1002 CHAPTER 28. HAUSDORFF MEASURE

Proof: By definition,Pjx+ e jx j ∈ S(A,ei)

if and only if|xi|< 2−1m(APi(Pjx+e jx j)).

Nowxi ∈ APi(Pjx+e jx j)

if and only ifxi ∈ APi(Pjx+(−x j)e j)

by the assumption on A which says that A is symmetric in the e j direction. Hence

Pjx+ e jx j ∈ S(A,ei)

if and only if|xi|< 2−1m(APi(Pjx+(−x j)e j))

if and only ifPjx+(−x j)e j ∈ S(A,ei).

This proves the lemma.

28.3.2 The Isodiametric InequalityThe next theorem is called the isodiametric inequality. It is the key result used to compareLebesgue and Hausdorff measures.

Theorem 28.3.5 Let A be any Lebesgue measurable set in Rn. Then

mn(A)≤ α(n)(r (A))n.

Proof: Suppose first that A is Borel. Let A1 = S(A,e1) and let Ak = S(Ak−1,ek). Thenby the preceding lemmas, An is a Borel set, diam(An)≤ diam(A), mn(An) = mn(A), and Anis symmetric. Thus x ∈ An if and only if −x ∈ An. It follows that

An ⊆ B(0,r (An)).

(If x ∈ An \B(0,r (An)), then −x ∈ An \B(0,r (An)) and so diam(An)≥ 2|x|>diam(An).)Therefore,

mn(An)≤ α(n)(r (An))n ≤ α(n)(r (A))n.

It remains to establish this inequality for arbitrary measurable sets. Letting A be such a set,let {Kn} be an increasing sequence of compact subsets of A such that

m(A) = limk→∞

m(Kk).

Then

m(A) = limk→∞

m(Kk)≤ lim supk→∞

α(n)(r (Kk))n

≤ α(n)(r (A))n.

This proves the theorem.