19.3 Technical Considerations
Let α(n) be the volume of the unit ball in ℝn. Thus the volume of B(0,r) in ℝn is α(n)rn
from the change of variables formula. There is a very important and interesting
inequality known as the isodiametric inequality which says that if A is any set in ℝn,
This inequality may seem obvious at first but it is not really. The reason it is not is that there
are sets which are not subsets of any sphere having the same diameter as the set. For
example, consider an equilateral triangle.
Lemma 19.3.1 Let f : ℝn−1 → [0,∞) be Borel measurable and let
Then S is a Borel set in ℝn.
Proof: Set sk be an increasing sequence of Borel measurable functions converging
pointwise to f.
Then (x,y) ∈ Sk if and only if f(x) > 0 and |y| < sk(x) ≤ f(x). It follows that Sk ⊆ Sk+1
But each Sk is a Borel set and so S is also a Borel set. This proves the lemma.
Let Pi be the projection onto
where the ek are the standard basis vectors in ℝn, ek being the vector having a 1 in the kth
slot and a 0 elsewhere. Thus Pix ≡∑
j≠ixjej. Also let
Lemma 19.3.2 Let A ⊆ ℝn be a Borel set. Then Pix → m(APix) is a Borel
measurable function defined on Pi(ℝn).
Proof: Let K be the π system consisting of sets of the form ∏
j=1nAj where Ai is Borel.
Also let G denote those Borel sets of ℝn such that if A ∈G then
where Rk = (−k,k)n. Thus K⊆G. If A ∈G
is Borel measurable because it is of the form
and these are Borel measurable functions of Pix. Also, if
is a disjoint sequence of sets in
and each function of Pix is Borel measurable. Thus by the lemma on π systems, Lemma
7.4.2, G = ℬ
and this proves the lemma.
Now let A ⊆ ℝn be Borel. Let Pi be the projection onto
and as just described,
Thus for x = (x1,
Since A is Borel, it follows from Lemma 19.3.1 that
is a Borel measurable function on Piℝn = ℝn−1.