136 
Two auxiliary propositions precede the proof of this theorem. 
Auxiliary proposition 3: If 0<q<1 and the segment S;(j = 
1,2,..) does not contain the origin, has one point in common with 
the segment dj dj 4 and has a length < y (b;-1—d;), where J; 1 > dj > 0 
; J;_1—d; » ’ 
and lim. 2—* = 0, the set formed by the segments Sj has at the 
i= CO jl 
origin a metric density Sq. 
Proof: We shall assume that for jo 4d; approaches the origin 
because else the auxiliary proposition would be evident. Let $ be 
an arbitrary point to the right of the origin and let w be the smallest 
value of 7 for which S; contains the point & or a point to the left 
of it. Then 
Oi = 0, + (Su—1 =a Ju) "6 =e length Su + (Oi Ju), 
hence 
dif isa lads, eet 
The subset of the segments S; lying between the origin and &, 
has therefore a measure which after division by & is not more than 
Gui = Gi 
§ 
ul 
Jd 
Iie Sig te Le @) 
(2) 
< ; en 5 AR Oy — On 
When § approaches the origin, w increases unlimitedly ; ————— 
ul 
approaches therefore to zero and according to (1) the lower limit of 
Sh. 5 
5 is not less than 1. The last term of (2) approaches to 0, so 
P= ik 
that the metric density of the set formed by the segments is at the 
origin <q. ; 
Auxiliary proposition 4: When lim. § f(§)=0 and when tu any 
50 
point to the right of the origin a number i ts conjugated, so that 
= lies in Gi or on its boundary, we have 
ke) 
appr. lim. Em; f (yi S—1) = 0. 
£=0 
Proof: For 80, t->o, hence according to the condition 
imposed on the net of cells 
1 
Hit E mi & B DEM 
mar na SN ae 1, (yi S—1)7 (qi S—1) 0. (3) 
1; Measure = 
1 
1 
Suppose O<q<1; let be the set of the points in te Aes 
i 
