Introduction to Hausdorff Measure and Dimension

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1 Introduction to Hausdorff Measure and Dimension Dynamics Learning Seminar, Liverpool) Poj Lertchoosakul 28 September Definition of Hausdorff Measure and Dimension Let X, d) be a metric space, let E X, and let s be a non-negative real number. For any δ > 0, define { Hδ s E) = inf diamu i )) s : } U i E, diamu i ) δ, where diamu i ) = sup{dx, y): x, y U i } is the diameter of U i, and where the infimum is taken over all countable δ-covers {U i } of E by sets U i with diamu i ) δ. Notice that as δ decreases, H s δ can only increase as there are fewer covers {U i} available, that is, if 0 < δ < δ, then H s δ E) H s δ E). We now define the s-dimensional Hausdorff outer measure of E to be H s E) = lim Hδ s E) = sup Hδ s E) [0, ]. δ 0 δ>0 This limit exists for any subset E of X, although the limiting value can be 0 or. Also, we can show that H s E) is either 0 or, except for possibly one value of s. In fact, it is clear that, for any E X and δ < 1, H s δ E) is non-increasing with s, and hence Hs E) is also non-increasing with s. Moreover, it is not hard to see that if H s 0 E) <, then H s E) = 0 for all s > s 0. Indeed, if s > s 0 and {U i } is a δ-cover of E, we have diamu i )) s δ s s 0 diamu i )) s 0, so, by taking infima, H s δ E) δs s 0 H s 0 δ E), and hence, by letting δ 0, Hs E) = 0. The Hausdorff dimension of E is defined by dim H E = inf{s 0: H s E) = 0}. That is, the Hausdorff dimension of E is the critical exponent such that H s E) = {, if s < dimh E, 0, if s > dim H E. 1

2 2 Some Properties of Hausdorff Measure Theorem 2.1 Outer Measure). The following are true for any metric space. i) Null Empty Set) H s ) = 0. ii) Monotonicity) If E F, then H s E) H s F ). iii) Countable Subadditivity) H s E j) Hs E j ). Proof. i) and ii) follow directly from the definition. iii) Assume that H s E j ) is finite for all j N; otherwise, it is trivial. Then it follows that Hδ se j) is also finite for all j N and any δ > 0. It suffices to show that Hδ s E j) Hs δ E j) for all δ > 0. Let ɛ > 0. Then, for any j, there exists a countable δ-cover {U i,j } of E j such that diamu i,j )) s < Hδ s E j) + ɛ 2 j. Since {U i,j }, is a countable δ-cover of E j, we have H s δ E j diamu i,j )) s < By taking δ 0, the result follows. H δ s E j) + ɛ ) 2 j = Hδ s E j) + ɛ. Let E be a subset of the metric space X. The set E is said to be H s -measurable if, for all A X, we have H s A) = H s A E) + H s A E c ). By Carathéodory s theorem, H s -measurable sets form a σ-algebra. Recall that a collection B of subsets of X is called a σ-algebra if it satisfies: 1. B is nonempty. 2. B is closed under complementation.. B is closed under countable unions. The Borel algebra on X is the smallest σ-algebra containing all open sets, and elements in the Borel algebra are called Borel sets. The restriction of H s to the σ-algebra of H s -measurable sets which includes open and closed sets, limsup and liminf sets, and G δ and F δ sets) is called the s-dimensional Hausdorff measure. Theorem 2.2 Countable Additivity). Let {E j } be a countable collection of disjoint H s -measurable subsets of X. Then we have H s E j = H s E j ). 2

