From Diameter Covers to Hausdorff Measure
Lebesgue measure describes size using volume, which is suited to subsets of Euclidean space of a fixed ambient dimension. Diameter covers give a different way to measure sets: cover a set by pieces whose diameters are small, and add a power of each piece’s diameter. Restricting the permitted diameters to a scale \(\delta\) makes the construction sensitive to the geometry at that scale. Letting \(\delta\) decrease to zero defines Hausdorff measure.
Let \((X,d)\) be a metric space and \(E\subseteq X\). No measurability assumption on \(E\) or on the covering sets is needed for the definitions below. For \(U\subseteq X\), define its diameter by \(\operatorname{diam}(U)=\sup\{d(x,y):x,y\in U\}\), with \(\operatorname{diam}(\varnothing)=0\). A singleton also has diameter zero. Fix a real number \(s\ge0\), called the exponent, and a scale \(\delta>0\).
The convention at \(s=0\) matters: a nonempty set of diameter zero costs one, not zero. This is the zero-dimensional version of counting how many covering pieces are required. For \(s>0\), a singleton costs zero because \(0^s=0\). The covers are countable, not arbitrary uncountable families; allowing uncountably many singletons would make the construction trivial even for large sets.
Why the Small-Scale Limit Exists
Proof. Every diameter-\(\delta_1\) cover is also a diameter-\(\delta_2\) cover, since \(\operatorname{diam}(U_i)\le\delta_1\le\delta_2\). Thus the family over which the first infimum is taken is contained in the family for the second. Taking an infimum over the smaller family cannot give a smaller value, so \(\mathcal H^s_{\delta_1}(E)\ge\mathcal H^s_{\delta_2}(E)\). This argument also applies to \(s=0\), since the cover costs themselves do not depend on \(\delta\).
Therefore, as \(\delta\) decreases, the contents are nondecreasing. Their supremum exists in \([0,\infty]\). To see that this supremum is the limit as \(\delta\downarrow0\), let \(L=\sup_{\delta>0}\mathcal H^s_\delta(E)\). If \(L<\infty\), then for every \(\varepsilon>0\), the definition of supremum gives some \(\delta_0>0\) with \(\mathcal H^s_{\delta_0}(E)>L-\varepsilon\). For \(0<\delta\le\delta_0\), monotonicity gives \(L-\varepsilon<\mathcal H^s_{\delta_0}(E)\le\mathcal H^s_\delta(E)\le L\). Hence the limit is \(L\). If \(L=+\infty\), then for every finite \(A\) some \(\delta_0>0\) has \(\mathcal H^s_{\delta_0}(E)>A\), and all smaller scales have content at least that large. The limit is then \(+\infty\). \(\square\)
This proposition is the reason the small-scale limit is well defined; no convergence theorem for integrals is involved. It does not yet assert that \(\mathcal H^s\) is an outer measure. Establishing the outer-measure properties is a separate step.
Worked Example: The Hausdorff Measure of an Interval
Take \(X=\mathbb R\) with its usual metric, \(E=[0,L]\) for \(L>0\), exponent \(s=1\), and normalization \(c_1=1\). We compute \(\mathcal H^1_\delta([0,L])\) for every \(\delta>0\).
For the upper bound, set \(N=\lceil L/\delta\rceil\), the least integer greater than or equal to \(L/\delta\). Then \(N\ge1\), \(N\ge L/\delta\), and so \(L/N\le\delta\). Divide \([0,L]\) into \(N\) closed subintervals of equal length \(L/N\). Their union is \([0,L]\), each has diameter \(L/N\le\delta\), and their total cost is \(N(L/N)=L\). Thus \(\mathcal H^1_\delta([0,L])\le L\).
For the lower bound, consider any countable diameter-\(\delta\) cover \((U_i)\) of \([0,L]\). Each nonempty \(U_i\) is bounded because its diameter is at most the finite number \(\delta\). It is contained in the interval from \(\inf U_i\) to \(\sup U_i\), whose length is \(\sup U_i-\inf U_i=\operatorname{diam}(U_i)\). By monotonicity and countable subadditivity of Lebesgue outer measure \(m^*\), and the identity \(m^*([0,L])=L\),
Taking the infimum over covers gives \(\mathcal H^1_\delta([0,L])\ge L\). Combining the bounds gives \(\mathcal H^1_\delta([0,L])=L\) for every \(\delta>0\); therefore \(\mathcal H^1([0,L])=L\). Check: when \(L=2\) and \(\delta=3/4\), \(N=\lceil 2/(3/4)\rceil=\lceil8/3\rceil=3\); the three intervals have length \(2/3\le3/4\), and their total cost is \(3(2/3)=2\), agreeing with the lower bound.
Normalization Changes the Scale, Not the Covering Rule
The constant \(c_s\) is a convention that fixes the numerical scale of the measure. If \(\widetilde{\mathcal H}^s_\delta\) denotes the construction with constant \(\widetilde c_s>0\), while \(\mathcal H^s_\delta\) uses \(c_s>0\), then for every \(E\) and \(\delta\),
Indeed, each cover cost is multiplied by the same positive constant. Multiplying all candidate costs by a positive constant multiplies their infimum by that constant; taking the small-scale limit preserves the same factor, including when the limit is infinite. A common convention is \(c_s=1\). For integer \(k\ge1\) in Euclidean space, another common normalization uses \(c_k=\omega_k/2^k\), where \(\omega_k\) is the \(k\)-dimensional Lebesgue measure of the unit ball in \(\mathbb R^k\). These conventions agree at \(k=1\), since the unit ball in \(\mathbb R\) is \([-1,1]\), so \(\omega_1=2\) and \(c_1=2/2=1\).
