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12 HILLEL FURSTENBERG The next notion of dimension we will discuss is more complicated but also more useful. It is called the Hausdorff dimension. We start with defining sets of Hausdorff α-measure 0. Definition. Fix α 0. Let X be a metric space and let A be any subset of X. We say that A is of Hausdorff α-measure 0 (Hα(A) = 0) if for every ε 0, there exists a countable family of balls Br i (xi) (xi ∈ X, ri 0) such that A ⊂ i Br i (xi) and ∑ i ri α ε. Note that if β α 0 and Hα(A) = 0, then Hβ(A) = 0 as well. Indeed, take ε ∈ (0, 1) and consider a covering of A by balls Br i (xi) with ri α ε. Since ε 1, it follows that ri 1 for all i, so rβ i ≤ rα, i whence ∑ i rβ i ≤ ∑i i rα i ε as well. Thus, for every A ⊂ X, the set {α 0 : Hα(A) = 0} is either empty, or a ray from some α0 ≥ 0 to +∞. In the second case, this α0 = inf{α 0 : Hα(A) = 0} is called the Hausdorff dimension of A and denoted by dim A. If the aforementioned set of α is empty, it is natural to say that dim A = ∞. Obviously, if A ⊂ B, then dim A ≤ dim B (every covering of B is a covering of A as well). Now suppose that the lower Minkowski dimension of A is less than β. It means that log Nr(A) log 1/r β along some sequence of values of r 0 tending to 0. For every r in this sequence, we have Nr(A) r−β, so for every β β, the set A can be covered by balls with the sum of β powers of radii not exceeding Nr(A)rβ ≤ rβ −β → 0 as r → 0. Thus, for every β M-dim A and every β β, we have Hβ (A) = 0, so dim A ≤ β. Since β is arbitrary here, we conclude that dim A ≤ M-dim A for all A. Since the Minkowski dimension of the unit cube Q ⊂ Rn equals n, we have dim A ≤ M-dim A ≤ M-dim Q = n for every A ⊂ Q. The following Lemma allows one to estimate the Hausdorff dimension of A from below. Lemma 1. Assume that there exists a measure μ on X such that μ(A) 0, and μ(Br(x)) ≤ Krα for all x ∈ X, r 0. Then dim A ≥ α. Proof. For every covering A ⊂ i Br i (xi), we have μ(A) ≤ i μ(Br i (xi)) ≤ K i ri α , so it is impossible to make ∑ i ri α less than K−1μ(A). Thus, we cannot have Hα(A) = 0, which means that dim A ≥ α. Now let us compute the Hausdorff dimension of the middle-third Cantor set C. On one hand, C is a subset of Cn and Cn consists of 2n intervals of length 3−n. Thus, M-dim C ≤ lim n→∞ log(2n) log(3n) = log 2 log 3 .