Number Theory: An Introduction to Mathematics

3 Proof of the Lattice Point Theorem; Other Results 337

Hence, by Proposition 10,

sup

x∈Rn

φ(x)≥

∫

Rn

Ψ(x)dx.

For example, letS⊆Rnbe a measurable set withλ(S)>m. Then there exists
aboundedmeasurable setS′⊆Swithλ(S′)>m.IfwetakeΨto be the indicator
function ofS′,then
∫

Rn

Ψ(x)dx=λ(S′)>m

and we conclude that there existsy∈Rnsuch that
∑

z∈Zn

Ψ(y+z)=φ(y)>m.

Since the only possible values of the summands on the left are 0 and 1, it follows that
there existm+1 distinct pointsz 1 ,...,zm+ 1 ∈Zn=Λsuch thaty+zj∈Sfor allj.
The proof of Proposition 9 can now be completed in the same way as before.
Let{Kα}be a family of subsets ofRn, where eachKαis theclosureof a nonempty
open setGα,i.e.Kαis the intersection of all closed sets containingGα. The family
{Kα}is said to be apackingofRnifα=α′impliesGα∩Gα′=∅and is said to be
acoveringofRnifRn=

⋃

αKα.ItissaidtobeatilingofR

nif it is both a packing

and a covering.
For example, ifΠis a fundamental parallelotope of a latticeΛ, then the family
{Π+a:a ∈Λ}is a tiling ofRn. More generally, ifGis a nonempty open sub-
set ofRnwith closureK, we may ask whether the family{K+a:a∈Λ}of all
Λ-translates ofKis either a packing or a covering ofRn. Some necessary conditions
may be derived with the aid of Proposition 9:

Proposition 11Let K be the closure of a bounded nonempty open set G⊆Rnand
letΛbe a lattice inRn.
If theΛ-translates of K are a covering ofRnthenλ(K)≥d(Λ), and the inequality
is strict if they are not also a packing.
If theΛ-translates of K are a packing ofRnthenλ(K)≤d(Λ), and the inequality
is strict if they are not also a covering.

Proof Suppose first that theΛ-translates ofKcoverRn. Then every point of a funda-
mental parallelotopeΠofΛhas the formx−a,wherex∈Kanda∈Λ. Hence

λ(K)=

∑

a∈Λ

λ(K∩(Π+a))

=

∑

a∈Λ

λ((K−a)∩Π)≥λ(Π)=d(Λ).

Suppose, in addition, that theΛ-translates ofKare not a packing ofRn. Then there
exist distinct pointsx 1 ,x 2 in the interiorGofKsuch thata=x 1 −x 2 ∈Λ.Let

Bε={x∈Rn:|x|≤ε}.