What Diameter Covers Reveal
A bounded interval in the line and a solid cube in Euclidean space have an evident geometric size, but Hausdorff measure asks a different question: what total cost is required to cover the set by pieces of small diameter? Comparing that cost with Lebesgue outer measure identifies the exponents at which the cost becomes infinite, stays positive and finite, or tends to zero. For intervals, this comparison also gives an exact value at the critical exponent. For cubes, the elementary estimates here give positive finite bounds at that exponent, not an exact formula.
Use the Euclidean metric on \(\mathbb{R}^d\), and let \(m\) denote Lebesgue measure on \(\mathbb{R}^d\), with \(m^*\) its outer measure. The covers below are countable covers by arbitrary subsets of \(\mathbb{R}^d\); cover sets need not be measurable. The Hausdorff measure defined by diameter costs is an outer measure, and its restriction to Borel sets is a measure. We will use \(m^*\) to compare a set with the pieces in a cover.
The Lebesgue Estimate for a Covering Piece
The key estimate does not require a cover set to be measurable. If \(U\subseteq\mathbb{R}^d\) is nonempty with finite diameter \(r\), then the range of each coordinate on \(U\) has length at most \(r\): the difference between two values of a coordinate is at most the Euclidean distance between the corresponding points. Thus \(U\) is contained in a box whose side lengths are at most \(r\). By monotonicity of Lebesgue outer measure and the formula for the measure of a box,
This also holds when \(r=0\), since then \(U\) contains just one point and has outer measure zero. If \(U\) has infinite diameter, the inequality with right side \(+\infty\) holds automatically. For \(d=1\), the same argument says \(m^*(U)\le\operatorname{diam}U\).
Proof. First consider the cube \(Q\). For \(0<s<d\), take any countable cover \((U_i)\) of \(Q\) with \(\operatorname{diam}U_i\le\delta\). Countable subadditivity of Lebesgue outer measure, the covering-piece estimate, and the inequality \(r^d\le\delta^{d-s}r^s\) for \(0\le r\le\delta\) give $$ a^d=m^*(Q)\le\sum_i m^*(U_i) \le\sum_i(\operatorname{diam}U_i)^d \le\delta^{d-s}\sum_i(\operatorname{diam}U_i)^s. $$ Taking the infimum over such covers shows \(\mathcal{H}^s_\delta(Q)\ge a^d\delta^{s-d}\). As \(\delta\downarrow0\), this lower bound tends to \(+\infty\), so \(\mathcal{H}^s(Q)=\infty\). For \(s=0\), if a cover has \(N\) nonempty sets, then the same estimate gives \(a^d\le N\delta^d\), since each set has diameter at most \(\delta\). Thus its cost \(N\) is at least \(a^d\delta^{-d}\). Taking the infimum and then letting \(\delta\downarrow0\) proves \(\mathcal{H}^0(Q)=\infty\).
For the upper bounds, divide each side of \(Q\) into \(N\) equal parts. The resulting \(N^d\) smaller cubes cover \(Q\), each with diameter \(\sqrt d\,a/N\). Choose \(N\) large enough that this diameter is at most \(\delta\). The total cost at exponent \(s>0\) is $$ N^d\left(\frac{\sqrt d\,a}{N}\right)^s =d^{s/2}a^sN^{d-s}. $$ If \(s>d\), this tends to zero as \(N\to\infty\), while the cover remains admissible for any fixed \(\delta>0\) once \(N\) is large enough. Therefore \(\mathcal{H}^s_\delta(Q)=0\), and hence \(\mathcal{H}^s(Q)=0\). At \(s=d\), the cost is \(d^{d/2}a^d\), giving \(\mathcal{H}^d_\delta(Q)\le d^{d/2}a^d\). The covering-piece estimate gives \(a^d\le\sum_i(\operatorname{diam}U_i)^d\) for every cover, so \(\mathcal{H}^d_\delta(Q)\ge a^d\). These bounds hold at every scale and therefore also hold for \(\mathcal{H}^d(Q)\). They establish the cube claims, including its dimension from the definition.
