof (in 'measure 9') of
"if $a_1,a_2\in\mathcal{F},a_1\cap a_2=\phi$ , then $(a_1\cup a_2)\in\mathcal{F}$ ",
you will know it contains that an outer measure becomes a measure.

Namely,
" $m^o(e)\leq m^o(e\cap a)+m^o(e\cap a^c)$ "(means $m^o$  is an outer measure),
and
" $m^o(e)\geq m^o(e\cap a)+m^o(e\cap a^c)$ ".

Therefore, we understand that
" $m^o(e) =m^o(e\cap a)+m^o(e\cap a^c)$ "
is satisfied.

However, this proposition is true on just only the already defined collection of sets.

The collection of sets

$\mathcal{F}=\left\{a\subset\Omega|\forall e\subset\Omega,m^o(e)=m^o(e\cap a)+m^o(e\cap a^c) \right\}$

is a $\sigma$ algebra and on the collection of sets, an  outer measure becomes a measure.

measures 10

On an arbitrary space $\Omega$ , if there is an outer measure $m^o:a\subset\Omega

\rightarrow \mathbb{R}$ , the collection of sets
$\mathcal{F}=\left\{a\subset\Omega|\forall e\subset\Omega,m^o(e)=m^o(e\cap a)+m^o(e\cap a^c) \right\}$
could be defined.

It has been proved that for the elements in this collection $\mathcal{F}$ ,
if $a_1,a_2\in\mathcal{F},a_1\cap a_2=\phi$ , then $(a_1\cup a_2)\in\mathcal{F}$ .

Although the proof is not easy, the number of set in the collection can be extended from 2 to $\infty$ . Namely, for $a_1,a_2,\cdots$  and $a_i\cap a_j=\phi$ , then $(\cup a_i)\in \mathcal{F}$ .

The collection $\mathcal{F}$ is called $\sigma$ algebra, and
on $\sigma$ algebra, an outer measure $m^o$  becomes a measure $m$ .

measures 9

[the preceding definition] You have to remember;
(1)$m^o$  is an outer measure,
(2)$\mathcal{F}=\left\{a\subset\Omega | \forall e\subset\Omega, m^o(e)=m^o(e\cap a)+m^o(e\cap a^c)   \right\}$

For $a_1,a_2\in\mathcal{F}$ , $a_1\cap a_2=\phi$ ,and  $\forall e\subset\Omega$ , we shall prove that
$m^o(e)=m^o(e\cap (a_1\cup a_2))+m^o(e\cap (a_1\cup a_2)^c)  $ .

If $A=a_1\cup a_2$ , then $A^c=(a_1\cup a_2)^c=a_1^c\cap a_2^c$ .
As $e=(e\cap A)\cup (e\cap A^c)=(e\cap(a_1\cup a_2))\cup (e\cap(a_1\cup a_2)^c)$ ,and
$m^o$  is an outer measure,
$m^o(e)\leq m^o(e\cap(a_1\cup a_2))+m^o(e\cap(a_1\cup a_2)^c)$ .

You know that as $a_1,a_2\in\mathcal{F}$ ,
$m^o(e)= m^o(e\cap a_1)+m^o(e\cap a_1^c)$  and $m^o(e)= m^o(e\cap a_2)+m^o(e\cap a_2^c)$ .

For any $e\subset\Omega $ , as this assertion is always true, please replace $e$  to $e\cap a_1^c$ in last equation . Hereby,
\begin{eqnarray*}
m^o(e\cap a_1^c)&=&m^o((e\cap a_1^c)\cap a_2)+m^o((e\cap a_1^c)\cap a_2^c)\\
&=&m^o(e\cap(a_1^c\cap a_2))+m^o(e\cap(a_1^c\cap a_2^c)) \\
&=& m^o(e\cap a_2)+m^o(e\cap(a_1\cup a_2)^c )
\end{eqnarray*}
where as $a_1\cap a_2=\phi$ , $e\cap(a_1^c\cap a_2)=e\cap a_2$ .
Therefore,
\[ m^o(e)=m^o(e\cap a_1)+m^o(e\cap a_2) +m^o(e\cap(a_1\cup a_2)^c) . \]

As $m^o$  is an outer measure,
\begin{eqnarray*}
m^o(e\cap a_1)+m(e\cap a_2)&\geq&m^o((e\cap a_1)\cup(e\cap a_2)) \\
&=&m^o(e\cap(a_1\cup a_2)) .
\end{eqnarray*}

We get
\[ m^o(e)\geq m^o(e\cap (a_1\cup a_2))+m^o(e\cap(a_1\cup a_2)^c) . \]

This assertion reverses an inequality sign on an outer masure.

