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2015/04/30

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.








2015/04/15

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