For the interval, the covering-piece estimate in dimension one gives, for \(0<s<1\) and every cover with diameters at most \(\delta\), $$ L=m^*(I)\le\sum_i\operatorname{diam}U_i \le\delta^{1-s}\sum_i(\operatorname{diam}U_i)^s. $$ Thus \(\mathcal{H}^s_\delta(I)\ge L\delta^{s-1}\), which tends to infinity as \(\delta\downarrow0\). At \(s=0\), a cover by \(N\) nonempty sets has \(L\le N\delta\), so its cost is at least \(L/\delta\), again tending to infinity. At \(s=1\), every cover has total diameter cost at least \(L\). Conversely, divide \(I\) into \(N\) closed subintervals of length \(L/N\), choosing \(N\) so that \(L/N\le\delta\). Their total cost is \(N(L/N)=L\). Hence \(\mathcal{H}^1_\delta(I)=L\) for every \(\delta>0\), and \(\mathcal{H}^1(I)=L\). Finally, for \(s>1\), the same subdivision has total cost \(N(L/N)^s=L^sN^{1-s}\), which tends to zero as \(N\to\infty\) while satisfying the diameter restriction. Thus \(\mathcal{H}^s(I)=0\). These results give \(\dim_H I=1\). \(\square\)
Worked Cover Calculations
Worked Example: An Interval of Length Three
Let \(I=[-1,2]\subseteq\mathbb{R}\), so its length is \(L=2-(-1)=3\). The theorem gives \(\mathcal{H}^1(I)=3\), \(\mathcal{H}^s(I)=\infty\) for \(0\le s<1\), and \(\mathcal{H}^s(I)=0\) for \(s>1\). Here are direct cover calculations at two exponents.
At \(s=\tfrac12\), any cover with diameters \(r_i\le\delta\) satisfies $$ 3\le\sum_i r_i\le\delta^{1/2}\sum_i r_i^{1/2}. $$ Therefore its cost is at least \(3\delta^{-1/2}\), which tends to infinity as \(\delta\downarrow0\). At \(s=2\), divide the interval into \(N\) pieces of length \(3/N\), with \(3/N\le\delta\). The total cost is $$ N\left(\frac{3}{N}\right)^2=\frac{9}{N}, $$ which tends to zero as \(N\to\infty\). Check: at \(s=1\), \(N\) pieces each have cost \(3/N\), so the total is \(N(3/N)=3\); the lower bound from Lebesgue outer measure is also \(3\).
Worked Example: A Cube in Three Dimensions
Take \(Q=[0,2]^3\subseteq\mathbb{R}^3\). Its volume is \(m(Q)=2^3=8\), and the critical exponent is \(s=3\). The theorem gives $$ 8\le\mathcal{H}^3(Q)\le 3^{3/2}\,2^3=24\sqrt3. $$ In particular, the critical measure is positive and finite.
For an explicit cover, divide each side into \(N=4\) pieces. There are \(4^3=64\) smaller cubes, each of side \(2/4=1/2\) and diameter \(\sqrt3/2\). Their total cost at \(s=3\) is $$ 64\left(\frac{\sqrt3}{2}\right)^3 =64\frac{3\sqrt3}{8} =24\sqrt3. $$ At \(s=2\), any cover with diameters at most \(\delta\) has cost at least \(8/\delta\), because \(8\le\delta\sum_i(\operatorname{diam}U_i)^2\). This tends to infinity as \(\delta\downarrow0\). At \(s=4\), the \(N^3\)-cube grid has total cost $$ N^3\left(\frac{2\sqrt3}{N}\right)^4 =\frac{144}{N}, $$ which tends to zero as \(N\to\infty\). Check: substituting \(N=4\) into this last formula gives \(144/4=36\), also obtained from \(64(\sqrt3/2)^4=64(9/16)=36\).
