Number Theory: An Introduction to Mathematics

2 Discrepancy 461

Proof Without loss of generality we may assumeξ 1 ≤···≤ξN. Writing

∫

I

f(t)dt=

∑N

n= 1

∫n/N

(n− 1 )/N

f(t)dt,

we obtain

∣

∣

∣

∣N

− 1

∑N

n= 1

f(ξn)−

∫

I

f(t)dt

∣

∣

∣

∣≤

∑N

n= 1

∫n/N

(n− 1 )/N

|f(ξn)−f(t)|dt

≤L

∑N

n= 1

∫n/N

(n− 1 )/N

|ξn−t|dt.

But for(n− 1 )/N≤t≤n/Nwe have

|ξn−t|≤max(|ξn−n/N|,|ξn−(n− 1 )/N|)≤D∗N,

by Proposition 11. The result follows.

As Koksma (1942) first showed, Proposition 14 can be sharpened in the following
way:

Proposition 15If the function f has bounded variation on the unit interval I , with
total variation V , then for any finite setξ 1 ,...,ξN∈I with discrepancy D∗N,

∣

∣

∣

∣N

− 1

∑N

n= 1

f(ξn)−

∫

I

f(t)dt

∣

∣

∣

∣≤VD

∗

N.

Proof Without loss of generality we may assumeξ 1 ≤···≤ξNand we putξ 0 =0,
ξN+ 1 =1. By integration and summation by parts we obtain

∑N

n= 0

∫ξn+ 1

ξn

(t−n/N)df(t)=

∫

I

tdf(t)−N−^1

∑N

n= 0

n(f(ξn+ 1 )−f(ξn))

=[tf(t)]^10 −

∫

I

f(t)dt−f( 1 )+N−^1

N∑− 1

n= 0

f(ξn+ 1 )

=N−^1

∑N

n= 1

f(ξn)−

∫

I

f(t)dt.

The result follows, since forξn≤t≤ξn+ 1 we have

|t−n/N|≤max(|ξn−n/N|,|ξn+ 1 −n/N|)≤D∗N.