3 Proof. If E 1 and E 2 are disjoint H s -measurable sets, then it follows by definition that H s E 1 E 2 ) = H s E 1 E 2 ) E 1 ) + H s E 1 E 2 ) E c 1) = H s E 1 ) + H s E 2 ). By using induction based on this idea, we see that H s k E j) = k Hs E j ) for all k N. Since, for any k N, we have k E j E j, we obtain the inequality Letting k, we have k ) H s E j H s E j = 1) H s E j k H s E j ). H s E j ). Moreover, the converse of 1) follows directly from the countable subadditivity property. Hence, the equality holds, as required. Theorem 2. Hausdorff and Lebesgue Measure). Let E be a Borel subset of R n. Then H n E) = 1 ω n E, where ω n is the volume of an n-dimensional ball of diameter 1, and where E denotes the Lebesgue measure of E. Proof [4]. It is easy to see that any Lebesgue null set of R n has n-dimensional Hausdorff measure zero, since it may be covered by balls of arbitrary small total content. Then n-dimensional Hausdorff measure is absolutely continuous with respect to Lebesgue measure, so dhn d = c for some locally integrable function c. As Hausdorff measure and Lebesgue measure are translation-invariant, c must also be translation-invariant and thus constant. Therefore, H n E) = c E for some c 0. To calculate the constant c, we compute the Hausdorff measure and Lebesgue measure of an n-dimensional ball of diameter 1, and this gives c = 1/ω n, as required. Theorem 2.4 Scaling). If E R n and λ > 0, then H s λe) = λ s H s E). Proof [2]. Note that λe = {λx: x E}. If {U i } is a δ-cover of E, then {λu i} is a λδ-cover of λe. It follows that 2) Hλδ s λe) diamλu i )) s = λ s diamu i )) s, for any δ-cover {U i } of E. By taking the infimum of the δ-covers {U i} of E in 2), we obtain Hλδ s λe) λs Hδ se). Letting δ 0 gives that Hs λe) λ s H s E). Replacing λ by 1/λ and E by λe gives the opposite inequality, as required.

4 Some Properties of Hausdorff Dimension Theorem.1. The Hausdorff dimension for sets in R n satisfies the following properties. i) If E F, then dim H E dim H F. ii) dim H E n. iii) If E > 0, then dim H E = n. iv) If E is countable, then dim H E = 0. v) If dim H E < n, then E = 0. vi) If E is open in R n, then dim H E = n. vii) dim H E j) = sup{dim H E j : j N}. Note that E is the Lebesgue measure of E. Proof. i) If E F, then H s E) H s F ) for s 0, which implies dim H E dim H F. ii), iii), iv), and v) can be deduced from the relationship between Hausdorff and Lebesgue measure in Theorem 2.. vi) If E is open, then it contains a ball of positive n-dimensional volume. vii) By the monotonicity property, dim H E j) dim H E j for each j. On the other hand, if s > dim H E j for all j, then H s E j ) = 0, and therefore H s E j) = 0, giving the opposite inequality. Theorem.2 Hausdorff Dimension Theorem). For any real r 0 and integer n r, there is a continuum fractals with Hausdorff dimension r in R n. Proof. The proof can be seen from []. 4 Some Calculation Examples of Hausdorff Dimension Example 4.1. Let F be the Middle Third Cantor set. Then dim H F = log 2/ log. Heuristic Solution [2]. The Cantor set F splits into a left part F L = F [0, 1/] and a right part F R = F [2/, 1]. Clearly, both parts are geometrically similar to F but scaled by a ratio 1/, and F = F L F R with this union disjoint. Thus, by the scaling property, we have H s F ) = H s F L ) + H s F R ) = ) 1 s H s F ) + ) 1 s H s F ) = 2 1 ) s H s F ) for any s. It follows that if, at the critical value s = dim H F, we have 0 < H s F ) <, then we must obtain s = log 2/ log. 4

5 Example 4.2. Let F be the set consisting of the numbers between 0 and 1 whose decimal expansions do not contain the digit 5. Then dim H F = log 9/ log 10. Heuristic Solution. The set F splits into 9 parts F k = F [k/10, k + 1)/10] for k = 0, 1, 2,, 4, 6, 7, 8, and 9, with these parts disjoint except possibly for endpoints which have s-dimensional Hausdorff measure 0 if s > 0. Then, by the scaling property, for s > 0, H s F k ) = 10 s H s F ) for all k. Now it follows that, for s > 0, we have H s F ) = k ) 1 s H s F k ) = 9 H s F ). 10 If we assume that, when s = dim H F, we have 0 < H s F ) <, then we must obtain s = dim H F = log 9/ log 10, as required. References [1] M. M. Dodson and S. Kristensen, Hausdorff dimension and Diophantine approximation, Proc. Sympos. Pure Math., 72, Part 1, Amer. Math. Soc., 2004 [2] K. Falconer, Fractal Geometry: Mathematical Foundations and Applications, 2 nd Edition, John Wiley & Sons Ltd, Chichester, 200. [] M. Soltanifar, On A Sequence of Cantor Fractals, Rose Hulman Undergraduate Mathematics Journal, Vol. 7, No. 1, [4] T. Tao, Notes 5: Hausdorff dimension optional), Lecture Note on Real Analysis, available at

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