From Hausdorff Measure to a Critical Exponent
Hausdorff measure assigns a cost to covers whose pieces are small in diameter. The exponent in that cost controls how strongly large pieces are penalized: increasing the exponent makes a cover of sufficiently small pieces cheaper, while decreasing it makes the same pieces more expensive. For a fixed set, this change produces a threshold. Below the threshold the Hausdorff measure is infinite; above it the measure is zero. At the threshold itself, the measure may be zero, finite and positive, or infinite.
Let \((X,d)\) be a metric space, and let \(E\subseteq X\). We use the diameter \(\operatorname{diam}(U)=\sup\{d(x,y):x,y\in U\}\), with \(\operatorname{diam}(\varnothing)=0\). The Hausdorff construction here is an outer measure on all subsets of \(X\); no measurability assumption on \(E\), finiteness assumption, or sigma-finiteness assumption is needed. We use the unnormalized convention for Hausdorff measure.
The diameter restriction is part of the definition at every scale. A cover that has low total cost but includes pieces larger than \(\delta\) is not an admissible \(\delta\)-cover. This restriction is what makes comparison of different exponents useful: on an admissible cover, every diameter is at most the same \(\delta\).
How the Exponent Changes the Measure
Proof. Fix a \(\delta\)-cover \((U_i)\) of \(E\), and write \(r_i=\operatorname{diam}(U_i)\). If \(r_i>0\), then \(r_i\le\delta\) gives \(r_i^t=r_i^s r_i^{t-s}\le\delta^{t-s}r_i^s\). If \(r_i=0\), then for \(s>0\) both costs are zero; for \(s=0\) the \(s\)-cost of a nonempty \(U_i\) is one and its \(t\)-cost is zero, so the same inequality holds. Empty sets contribute zero. Summing over \(i\) shows that the \(t\)-cost of this cover is at most \(\delta^{t-s}\) times its \(s\)-cost. Taking the infimum over all \(\delta\)-covers proves the displayed inequality.
If \(\mathcal{H}^s(E)=0\), then \(\mathcal{H}^s_\delta(E)\to0\) as \(\delta\downarrow0\). The inequality implies \(\mathcal{H}^t_\delta(E)\le\delta^{t-s}\mathcal{H}^s_\delta(E)\), so \(\mathcal{H}^t_\delta(E)\to0\) and \(\mathcal{H}^t(E)=0\). For the other assertion, suppose \(\mathcal{H}^t(E)<\infty\), where \(t<s\). Applying the proved inequality with \(t\) as the lower exponent gives \(\mathcal{H}^s_\delta(E)\le\delta^{s-t}\mathcal{H}^t_\delta(E)\le\delta^{s-t}\mathcal{H}^t(E)\). The last expression tends to zero, hence \(\mathcal{H}^s(E)=0\). \(\square\)
Thus, positive measure at an exponent forces infinite measure at every lower exponent: if \(\mathcal{H}^s(E)>0\) and \(t<s\), then \(\mathcal{H}^t(E)\) cannot be finite, by the theorem, so it is infinite. The comparison concerns different exponents on the same set; it does not say that the measure at the critical exponent must take a particular value.
Proof. The set of exponents \(r\ge0\) for which \(\mathcal{H}^r(E)=0\) is upward closed by the theorem. If \(s>D\), the definition of the infimum gives some such exponent \(r<s\); upward closure then gives \(\mathcal{H}^s(E)=0\). If \(s<D\) and \(\mathcal{H}^s(E)\) were finite, the theorem would imply \(\mathcal{H}^r(E)=0\) for every \(r>s\). Taking \(r\) arbitrarily close to \(s\) from above would force \(D\le s\), a contradiction. Therefore \(\mathcal{H}^s(E)\) is not finite and must equal \(+\infty\). \(\square\)
Basic Dimensions from Covers
Worked Example: A Single Point
Take \(X=\mathbb{R}\) with its usual metric and \(E=\{x_0\}\), for any \(x_0\in\mathbb{R}\). For every \(\delta>0\), the one-set cover \(U_1=\{x_0\}\) is admissible and has diameter zero. For every \(s>0\), its cost is \(0^s=0\), so \(\mathcal{H}^s_\delta(E)=0\) and \(\mathcal{H}^s(E)=0\). At exponent zero, any cover of this nonempty set has at least one nonempty member, so its cost is at least one; the singleton cover has cost one. Thus \(\mathcal{H}^0(E)=1\). The zero-measure exponents are exactly the positive exponents, and their infimum is \(0\). Hence \(\dim_H\{x_0\}=0\).
Check: the result does not depend on the location of \(x_0\), since the diameter of a singleton is always zero.
Worked Example: A Finite Set with Three Points
Let \(F=\{-1,2,4\}\subset\mathbb{R}\). For any \(s>0\) and any \(\delta>0\), cover \(F\) by its three singleton sets. Each has diameter zero and therefore cost zero, so \(\mathcal{H}^s_\delta(F)=0\) and \(\mathcal{H}^s(F)=0\). At \(s=0\), choose \(\delta=1\). Distances between distinct points are \(3\), \(2\), and \(5\), all greater than \(1\). No admissible covering set can contain two of the points, so at least three nonempty sets are required. The three singleton sets attain this cost, giving \(\mathcal{H}^0_1(F)=3\); the same argument applies for every \(\delta<2\), and thus \(\mathcal{H}^0(F)=3\). Therefore \(\dim_H F=0\).