Worked Example: The Unit Square
Let \(Q=[0,1]^2\subseteq\mathbb{R}^2\). Its area is \(1\), and its critical exponent is \(2\). The theorem gives \(1\le\mathcal{H}^2(Q)\le 2\). The upper bound follows by dividing the square into \(N^2\) squares of side \(1/N\): each has diameter \(\sqrt2/N\), so at exponent \(2\) the total cost is $$ N^2\left(\frac{\sqrt2}{N}\right)^2=2. $$ The lower bound is the area estimate \(1=m(Q)\le\sum_i(\operatorname{diam}U_i)^2\) for every cover.
At \(s=1\), every cover with diameters at most \(\delta\) has cost at least \(1/\delta\), since \(1\le\delta\sum_i\operatorname{diam}U_i\); consequently \(\mathcal{H}^1(Q)=\infty\). At \(s=3\), the grid cost is $$ N^2\left(\frac{\sqrt2}{N}\right)^3=\frac{2\sqrt2}{N}, $$ which tends to zero, so \(\mathcal{H}^3(Q)=0\). Check: for \(N=2\), there are \(2^2=4\) squares of diameter \(\sqrt2/2\), and their \(s=3\) cost is \(4(\sqrt2/2)^3=\sqrt2/2\), matching \(2\sqrt2/2=\sqrt2\)? Recomputing, \(4(\sqrt2/2)^3=4(2\sqrt2/8)=\sqrt2\). Thus the grid formula gives \(2\sqrt2/2=\sqrt2\), as required.
What the Critical Estimate Does—and Does Not—Say
For an interval, the lower bound from Lebesgue outer measure and the interval subdivision cover meet exactly at exponent \(1\). That is why the unnormalized Hausdorff measure in this convention equals the interval’s length. For a cube in dimension \(d>1\), the same lower-bound argument gives its volume, but the elementary grid cover only gives the upper bound \(d^{d/2}\) times its volume. Those two estimates establish positive finite measure and determine the dimension; they do not prove that the critical Hausdorff measure equals the volume.
A common mistake is to conclude from \(m(Q)=a^d\) that \(\mathcal{H}^s(Q)=a^d\) for every exponent \(s\), or even that \(\mathcal{H}^d(Q)=a^d\). The covering-piece estimate only compares \(m^*(U)\) with \((\operatorname{diam}U)^d\), so its exponent is \(d\); for \(s\ne d\), the scale factor \(\delta^{d-s}\) changes the conclusion. At \(s=d\), it proves a lower bound, not an exact value. The grid cover supplies an upper bound, and the bounds need not coincide.
For intervals and cubes, the decisive comparison is therefore scale-sensitive: below the dimension, a fixed positive Lebesgue measure forces the covering cost to grow without bound; above the dimension, increasingly fine grids make the cost vanish. At the dimension, the same methods show exactly \(\mathcal{H}^1(I)=m(I)\) for intervals and show \(0<\mathcal{H}^d(Q)<\infty\) for cubes.
Check Your Understanding
Use countable diameter covers and the Lebesgue outer-measure estimate in each question.
- For \(I=[1,5]\subseteq\mathbb{R}\), compute \(\mathcal{H}^1(I)\) and state what happens for \(s=\tfrac12\) and \(s=2\).
- For a cube of side length \(a>0\) in \(\mathbb{R}^d\), derive the lower bound on \(\mathcal{H}^s_\delta(Q)\) when \(0<s<d\).
- Explain why the \(N^d\)-cube grid gives a total cost \(d^{s/2}a^sN^{d-s}\).
- What do the elementary estimates establish about \(\mathcal{H}^d([0,1]^2)\), and why do they not establish its exact value?
- For a cover of an interval of length \(L>0\) by sets of diameter at most \(\delta\), derive a lower bound for the number of nonempty cover sets when \(s=0\).