Gruber P. Convex and Discrete Geometry

62 Convex Bodies
For the proof of the theorem it is sufficient to show the following:
(1) Let x ∈ cl conv R(M). Then x ∈ R(M).
For x = o this is trivial. Thus we may suppose that x �= o. Then the proof of (1) is
split into several steps.
First, let
N ={A ∈ M : x ∈ cl conv R(A)}.
Order N by set inclusion (up to sets of |µ|-measure 0). Then
(2) N has a minimal element.
In order to prove (2), the following will be shown first:
(3) Let {Ai, i = 1, 2,...} be a countable decreasing chain in N . Then A0 =
�
i Ai is a lower bound of this chain in N .
Clearly, A0 is a lower bound of the chain. We have to show that A0 ∈ N ,
i.e. x ∈ cl conv R(A0). For this it is sufficient to prove that cl conv R(A0) ⊇
�
i cl conv R(Ai). For each measurable subset Ai ∩ B of Ai, where B ∈ M, the
definition of |µ| implies that
�µ(A0 ∩ B) − µ(Ai ∩ B)� =�µ � (Ai\A0) ∩ B � �≤|µ|(Ai\A0) = εi,
say. Thus R(A0) + εi Bd ⊇ R(Ai). Hence cl conv R(A0) + εi Bd ⊇ cl conv R(Ai).
Since εi → 0 (note that A0 = �
i Ai and |µ| is a finite measure), we see that
cl conv R(A0) ⊇ �
i cl conv R(Ai). Since x ∈ cl conv R(Ai) for each i, itfinally
follows that x ∈ cl conv R(A0), orA0 ∈ N . The proof of (3) is complete. Using (3),
we next show the following refinement of (3):
(4) Let {Aι,ι∈ I } be a chain in N . Then it has a lower bound in N .
If this chain has a smallest element, we are finished. Otherwise choose a decreasing
countable sub-chain {Aιi , i = 1, 2,...} such that inf{|µ|(Aιi ) : i = 1, 2,...}=
inf{|µ|(Aι) : ι ∈ I }. Then A0 = �
i Aιi ∈ N by (3) and |µ|(A0) = inf{|µ|(Aι) : ι ∈
I }. We have to show that A0 ⊆ Aι for each ι ∈ I (uptoasetof|µ|-measure 0). Let
ι ∈ I . If there is an i such that ιi