Gruber P. Convex and Discrete Geometry

The Reverse Normal Image
10 Problems of Minkowski and Weyl and Some Dynamics 189
Let C ∈ Cp be given. Given a Borel set B ⊆ Sd−1 ,thereverse normal or reverse
(B) of B in bd C is defined by
spherical image n −1
C
n −1
C (B) = � x ∈ bd C : there is an exterior normal
unit vector of bd C at x contained in B � .
Let µd−1 be the (d − 1)-dimensional Hausdorff measure in E d . A subset of bd C is
measurable if it is measurable with respect to µd−1| bd C.
Proposition 10.1. The reverse normal image of a Borel set in S d−1 is measurable in
bd C.
Proof. Let M be the family of all sets B in Sd−1 for which n −1
C (B) is measurable.
We have to show that M contains all Borel sets in Sd−1 . This will be done in three
steps.
First, the following will be shown:
(1) Let B ⊆ Sd−1 be compact. Then n −1
C (B) is compact in bd C and thus
B ∈ M.
To see this, let x1, x2, ··· ∈ n −1
C (B) such that x1, x2, ··· → x ∈ bd C, say.We
have to show that x ∈ n −1
C (B). Choose u1, u2, ··· ∈ B ⊆ S d−1 for which xn ∈
n −1
C (un) for n = 1, 2,... By considering a suitable subsequence and renumbering,
if necessary, we may suppose that u1, u2, ···→u ∈ S d−1 , say. Since B is compact,
u ∈ B. The body C is contained in each of the support halfspaces {z : un · z ≤
un · xn}. Since un → u, xn → x, it follows that C is contained in the halfspace
{z : u · z ≤ u · x} and x is a common boundary point of this halfspace and of C.
Hence this halfspace supports C and thus x ∈ n −1
C (u) ⊆ n−1
C (B), concluding the
proof of (1).
Second, we claim the following:
(2) Let B ∈ M. Then B c = S d−1 \B ∈ M.
At each point of n −1
C
(B) ∩ n−1
C (Bc ) there are two different exterior unit normal vec-
(B) ∩ n−1
C (Bc )
� −1
is singular. The theorem of Anderson and Klee 5.1 then shows that µd−1 nC (B) ∩
n −1
C (Bc ) � = 0. The set n −1
C (Bc ) thus differs from the set Sd−1 \n −1
C (B) which is
measurable by the assumption in (2), by a set of measure 0 and therefore is measurable
itself. The proof of (2) is complete.
Third, the following statement holds:
tors of bd C, one in B, the other one in B c . Hence each point of n −1
C
(3) Let B1, B2, ···∈M. Then B = B1 ∪ B2 ∪···∈M.
This follows from the identity n −1
C (B) = n−1
C (B1) ∪ n −1
C (B2) ∪···
Having proved (1) – (3), it follows that M is a σ -algebra of subsets of Sd−1 containing all compact sets in Sd−1 . It thus contains the smallest such σ -algebra,
that is the family of all Borel sets in Sd−1 . ⊓⊔