Worked Example: Changing the Normalization for an Interval
For \([0,2]\subset\mathbb R\) and \(s=1\), the calculation above gives unnormalized content \(2\) and unnormalized Hausdorff measure \(2\). If instead the normalization is \(\widetilde c_1=1/2\), every cover cost is multiplied by \(1/2\). Thus \(\widetilde{\mathcal H}^1_\delta([0,2])=(1/2)2=1\) at every \(\delta>0\), and \(\widetilde{\mathcal H}^1([0,2])=1\).
The result changes because the chosen unit of measure changed, not because the permitted covers changed. In particular, one must state the normalization before comparing numerical values. Check: the ratio of the new value to the old one is \(1/1? \) More directly, the values are \(1\) and \(2\), so their ratio is \(1/2=\widetilde c_1/c_1\), as required.
Zero-Dimensional and Countable Examples
Worked Example: A Three-Point Set with Exponent Zero
Let \(E=\{0,1,3\}\subset\mathbb R\), take \(s=0\), and use \(c_0=1\). A nonempty covering set costs one, regardless of its diameter, while an empty set costs zero.
If \(0<\delta<1\), no set of diameter at most \(\delta\) can contain two distinct points of \(E\), because the smallest distance between distinct points is \(1\). At least three nonempty covering sets are required. The three singleton sets form a cover with cost \(3\), so \(\mathcal H^0_\delta(E)=3\).
If \(1\le\delta<3\), the set \(\{0,1\}\) has diameter \(1\le\delta\), and \(\{3\}\) has diameter zero, so two sets cover \(E\). One set cannot cover \(E\), because \(\operatorname{diam}(E)=3>\delta\). Hence \(\mathcal H^0_\delta(E)=2\). If \(\delta\ge3\), \(E\) itself is a one-set cover, so \(\mathcal H^0_\delta(E)=1\). As \(\delta\downarrow0\), the content is eventually \(3\), and \(\mathcal H^0(E)=3\). Check: at \(\delta=1/2\), the three singleton sets cost \(1+1+1=3\); at \(\delta=3\), the single set \(E\) costs \(1\).
Worked Example: Countable Sets at Positive Exponent
Let \(E=\{x_1,x_2,\ldots\}\) be a countable subset of any metric space, and fix \(s>0\). For each \(\delta>0\), the singleton sets \(\{x_i\}\) form a countable cover of \(E\). Each has diameter zero and cost \(0^s=0\), so the total cost is \(0\). Since cover costs are nonnegative, their infimum is also \(0\): \(\mathcal H^s_\delta(E)=0\). This holds at every scale, hence \(\mathcal H^s(E)=0\). The same reasoning applies to finite sets.
At exponent zero, the cost rule is different. For a finite set of \(q\) distinct points, choose \(\delta\) smaller than the minimum distance between distinct points; when \(q=1\), any \(\delta>0\) suffices. Then no covering set can contain two of the points, so at least \(q\) nonempty sets are needed, and the singleton cover uses exactly \(q\). Thus the small-scale value is \(q\). For a countably infinite set, given any positive integer \(q\), choose \(q\) distinct points. Their minimum pairwise distance is positive, so at all sufficiently small scales these \(q\) points require at least \(q\) covering sets. Since this holds for every \(q\), the zero-dimensional content tends to \(+\infty\). These conclusions follow from the covering definition and do not require the set to be measurable.
A Common Mistake: Holding the Number of Pieces Fixed
For \([0,1]\), it is incorrect to argue that its one-dimensional Hausdorff measure is zero because each set in a diameter-\(\delta\) cover costs at most \(\delta\), and \(\delta\) tends to zero. The number of sets in a cover cannot be held fixed as the permitted diameter shrinks. For example, at scale \(\delta=1/n\), partition \([0,1]\) into \(n\) intervals of length \(1/n\). The total cost is \(n(1/n)=1\), not \(1/n\). The sum in the definition includes every covering set; an upper bound on each individual term is not an upper bound on the whole sum unless the number of terms is also controlled. The interval calculation proves the matching lower bound, so the content is exactly \(1\) at every scale under the normalization \(c_1=1\).
Another error is to treat the normalization as if it changed which covers are admissible. It does not: the condition \(\operatorname{diam}(U_i)\le\delta\) is unchanged. The constant multiplies the cost after the covering pieces have been chosen and therefore scales all resulting contents and measures by the same factor.
Check Your Understanding
Use the definitions and the scale monotonicity proposition to answer the following.
- For \(E=[0,3]\subset\mathbb R\) with \(s=1\) and \(c_1=1\), compute \(\mathcal H^1_\delta(E)\) for an arbitrary \(\delta>0\).
- Explain why \(\mathcal H^s(E)=0\) for a countable set \(E\) whenever \(s>0\), and identify the property of singleton covers used.
- For \(E=\{0,2\}\) and \(s=0\), compute the content when \(0<\delta<2\) and when \(\delta\ge2\).
- If the normalization constant is multiplied by \(5\), how do the \(\delta\)-contents and the limiting Hausdorff measure change?
- Why does the fact that every covering set costs at most \(\delta\) fail to prove that the one-dimensional measure of an interval is zero?