Gruber P. Convex and Discrete Geometry

402 Geometry of Numbers
(11) Sn + Tn is compact and thus measurable and
m(Sn + Tn) ≥ m(Tn) = τnd(L) for n = 1, 2,...
It follows from (3) and (4) that
0 ≤ σn ≤ 1, σn ≤ σn−1(1 − σn−1) for n = 2, 3,...
and therefore, σn → 0asn →∞. Hence (4) yields τn → σ + τ as n →∞.
Combine this with (11), (4) and (2) to get
m(S + T) ≥ (σ + τ)d(L) = m(S) + m(T),
as required. The proof of (ii) for compact sets S, T is complete.
If S, T and S + T are measurable, then there are non-decreasing sequences
S1 ⊆ S2 ⊆ ··· ⊆ S, T1 ⊆ T2 ⊆ ··· ⊆ T of non-empty compact sets the
measures of which tend to m(S) and m(T), respectively. Applying (ii) to the nonempty
compact sets Sn, Tn and letting n →∞, yields (ii) for S, T. ⊓⊔
Proof of Macbeath and Kneser
As an essential tool for the second proof, we state, without proof, the following result
on finite Abelian groups. For a proof see Kneser [601].
Lemma 26.2. Let A, B be non-empty subsets of a finite Abelian group G. Then the
following claims hold:
(i) If #A + #B > #G then A + B = G.
(ii) If #A + #B ≤ #G then #(A + B) ≥ #A + #B − #H,
where H is a proper sub-group of G.
Proof (of the sum theorem for S, T, S + T Jordan measurable).
(i) The simple proof of statement (i) is as earlier.
(ii) For the proof of Statement (ii) of the sum theorem, it is sufficient to show the
following proposition:
(12) Let S, T ⊆[0, 1) d be Jordan measurable such that V (S) + V (T )