ITERATED INTEGRALS 
65 
Let 21 denote all the points of the unit square. Let us denote 
the discrete point set used in Ex. 1 by ®. We define / now as 
follows : / shall have in T) the values assigned to it at these points 
in Ex. 1. At the other points A = 21 — £),/ shall have the value 1. 
Then f* t* f* f* 
I = I + I =1=1. (3 
j 21 JA A A 
On the other hand S8 af} consists of the irrational points in 58 and 
a finite number of other points. Thus 
f f f= 1. 
(4 
Hence again the two 3), 4) exist and are equal. 
Let us look at the results we get if we use integrals of types I 
and II. We will denote them by Q and F as in 62. 
We see at once that 
G% — V% = Pn = 1. 
Let us now calculate the iterated integrals 
<V CV, (5 
and Fg Fig. (6 
We observe that 
Cq = 1 for x irrational 
= + go for x rational. 
Thus the integral 5) either is not defined at all since the field 
93s does not exist, or if we interpret the definition as liberally as 
possible, its value is 0. In neither case is 
0% = (7© (7g. 
Let us now look at the integral 6). We see at once that 
VvVv 
does not exist, as V& = 1 for rational x, and = +oo for irrational 
x. On the other hand 
F S8 Fs = 1 
Hence in this case 
TV 
TV Fe =4-oo. 
TV F(£.