Gruber P. Convex and Discrete Geometry

162 Convex Bodies
Since the measure of a Borel set is the supremum of the measures of its compact
subsets, it is sufficient to prove (3) for compact sets R, S, T . Then, by shifting S, T
suitably, we may suppose that 0 is the right endpoint of S and the left endpoint of T .
The set (1 − λ)S + λT then includes the sets (1 − λ)S and λT whichhaveonly0in
common. Hence |R| ≥|(1 − λ)S + λT |≥(1 − λ)|S|+λ|T |, concluding the proof
of (3).
Now, let d = 1. Then f, g, h are non-negative Borel functions on R satisfying
(1). Since the integral of a non-negative measurable function on R is the supremum
of the integrals of its bounded, non-negative measurable minorants, we may assume
that f, g, h are bounded. Excluding the case where f or g is a zero function, and
taking into account that the assumption in (1) and the inequality (2) have the same
homogeneity, we may assume that sup f = sup g = 1. By Fubini’s theorem we then
can write the integrals of f and g in the form
�
R
f (x) dx =
�1
0
|{x : f (x) ≥ t}| dt,
�
R
g(x) dx =
If f (x) ≥ t and g(y) ≥ t, then h((1 − λ)x + λy) ≥ t by (1). Thus,
�1
0
|{y : g(y) ≥ t}| dt.
{z : h(z) ≥ t} ⊇(1 − λ){x : f (x) ≥ t}+λ{y : g(y) ≥ t}.
For 0 ≤ t < 1 the sets on the right-hand side are non-empty Borel sets in R. Thus
|{z : h(z) ≥ t}| ≥ (1 − λ)|{x : f (x) ≥ t}| + λ|{y : g(y) ≥ t}|
by (3). Integration from 0 to 1 and the inequality between the arithmetic and the
geometric mean, see Corollary 1.2, then yield (2) in case d = 1:
�
R
h(z) dz ≥
�1
0
|{z : h(z) ≥ t}| dt
�1
�1
≥ (1 − λ) |{x : f (x) ≥ t}| dt + λ |{y : g(y) ≥ t}| dt
0
�
= (1 − λ)
R
�
f (x) dx + λ
R
g(y) dy
R
R
�
≥
� �1−λ� f (x) dx
� �λ g(y) dy .
Assume finally, that d > 1 and that the theorem holds for d − 1. Let f, g, h be
non-negative Borel functions on E d satisfying (1). Embed E d−1 into E d as usual, i.e.
E d = E d−1 × R. Letu = (0,...,0, 1) ∈ E d .Lets, t ∈ R and r = (1 − λ)s + λt.
By (1), the non-negative measurable functions f (x + su), g(y + tu), h(z + ru) on
E d−1 satisfy the following condition:
0