Since, if $'\geq'$  and $'\leq'$  happen at same time in a equation, then just only $'='$ is available,
\[ m^o(e)= m^o(e\cap(a_1\cup a_2))+m^o(e\cap(a_1\cup a_2)^c) . \]
Hence, $a_1,a_2\in\mathcal{F}\quad (a_1\cap a_2=\phi)$ , then
\[  (a_1\cup a_2)\in \mathcal{F} \]

measures 8

Given a map of measurement $m^o$ in a $\mathcal{C}$ which is a collection of sets in $\Omega$ .
$m^o: a\in\mathcal{C}\rightarrow x\in\mathbb{R}$

Suppose $m^o$  satisfies; 
  
(1)for any $a\in\mathcal{C}$ ,  $m^o(a)\geq 0$ ,

(2) $m^o(\phi)=0$ ,

(3) if $a_1\subset a_2 (\in \mathcal{C})$ , then $m^o(a_1)\leq m^o(a_2)$ , 
(4) if $a_1,a_2,\cdots\in\mathcal{C}$  ,  $a_i\cap a_j=\phi (i\ne j)$ , then $m^o(\cup a_i)\leq \sum m^o(a_i)$  ($i,j=1,2,\cdots $)  . 

We calls $m^o$  an outer measure. By using $m^o$ , a following collection $\mathcal{F}$  of sets can be defined. 

$\mathcal{F}=\left\{ a\subset\Omega | \mbox{any } e\subset\Omega, m^o(e)=m^o(e\cap a)+m^o(e\cap a^c)  \right\}\subset\mathcal{C}$ , 

where $e$  is an arbitrary set in $\Omega$  and $a^c$  is the complementary set of $a$  in $\Omega$. 

What properties $\mathcal{F}$  has?

(i)$\Omega, \phi\in \mathcal{F}$
Any collection of sets and $\Omega$  have the empty set $\phi$ .If $a$  is $\phi$ , then $a^c$  is $\Omega$ . 
Therefore, when $a$  is $\phi$ , for any $e$ , 
$m^o(e)=m^o(e\cap \phi)+m^o(e\cap \Omega)$  is always true. 

(ii)if $a\in \mathcal{F}$ , then $a^c\in\mathcal{F}$ . 
As for any set $b\in\mathcal{F}$  $(b^c)^c=b$ , $m^o(b)=m((b^c)^c)$ . 
Therefore, when $a$  is $b^c$ ,  for any $e$ , 
$m^o(e)=m^o(e\cap b^c)+m^o(e\cap (b^c)^c)$  is always true. 

(iii)if $a_1,a_2\in\mathcal{F}$ , and $a_1\cap a_2=\phi$ , then  $a_1\cup a_2\in \mathcal{F}$ . 
 (　→　We will prove this proposition in next post. )

measures 7

The properties of a mesure $m$  are, for a $\mathcal{F}$  of $\Omega$ ,

(1) for any $a\in\mathcal{F}$ ,  $m(a)\geq 0$ ,

(2) $m(\phi)=0$ ,

(3) if $a_1,a_2,\cdots\in\mathcal{F}$  ,  $a_i\cap a_j=\phi (i\ne j)$ , then $m(\cup a_i)=\sum m(a_i)$  ($i,j=1,2,\cdots $)  . 

Here, we will weaken the condition (3) to 

(3'-1) if $a_1\subset a_2 (\in \mathcal{F})$ , then $m(a_1)\leq m(a_2)$ , 
(3'-2) if $a_1,a_2,\cdots\in\mathcal{F}$  ,  $a_i\cap a_j=\phi (i\ne j)$ , then $m(\cup a_i)\leq \sum m(a_i)$  ($i,j=1,2,\cdots $)  . 

This means that we will accept to measure a set to large. 
(This means to allow measurement of a given set by approximating it from the outside of its perimeter.)
Such a ruler is called an outer measure $m^o$. 

An outer measure $m^o$  is more natural than the preceding measure $m$  satisfying (3).  

