276 Dr Cramer, On the distribution of primes 



Proof. To prove this lemma, it is only necessary to show that 

 2 7) (n) < ^ e* 



t<\ogn<t+y ^ 



if t belongs to the complementary set S^. This follows immediately 

 from Lemma 3, for we have, since r] (n) > 0, 



1 ^0^) TT + 1 



if X is sufficiently great and t belongs to S^. Hence 



t<\osn<t+y ^ t 



for all sufficiently large values of x. 



Lemma 5. Let S he a set of points of measure M, situated in 

 a finite interval ab. The?! it is possible to divide ah into sub-intervals 

 of length 8 {the two extreme intervals being possihkj less than 8) in 

 such a way that not more than M/8 of the points of division belong to S. 



^ Proof Consider any such division of a6, and denote by a^ an 

 arbitrary sub-mterval of length 8. Let ^ be a point in a/3 and 

 denote by <^ {x) the number of points of S which are " congruent " 

 to a; according to the adopted division of ab. Then it is clear that 



(x) dx = M. 



^Y\ <^^r?. T'^ ^^ ^* ^^^'* ^^^ P^i^'^ ^« in «/5, such that 

 <HXo)^M/b. btartmg the division of ah from this point, we see 

 that the conditions stated in the lemma are fulfilled. 



Lemma 6. Let us denote bg a^ the set of points t, belonqinq to 

 the interval {x-l,x-^ 1), such that 



7r(e*+«3/)-7r(e«-«2/)£^e«, 



where c is a positive constant Here ir (v) denotes as usual the 

 number of primes less than or equal to v. Then it is possible to qive 

 such a value to c that the measure /., of a, satisfies the relation 



( - — 

 fjix = \x^e 2 



• , ^T-f ^o^ shall prove this lemma by first excluding from the 

 interval (^-2, x + 2) a certain set, the measure of which satisfies 



