138 
With a view to this we shall first prove 
appr.lim. F(8)=0, where F (= ES" mj fG; ES (Tye 
—&=—0 y=! 
As f(£) does not increase for increasing £ when it does not pass 
the origin, the terms of the latter sum are 2 0. 
1 
If g represents an arbitrary number between O and 3, ae 0; 
J 
dj 
and /; = the length of the segment (dj, dj»), ae 2 hence ql; <1; = 
| Î 
(a 1) 0;<9;, so that if each of the segments (d;, d;—1) is 
4) 
produced at both ends with an interval of length q/j, none of these 
productions contains the origin. According to the auxiliary proposition 
2 the metric density of the set formed by these productions, is at 
the origin at most 2g. Now let 8 be an arbitrary point to the right 
of the origin and let # be an arbitrary measurable set to the right 
of @ which has no point in common with any of these productions. 
Let us first consider the terms in F'(£) for which § 2 d;; the accent 
excludes the case 0; SE < dj; and the production dj, <8 << d;_1+q] 
does not occur in Z, hence 
(+98; bj a 
a En En HOEN) ,, w<{ = ue dé -[= ie os (5) 
DAE Sis es God, Ls rl ge. zals 
In the remaining terms of F'(8), §< d;, hence for these, because 
the production J; — ql; <$<d; does not occur in £, we have 
ale KEO Seren es \ 
far amen lk Oj =5) ee : LN FO dé hea 
She 
8 ie att 10.8 
(+90; 1; 1-8, 48 FD, i, (6) 
ie pare ie aa f cs fre 
gol, 18 09,1; osb 
0 
for from ES (1+4q) 6;-14; = (149) (1 — 3 <3 ee 2 there follows 
g 
1 
1—§ > 
As lm 6;1;=0 and kek § f(§)=0 we can choose v,; (only 
== 
dependent on the Bert me and the net of cells) in such a way 