Using the outer measure $m^o$ , we will define the family of sets such that 

$\left\{ a\subset\Omega | \mbox{any } e\subset\Omega, m^o(e)=m^o(e\cap a)+m^o(e\cap a^c)  \right\}$ , 

where $a^c$  is the complementary set of $a$  in $\Omega$. 

C.Caratheodory proved these sets were measurable. Hence, it is called Lebesgue measurable sets.

measures 6

Infinite operations in measurements need the conditions to rulers and things that we want to measure.

At first, rulers $m$  have to satisfy,

(1) for any $a\in\mathcal{F}$ ,  $m(a)\geq 0$ ,

(2) $m(\phi)=0$ ,

(3) if $a_1,a_2,\cdots\in\mathcal{F}$  ,  $a_i\cap a_j=\phi (i\ne j)$ , then $m(\cup a_i)=\sum m(a_i)$  ($i,j=1,2,\cdots $) . 

, and measured things $\mathcal{F}$ have to satisfy,

(1)$\Omega, \phi\in\mathcal{F}$ ,
(2)if $a\in\mathcal{F}$ , $a^c\in\mathcal{F}$
(3)if $a_1,a_2,\cdots\in\mathcal{F}$  , then $(\cup a_i) \in \mathcal{F}$ . 

C. Caratheodory proved these conditions were closely linked.

measures 5

At first, infinite operations require that a union of sets is in the $\sigma$  algebra.
That is to say,
if all $I^j\in \mathcal{F}$ , then $\cup I^j\in\mathcal{F}$ .

In addition, as we want to measure the area of a figure $C$ , $m(\cup I^j)$  must exist.
Hence, the following thing will be put in axioms.

If all $I^j\in \mathcal{F}$ , and any $I^k\cap I^l=\phi$ , then $m(\cup I^j)=\sum m(I^j)$ .

It means, for example, for any $I^k\cap I^l=\phi$ ,
if $m(I^1)=1$ ,
$m(I^2)=0.4$ ,
$m(I^3)=0.01$ ,
$m(I^4)=0.004$ ,
$m(I^5)=0.0002$ ,
$m(I^6)=0.00001$
$\cdots$ , $\cdots$ , $\cdots$　
, then $m(\cup I^j)=\sum m(I^j)=\sqrt{2}$ .

This proposition shows a precise and adequate measurement for infinite operations.

If $m(\cup I^k)\leq m(C)\leq m(\cup I^l)$  and $m(\cup I^k)=m(\cup I^l), \mbox{ when } k,l\rightarrow \infty $ ,
we will get $m(\cup I^k)=m(\cup I^l)=m(C)$ .

measures 4

For measuring various kinds of figures $C$  in $\Omega=([0,1]\times [0,1])\subset \mathbb{R}^2$ , infinite operations are required;
\[  \sup\cup I^l \subset C\subset \inf\cup I^k  \]
,where $I$s  are squares whose area is precisely known and measurable.

You might think this way is a matter of course.  However, it is not always promised that the infinite operations for subsets in $\Omega$  will bring the desired outcomes.
Then, the following things are put in axioms.

If all $I^j\subset \Omega$ , then $\cup I^j\subset \Omega$  $(i=1,2,\cdots)$ .

In other words,

If $I^j\in \mathcal{F}$ , then $\cup I^j\in\mathcal{F}$   $(i=1,2,\cdots)$ ,
where $\mathcal{F}$  is a family of subsets in $\Omega$ .

It indicates $\cup I^j$  is measurable and $m(\cup I^j)$  exists, including $\pm\infty$ .
 

measures 3

There is an arbitrary figure $C$ in $\Omega=( \left[0,1\right]\times\left[0,1\right] )\subset \mathbb{R}^2$ . Then, we want to measure the area of $C$ . If the set function $m$  is a measure of the area in $\Omega$ ,
\[  m : C\in\Omega\rightarrow x\in\mathbb{R}, \]
it is clear that $m(C)\leq 1$ .

However, as we want to measure more precisely any figures $C$, we keep making many various kinds of  small squares $I^k$ whose horizontal size is $c_h$  and vertical size is $c_v$ . Then,  $m(I^k)=c_h^k\times c_h^k,\quad (0\lt c_h,c_v\lt 1,k=1,2,\cdots)$ .

Covering $C$  by some $I^k$ , we can put $C\subset \cup I^k$ and covering some $I^l$  by $C$, we can put $\cup I^l\subset C$ . Hence,
\[  \cup I^l\subset C\subset \cup I^k  \]

Reducing the size of squares and increasing the number of squares,
\[  \sup\cup I^l\subset C\subset \inf\cup I^k  \]
Please note that $\sup\cup I^l$  and $\inf\cup I^k$  must be measurable.

