Multivariable Advanced Calculus

420 HAUSDORFF MEASURES AND AREA FORMULAand diam(C i j ) ≤ δ. Then H s δ(E) ≤It follows that since ε > 0 is arbitrary,≤∞∑i=1 j=1∞∑β(s)(r(Cj)) i s∞∑Hδ(E s i ) + ε/2 ii=1≤ ε +H s δ(E) ≤∞∑Hδ(E s i ).i=1∞∑Hδ(E s i )i=1which shows Hδ s is an outer measure. Now notice that Hs δ(E) is increasing as δ → 0.Picking a sequence δ k decreasing to 0, the monotone convergence theorem impliesH s (E) ≤∞∑H s (E i ).i=1This proves the lemma. The outer measure H s is called s dimensional Hausdorff measure when restricted tothe σ algebra of H s measurable sets. Recall these are the sets E such that for all S,H s (S) = H s (S ∩ E) + H s (S \ E) .Next I will show the σ algebra of H s measurable sets includes the Borel sets. Thisis done by the following very interesting condition known as Caratheodory’s criterion.15.1.1 Properties Of Hausdorff MeasureDefinition 15.1.3 For two sets A, B in a metric space, definedist (A, B) ≡ inf {||x − y|| : x ∈ A, y ∈ B} .Theorem 15.1.4 Let µ be an outer measure on the subsets of X, a closed subsetof a normed vector space and supposeµ(A ∪ B) = µ(A) + µ(B)whenever dist(A, B) > 0, then the σ algebra of measurable sets contains the Borel sets.Proof: It suffices to show that closed sets are in F, the σ-algebra of measurablesets, because then the open sets are also in F and consequently F contains the Borelsets. Let K be closed and let S be a subset of Ω. Is µ(S) ≥ µ(S ∩ K) + µ(S \ K)? Itsuffices to assume µ(S) < ∞. LetK n ≡ {x : dist(x, K) ≤ 1 n }By Lemma 7.4.4 on Page 161, x → dist (x, K) is continuous and so K n is closed. Bythe assumption of the theorem,µ(S) ≥ µ((S ∩ K) ∪ (S \ K n )) = µ(S ∩ K) + µ(S \ K n ) (15.1)