Gruber P. Convex and Discrete Geometry

400 Geometry of Numbers
It is well known that convolutions are continuous. This implies that the expression
m � (U + y) ∩ V � is continuous as a function of y. Taking into account the fact that
Ed /L is compact, this expression is integrable. Then
�
m � (U + y) ∩ V � dm(y)
E d /L
=
=
=
�
�
Ed /L Ed /L
�
�
Ed /L Ed /L
�
E d /L
1U(x − y)1V(x) dm(x) dm(y)
1U(x − y)1V(x) dm(y) dm(x)
m(U)1V(x) dm(x) = m(U) m(V),
by Fubini’s theorem, since Ed /L is Abelian and thus unimodular. Since m � (U +
y) ∩ V � is continuous, as a function of y, on the compact connected space Ed /L of
measure d(L), there is a coset z ∈ Ed /L such that:
�
m � (U + y) ∩ V � dm(y) = m � (U + z) ∩ V � d(L). �
E d /L
Proof (of the sum theorem). (i) If Proposition (i) does not hold, choose x ∈ E d /L
such that x �∈ S + T. Then x − s �∈ T for s ∈ S, or
(x − S) ∩ T =∅.
This yields a contradiction and thus concludes the proof of (i):
m(S) + m(T) = m(x − S) + m(T) = m � (x − S) ∪ T � ≤ m(E d /L) = d(L).
(ii) First, we show the following.
(1) Let U, V ⊆ E d /L where U ∩ V �= ∅. Then U + V ⊇ U ∩ V + U ∪ V.
Let x ∈ U ∩ V and y ∈ U ∪ V. Ify belongs to U, then we may regard x as belonging
to V, since it belongs to both U and V. Hence x + y = y + x ∈ U + V. Similarly, if
y belongs to V, weregardx as belonging to U.
The crucial task in the proof of (ii) is to show it for compact sets S and T.Let
(2) m(S) = σ d(L), m(T) = τd(L).
By assumption, σ + τ ≤ 1. If σ = 0orτ = 0, then (ii) holds trivially. After
exchanging S and T and renaming, if necessary, we may thus assume that
(3) 0