If $l,k\rightarrow \infty$  and
\[  m(\sup\cup I^l)=m(\inf\cup I^k)=m^* , \]
then we will define $m(C)=m^*$ .

You will remember the theorem of Darboux in integral.

measures 2

Let consider a square of the vertical and horizontal size 1.
Although it is the very small square, if the elements of the square are the combinations of two real numbers $(c_1,c_2)$ , the number of the elements is $\aleph$ ,
and many subsets are included in the square.
\[  \Omega=\left\{ (c_1,c_2)\in \mathbb{R}^2 | 0\leq c_1,c_2\leq 1 \right\}  \]

A simplest subset in $\Omega$  is a arbitrary figure. We want to measure the area of the subset.
Therefore, making many small quadrangles whose area is trivial, we will try to cover the figure with them.

If the figure is also a quadrangle, the figure will be precisely covered with some small quadrangles.  Then, the area of the figure is measured by calculating the small using quadrangles.

However, we can not cover the figure whose shape is curves with any ordinary quadrangles.
Shortage or surplus occurs at the corner.

Infinite operations will be required.

measures

A measure $m$  is a set function which maps a subset in the measure spaces to a real number. 
\[ m: a\in\mathcal{F}\rightarrow m(a)=x\in\mathbb{R} \]
, where $\mathcal{F}$  is a family of subsets of the measure space. 

What conditions are needed for a measure $m$ as a set function?

On standard definitions (or finally reached definitions), it is as follows;

(1) for any $a\in\mathcal{F}$ ,  $m(a)\geq 0$ ,

(2) $m(\phi)=0$ ,

(3) if $a_1,a_2,\cdots\in\mathcal{F}$  ,  $a_i\cap a_j=\phi (i\ne j)$ , then $m(\cup a_i)=\sum m(a_i)$  ($i,j=1,2,\cdots $) 

You may think it is very natural. Perhaps it is fairly accurate in terms of  finite operations or various sets having a number of good shapes. 

However, there are very strange sets (or subsets) in a measure space.  

Therefore, these conditions are closely related with the ones which the families $\mathcal{F}$ of subsets satisfy. 

measure spaces 2 (set functions)

You might want to know a reason why we need to prepare a set function. 

Saying brief words, because a measure is a set function and a measure space is based on a measure (e.g. a set function). However it might not be easy to understand these relations.

As a set function is the mapping from a set to a real number,

the input values or the arguments of the function are sets.

Therefore, the domain of the set function must be a family of sets.

In an elementary level, a function will be defined on a space (or a universal set), 

after creating the space (e.g. $\mathbb{R}^n$). 

We can not give a value of an element of a space to the set function, and

we must give a subset of a space. 

It is not enough to just only define the ordinary space for a set function.

That is to say, for operating a set function, we have to define a space and subsets in the space.

Measure spaces need a set function and a family of sets in the space.  

Usually, a measure space is written by the triplet $(\Omega, \mathcal{F},m)$ , where 

$\Omega$  is a universal set or a whole space, 

$\mathcal{F}$  is a family of subsets in $\Omega$ , and 

$m$  is a measure (a set function). 

measure spaces (set functions)

For creating a measure space, some preparations will be required.
These preparations, although using terms are almost known,
if you are not familiar with, may be difficult to understand.


First of all, we shall define a set function.
For an arbitrary set $A$ , if a function gives a real number $x$,
the function is called a set function.

$f: A\rightarrow x\in\mathbb{R}$

(1) for an interval $A=[a,b]\subset \mathbb{R}$ ,
the function $f(A)=|b-a|$ is a set function.

(2) for a finite set $A=\left\{a_1,a_2,\cdots a_n\right\}$ ,
the function $f(A)=n$  which gives a number of elements of $A$
is a set function. This function is called number measure and
usually expressed by $\#(A)$.

Please note that a set function is not a correspondence of values to a value,
and it is a correspondence of a set to a real number.
Namely, a set function is not able to get a real value by giving a value of the set.

coffee break 10 ( Happy new year )

New year 2015 has begun.

I would like to continue to pick up a variety of mathematical issues this year.

If you can look at some posts occasionally, I will be so glad.

I want to start from a measure space in next post.

Thanks a lot