28.3. TECHNICAL CONSIDERATIONS 1001

Now suppose x1 and x2 ∈ S(A,ei)

x1 = Pix1 + y1ei, x2 = Pix2 + y2ei.

For x ∈ A definel(x) =sup{y : Pix+yei ∈ A}.

g(x) = inf{y : Pix+yei ∈ A}.

Then it is clear thatl(x1)−g(x1)≥ m(APix1)≥ 2|y1|, (28.3.8)

l(x2)−g(x2)≥ m(APix2)≥ 2|y2|. (28.3.9)

Claim: |y1− y2| ≤ |l(x1)−g(x2)| or |y1− y2| ≤ |l(x2)−g(x1)|.Proof of Claim: If not,

2|y1− y2|> |l(x1)−g(x2)|+ |l(x2)−g(x1)|

≥ |l(x1)−g(x1)+ l(x2)−g(x2)|

= l(x1)−g(x1)+ l(x2)−g(x2).

≥ 2 |y1|+2 |y2|

by 28.3.8 and 28.3.9 contradicting the triangle inequality.Now suppose |y1− y2| ≤ |l(x1)−g(x2)|. From the claim,

|x1−x2| = (|Pix1−Pix2|2 + |y1− y2|2)1/2

≤ (|Pix1−Pix2|2 + |l(x1)−g(x2)|2)1/2

≤ (|Pix1−Pix2|2 +(|z1− z2|+2ε)2)1/2

≤ diam(A)+O(√

ε)

where z1 and z2 are such that Pix1 + z1ei ∈ A, Pix2 + z2ei ∈ A, and

|z1− l(x1)|< ε and |z2−g(x2)|< ε.

If |y1− y2| ≤ |l(x2)−g(x1)|, then we use the same argument but let

|z1−g(x1)|< ε and |z2− l(x2)|< ε,

Since x1,x2 are arbitrary elements of S(A,ei) and ε is arbitrary, this proves 28.3.7.The next lemma says that if A is already symmetric with respect to the jth direction,

then this symmetry is not destroyed by taking S (A,ei).

Lemma 28.3.4 Suppose A is a Borel set in Rn such that Pjx+ e jx j ∈ A if and only ifPjx+(−x j)e j ∈ A. Then if i ̸= j, Pjx+ e jx j ∈ S(A,ei) if and only if Pjx+(−x j)e j ∈ S(A,ei).