Check: the three pairwise distances are \(|2-(-1)|=3\), \(|4-2|=2\), and \(|4-(-1)|=5\), confirming that no pair can share a set of diameter at most \(1\).
Worked Example: A Nondegenerate Closed Interval
Let \(E=[a,b]\subset\mathbb{R}\), where \(a<b\), and put \(L=b-a>0\). We show directly that \(\dim_H E=1\). We use the elementary length property of Lebesgue outer measure on \(\mathbb{R}\): \(m^*([a,b])=L\), and \(m^*\) is countably subadditive. For any bounded set \(U_i\subset\mathbb{R}\), its interval hull has length \(\operatorname{diam}(U_i)\), so \(m^*(U_i)\le\operatorname{diam}(U_i)\).
Fix \(0<s<1\), \(\delta>0\), and any admissible \(\delta\)-cover \((U_i)\) of \([a,b]\). Countable subadditivity gives \(L=m^*([a,b])\le\sum_i m^*(U_i)\le\sum_i\operatorname{diam}(U_i)\). Writing \(r_i=\operatorname{diam}(U_i)\), for \(r_i>0\) and \(r_i\le\delta\) we have \(r_i^s=r_i r_i^{s-1}\ge r_i\delta^{s-1}\), since \(s-1<0\). The inequality also holds when \(r_i=0\). Therefore every admissible cover has \(s\)-cost at least \(L\delta^{s-1}\), and \(\mathcal{H}^s_\delta([a,b])\ge L\delta^{s-1}\). As \(\delta\downarrow0\), this lower bound tends to infinity. Hence \(\mathcal{H}^s([a,b])=\infty\) for every \(0<s<1\).
For \(s=0\), each nonempty covering set has cost one and, when its diameter \(r_i\le\delta\), its cost is at least \(r_i/\delta\). The same length bound shows that every \(\delta\)-cover has cost at least \(L/\delta\). Thus \(\mathcal{H}^0_\delta([a,b])\ge L/\delta\), which tends to infinity as \(\delta\downarrow0\).
Now fix \(s>1\) and \(\delta>0\). For each positive integer \(N\) large enough that \(L/N\le\delta\), cover \([a,b]\) by the \(N\) closed subintervals of length \(L/N\). Their total \(s\)-cost is \(N(L/N)^s=L^sN^{1-s}\). As \(N\to\infty\), this tends to zero because \(1-s<0\). Since \(\mathcal{H}^s_\delta\) is an infimum over admissible covers, it equals zero; consequently \(\mathcal{H}^s([a,b])=0\). The exponents below \(1\) give infinite measure and those above \(1\) give zero measure. The threshold behavior proves \(\dim_H[a,b]=1\).
Check: when \(L=3\) and \(s=\tfrac12\), the lower bound at \(\delta=\tfrac1{16}\) is \(3(1/16)^{-1/2}=3\cdot4=12\). When \(s=2\), the partition into \(N\) intervals has total cost \(N(3/N)^2=9/N\), which tends to zero as \(N\to\infty\). These values agree with the two sides of the threshold \(1\).
A Common Covering Error
A tempting but incorrect argument for \([0,1]\) is: “For any \(s>0\), cover the interval by the single set \([0,1]\); its cost is \(1^s=1\), so the Hausdorff measure is zero.” The calculation uses a set of diameter \(1\), however, and this is not an admissible cover when the scale is \(\delta<1\). Hausdorff measure is obtained only after imposing arbitrarily small diameter restrictions.
The correction is to keep \(\delta\) fixed while choosing admissible covers. For \(s>1\), partitioning into \(N\) intervals gives cost \(N^{1-s}\), which can be made arbitrarily small while each interval has length at most \(\delta\). For \(0<s<1\), the length estimate gives the lower bound \(\delta^{s-1}\), which grows without bound as \(\delta\downarrow0\). The scale restriction is therefore essential on both sides of the threshold.
The exponent comparison is useful beyond these basic examples because it identifies dimension without requiring an exact formula for the Hausdorff measure at the critical exponent. To determine a dimension, it is enough to establish zero measure above a candidate exponent and infinite measure below it. The value exactly at the candidate exponent is a separate question.
Check Your Understanding
Use admissible covers at a fixed scale, and distinguish conclusions above, below, and at the critical exponent.
- For a nonempty singleton in a metric space, compute \(\mathcal{H}^s\) for \(s>0\) and explain why its dimension is zero.
- Suppose \(\mathcal{H}^{3/2}(E)=0\). What can be concluded about \(\mathcal{H}^2(E)\), and which exponent-comparison inequality applies?
- For \(E=[2,5]\), use the interval lower-bound argument to show that \(\mathcal{H}^{1/2}(E)=\infty\).
- Why does covering \([0,1]\) by one set not establish that its Hausdorff measure is zero?
- Does knowing \(\dim_H E=D\) alone determine \(\mathcal{H}^D(E)\)? Explain what the threshold theorem does and does not say.