INTEGRALS 
ITERABLE FIELDS 
21 
S, let it be so on reraov- 
g field A have the com- 
f; 
ipposing 1) is correct if 
ie set ®. For consider 
D lie in the unit square, 
the irrational ordinates, 
hat, when 
tively prime, 
or 
the limited point set SÎ. 
norm d. All the points 
3e are inner points of Si. 
/ (1 
‘j 
(2 
To show that 1) exist we need only to show that for each e > 0 
there exists a d 0 such that for any rectangular divisions D', D" of 
norms < d 0 
j -5 
< e. 
To this end, we denote by E the division formed by superimpos 
ing D" on E'. Then E is a rectangular division of norm < d 0 . 
^ % E -% D . = A\ 
If d 0 is sufficiently small, ^ ^ 
an arbitrarily small positive number. Then 
A =(f -/ W/ -T H/1 + 1/1« 
J* E ) Ahl D „ J \Aa'\ 
if -r] is taken small enough. 
2. The integrals 
17, f/, 
heretofore considered may be called the outer, lower and upper in 
tegrals, in contradistinction. 
3. Let f be limited in the limited metric field Si. Then the inner 
and outer lower (upper) integrals are equal. 
For SIq is an unmixed part of Si such that 
Cont Si a = Si, as d == 0. 
Then by 6, 1, nr 
lim f f = f /. 
rf=o ^21 
But the limit on the left is by definition 
r* 
I 
Ln 
4. When Si has no inner points, 
/7=o- 
integrals respectively.