Gruber P. Convex and Discrete Geometry

4 Support and Separation 63
Here H(u) is the support hyperplane of cl conv R(A) with exterior normal vector
u. The absolute continuity of ν with respect to |µ| is obvious. Clearly, H(u) is the
hyperplane
u · y = sup � u · z : z ∈ cl conv R(A) � = sup � u · z : z ∈ R(A) �
= sup � ν(C) : C ∈ M, C ⊆ A �
= sup � ν(A + ∩ C) + ν(A 0 ∩ C) + ν(A − ∩ C) : C ∈ M, C ⊆ A � = ν(A + ).
For the proof of the inclusion H(u) ∩ cl conv R(A) ⊇ µ(A + ) + cl conv R(A 0 ),it
is sufficient to show the following: Let B ∈ M, B ⊆ A 0 . Then µ(A + ) + µ(B) ∈
H(u) ∩ cl conv R(A). Clearly, A + ∪ B ∈ M and A + ∪ B ⊆ A. Hence
µ(A + ) + µ(B) = µ(A + ∪ B) ∈ R(A) ⊆ cl conv R(A),
u · � µ(A + ) + µ(B) � = ν(A + ) + ν(B) = ν(A + ).
Thus µ(A + ) + µ(B) ∈ H(u) ∩ cl conv R(A), concluding the proof of the first
inclusion. Next, the reverse inclusion H(u)∩cl conv R(A) ⊆ µ(A + )+cl conv R(A 0 )
will be shown. Since µ is a finite vector-valued measure, R(A) is bounded. Thus
cl conv R(A) = conv cl R(A) by Proposition 3.2. For the proof of the reverse
inclusion, it is thus sufficient to show the following: let y ∈ H(u) ∩ cl R(A). Then
y ∈ µ(A + ) + cl conv R(A 0 ). Clearly, yi → y � ∈ H(u) � as i →∞for suitable
yi = µ(Ai) ∈ R(A). Thus u · yi = u · µ(Ai) = ν(Ai) → u · y and thus
ν(Ai) = ν(A + ∩ Ai) + ν(A 0 ∩ Ai) + ν(A − ∩ Ai)
= ν(A + ∩ Ai) + ν(A − ∩ Ai) → u · y = ν(A + ).
Hence ν(A + ∩ Ai) → ν(A + ) and ν(A − ∩ Ai) → 0orν(A + \Ai) → 0 and
ν(A − ∩ Ai) → 0. Since |µ| is absolutely continuous with respect to |ν| on A +
and also on A − , it thus follows that |µ|(A + \Ai) → 0 and |µ|(A − ∩ Ai) → 0. Hence
µ(A + \Ai), µ(A − ∩ Ai) → o and therefore
y = lim µ(Ai) = lim � µ(A + ∩ Ai) + µ(A 0 ∩ Ai) + µ(A − ∩ Ai) �
= lim � µ(A + ) − µ(A + \Ai) + µ(A 0 ∩ Ai) + µ(A − ∩ Ai) �
= µ(A + ) + lim µ(A 0 ∩ Ai) ∈ µ(A + ) + cl R(A 0 )
⊆ µ(A + ) + cl conv R(A 0 ).
This concludes the proof of the reverse inclusion. The proof of (5) is complete.
Third, we shall prove the following:
(6) Let A ∈ M be minimal such that x ∈ cl conv R(A). Then x ∈ R(A).
Note that x �= o and o, x ∈ cl conv R(A). We distinguish two cases. First case:
x ∈ relint cl conv R(A). By Lemma 3.1 the point x is in the relative interior of a convex
polytope with vertices µ(A1),...,µ(Ak) ∈ R(A), say.IfB ∈ M, B ⊆ A,
is such that |µ|(B) is sufficiently small (such B exist since the measure |µ| is
non-atomic), then �µ(A1\B) − µ(A1)�,...,�µ(Ak\B) − µ(Ak)� ≤|µ|(B) are
so small that x is still in the relative interior of the convex polytope with vertices
µ(A1\B),...,µ(Ak\B) ∈ R(A\B) ⊆ R(A). Hence x ∈ cl conv R(A\B),