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Gruber P. Convex and Discrete Geometry

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26 The Torus Group E d /L 399<br />

theory, see Mann [687, 688] <strong>and</strong> Danilov [235], <strong>and</strong> in other areas of mathematics.<br />

Minkowski’s theorem on mixed volumes 6.5 <strong>and</strong> the Brunn–Minkowski theorem 8.1<br />

are also α + β type results.<br />

The sum theorem of Macbeath <strong>and</strong> Kneser, which will be proved later, gives a<br />

satisfying answer to this question for the torus group. Let L be a lattice in E d .<br />

The Sum Theorem of Kneser–Macbeath<br />

Macbeath [673] <strong>and</strong> Kneser [601, 602] proved the following result.<br />

Theorem 26.1. Let S, T be non-empty sets in E d /L such that S, T <strong>and</strong> S + T are<br />

measurable. Then the following claims hold:<br />

(i) If m(S) + m(T) >d(L), then S + T = E d /L.<br />

(ii) If m(S) + m(T) ≤ d(L), then m(S + T) ≥ m(S) + m(T).<br />

Note that the Lebesgue measurability of S <strong>and</strong> T does not necessarily imply<br />

the Lebesgue measurability of S + T. IfS <strong>and</strong> T are Borel, their sum is Lebesgue<br />

measurable, but need not be Borel. For references, see the Remark in Sect. 8.2.<br />

If S + T is not measurable, a result similar to the sum theorem of Kneser <strong>and</strong><br />

Macbeath holds, where m(S + T) is replaced by the corresponding inner measure of<br />

S + T.<br />

We give two proofs. The first one is due to Kneser [602], see also Cassels [195].<br />

The second proof is due to Macbeath [673] <strong>and</strong> Kneser [601]. In the second proof,<br />

we consider only the case where S, T <strong>and</strong> S + T all are Jordan measurable sets.<br />

Proof of Kneser<br />

As a preparation for the first proof, we show the following result; versions of it<br />

are useful tools in integral geometry. See, e.g. Hadwiger [468], Santaló [881] or<br />

Schneider <strong>and</strong> Weil [911].<br />

Lemma 26.1. Let U, V ⊆ E d /L be measurable. Then there is z ∈ E d /L such that:<br />

m � (U + z) ∩ V � d(L) = m(U) m(V).<br />

Proof. Let 1U <strong>and</strong> 1V be the characteristic functions of U <strong>and</strong> V, respectively. Then<br />

m � (U + y) ∩ V � =<br />

=<br />

�<br />

E d /L<br />

�<br />

E d /L<br />

1U+y(x)1V(x) dm(x)<br />

1U(x − y)1V(x) dm(x).

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