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Then μ [a, b) = μ [a1 , b1 ) × [a , b ) = μ1 [a1 , b1 ) = vol1 [a1 , b1 ) μ1 [0, 1) = vol1 [a1 , b1 ) μ [0, 1) × [a , b ) . A simple induction argument now gives n μ [a, b) = μ [0, 1)n vol1 [aj , bj ) = αn voln [a, b) . j=1 (ii) Suppose A ∈ B n [or A ∈ L(n)] and let (Ik ) be a sequence in J (n) that covers A. It follows from (i) that μ(A) ≤ k μ(Ik ) = αn k λn (Ik ) . 4 that μ(A) ≤ αn λ∗n (A) = αn λn (A) . (iii) Now suppose B ∈ B n [or B ∈ L(n)] is bounded. There exists I ∈ J (n) such that B ⊂ I ⊂ I.

Proof (i) We ﬁrst assume that f : N → Rm is (globally) Lipschitz continuous. Then there is an L > 0 such that |f (x) − f (y)|∞ ≤ L |x − y|∞ for x, y ∈ N . 4) Suppose 0 < ε < Lm . 4 ∞ a sequence (Ik ) in J (n) that covers N and satisﬁes k=0 λn (Ik ) < ε/Lm . We can take the edge lengths to be e rational. By subdivision, we can also assume without losing generality that every Ik is a cube, of side length ak , say. 4), then, f (N ∩ Ik ) is contained in a cube Jk ⊂ J(m) of side length Lak . The n-volume of these cubes is for k ∈ N .

Verify that Ff is a measure-generating function for which μ∗Ff [a, b) = −∞ < a < b < ∞. 4 b a f (ξ) dξ when Suppose A ⊂ Rn . Prove: (a) Hs∗ (A) = lim inf ε→0+ (b) If f : A → R m ∞ k=0 diam(Ak ) s ; Ak ⊂ Rn , diam(Ak ) ≤ ε, k ∈ N, A ⊂ k Ak . is Lipschitz continuous with Lipschitz constant λ, then Hs∗ f (A) ≤ λs Hs∗ (A) . (c) For every isometry ϕ : Rn → Rn , we have Hs∗ ϕ(A) = Hs∗ (A). 2 (d) Suppose n > n and Hs∗ is the Hausdorﬀ outer measure on Rn . Then Hs∗ (A) = Hs∗ (A). That is, the Hausdorﬀ outer measure is independent of the dimension of the ambient Rn .