Multivariable Advanced Calculus

430 HAUSDORFF MEASURES AND AREA FORMULABy the Vitali covering theorem, there exists a sequence of disjoint balls, {B i } suchthat B (0, 1) = (∪ ∞ i=1 B i) ∪ N where m n (N) = 0. Then Hδ n (N) = 0 can be concludedbecause Hδ n ≤ Hn and Lemma 15.1.7. Using m n (B (0, 1)) = H n (B (0, 1)) again,H n δ (B (0, 1)) = H n δ (∪ i B i ) ≤= β (n)α (n)∞∑i=1∞∑β (n) r (B i ) ni=1α (n) r (B i ) n = β (n)α (n)∞∑m n (B i )= β (n)α (n) m n (∪ i B i ) = β (n)α (n) m n (B (0, 1)) = β (n)α (n) Hn (B (0, 1))which implies α (n) ≤ β (n) and so the two are equal. This proves that if α (n) = β (n) ,then the H n = m n on the measurable sets of R n .This gives another way to think of Lebesgue measure which is a particularly niceway because it is coordinate free, depending only on the notion of distance.For s < n, note that H s is not a Radon measure because it will not generallybe finite on compact sets. For example, let n = 2 and consider H 1 (L) where L is aline segment joining (0, 0) to (1, 0). Then H 1 (L) is no smaller than H 1 (L) when L isconsidered a subset of R 1 , n = 1. Thus by what was just shown, H 1 (L) ≥ 1. HenceH 1 ([0, 1] × [0, 1]) = ∞. The situation is this: L is a one-dimensional object inside R 2and H 1 is giving a one-dimensional measure of this object. In fact, Hausdorff measurescan make such heuristic remarks as these precise. Define the Hausdorff dimension of aset, A, asi=1dim(A) = inf{s : H s (A) = 0}15.2.4 A Formula For α (n) ∗What is α(n)? Recall the gamma function which makes sense for all p > 0.Γ (p) ≡∫ ∞Lemma 15.2.6 The following identities hold.(∫ 1Γ(p)Γ(q) =0e −t t p−1 dt.pΓ(p) = Γ(p + 1),0)x p−1 (1 − x) q−1 dx Γ(p + q),( 1Γ =2)√ πProof: Using integration by parts,Γ (p + 1) =∫ ∞0= pΓ (p)e −t t p dt = −e −t t p | ∞ 0+ p∫ ∞0e −t t p−1 dt