From Ball Mass to Hausdorff Measure
Hausdorff measure is defined by measuring sets through covers of small diameter. A useful way to prove that a set cannot have small Hausdorff measure is to put a finite measure on it and show that no small ball can carry too much mass. If the mass of every radius-\(r\) ball is at most a constant times \(r^s\), then a cover by sets of small diameter must account for all the mass using at least that much total \(s\)-power of diameter.
Let \((X,d)\) be a metric space, let \(\mathcal{B}_X\) be its Borel sigma-algebra, and let \(\mu\) be a finite Borel measure on \(X\). For \(s>0\), define the \(s\)-dimensional Hausdorff content at scale \(\delta>0\) by \(\mathcal{H}^s_\delta(E)=\inf\sum_i(\operatorname{diam}U_i)^s\), where the infimum is over countable covers \(E\subseteq\bigcup_i U_i\) by subsets \(U_i\subseteq X\) with \(\operatorname{diam}U_i\le\delta\). The Hausdorff measure is \(\mathcal{H}^s(E)=\lim_{\delta\downarrow0}\mathcal{H}^s_\delta(E)\). This convention uses the \(s\)-power of diameter without an additional normalizing factor.
The support condition in this statement means that all the measure is concentrated on \(E\): in particular, \(\mu(E)=\mu(X)\). It is enough to know the ball estimate for centers in \(E\), since the proof only uses such centers, but the stated all-centers condition is a convenient standard form. The theorem allows \(\mu(E)=0\), in which case its conclusion is the trivial lower bound \(\mathcal{H}^s(E)\ge0\). The restriction \(s>0\) matters when we shrink a ball around a set of diameter zero.
Proof by Arbitrary Diameter Covers
The key geometric fact is that a set of diameter \(d\) which contains a point \(x\) lies inside every ball centered at \(x\) whose radius is strictly greater than \(d\). It need not lie in a ball of radius \(d/2\). The slight enlargement of the radius can be made as small as needed because \(r\mapsto r^s\) is continuous for \(s>0\).
Proof. Fix \(\delta>0\) and a countable cover \(E\subseteq\bigcup_i U_i\) with \(\operatorname{diam}U_i=d_i\le\delta\). If \(\sum_i d_i^s=\infty\), then \(\sum_i d_i^s\ge\mu(E)/C\) already holds. Suppose instead that this sum is finite, and discard any empty members of the cover. For each remaining \(U_i\) that meets \(E\), choose \(x_i\in U_i\cap E\). Given \(\varepsilon>0\), continuity of \(r^s\) allows a choice \(r_i>d_i\) such that \(r_i^s\le d_i^s+\varepsilon 2^{-i}/C\). This also works when \(d_i=0\), because \(r^s\to0\) as \(r\downarrow0\). For every \(y\in U_i\), \(d(y,x_i)\le d_i<r_i\), so \(U_i\subseteq B(x_i,r_i)\). Thus these balls cover \(E\). Countable subadditivity and the assumed ball bound give \(\mu(E)\le\sum_i\mu(B(x_i,r_i))\le C\sum_i r_i^s\le C\sum_i d_i^s+\varepsilon\). Letting \(\varepsilon\downarrow0\) yields \(\sum_i d_i^s\ge\mu(E)/C\). This holds for every admissible cover, so \(\mathcal{H}^s_\delta(E)\ge\mu(E)/C\). Taking \(\delta\downarrow0\) proves the result. \(\square\)
Notice that the estimate does not require the covering sets \(U_i\) to be measurable: we bound their intersections with \(E\) by measurable balls instead. Finiteness of \(\mu\) makes the mass \(\mu(E)\) finite; the proof’s cover estimate itself uses countable subadditivity and the ball-growth hypothesis.
Worked Applications
Worked Example: Uniform Measure on the Unit Interval
Let \(X=\mathbb{R}\) with its usual distance and Borel sigma-algebra, let \(E=[0,1]\), and define the finite Borel measure \(\mu(A)=\tfrac12 m(A\cap[0,1])\). Its total mass is \(\mu(E)=\tfrac12 m([0,1])=\tfrac12\). For any \(x\in\mathbb{R}\) and \(r>0\), the interval \(B(x,r)=(x-r,x+r)\) has length \(2r\), so \(\mu(B(x,r))=\tfrac12 m(B(x,r)\cap[0,1])\le\tfrac12(2r)=r\). The theorem applies with \(s=1\) and \(C=1\); therefore \(\mathcal{H}^1([0,1])\ge\tfrac12\).
As an independent check of the scale, partition \([0,1]\) into \(N\) closed intervals of length \(1/N\), where \(N\) is any integer with \(1/N\le\delta\). Their diameters sum to \(N(1/N)=1\), so \(\mathcal{H}^1_\delta([0,1])\le1\), and hence \(\mathcal{H}^1([0,1])\le1\). The lower bound \(\tfrac12\) is consistent with this cover estimate. (The mass distribution principle supplies a lower bound, not necessarily an exact value.)
Worked Example: Area Measure on a Square
Take \(X=\mathbb{R}^2\), \(E=[0,1]^2\), and let \(\mu(A)=m(A\cap E)\), where \(m\) is Lebesgue measure on \(\mathbb{R}^2\). This is a finite Borel measure with \(\mu(E)=1\). For any center \(x=(x_1,x_2)\) and radius \(r>0\), the ball \(B(x,r)\) is contained in the coordinate square \((x_1-r,x_1+r)\times(x_2-r,x_2+r)\). That square has area \((2r)(2r)=4r^2\), and therefore \(\mu(B(x,r))\le4r^2\). With \(s=2\), \(C=4\), the theorem gives \(\mathcal{H}^2(E)\ge1/4\).
For a cover check, divide \(E\) into \(N^2\) closed squares of side \(1/N\). Each has diameter \(\sqrt{(1/N)^2+(1/N)^2}=\sqrt{2}/N\), and the sum of squared diameters is \(N^2(2/N^2)=2\). Choosing \(N\) large enough that \(\sqrt{2}/N\le\delta\) gives \(\mathcal{H}^2_\delta(E)\le2\). Thus the theorem’s lower bound \(1/4\) is compatible with an explicit small-diameter cover.
Worked Example: A Weighted Measure on the Interval
Again take \(X=\mathbb{R}\) and \(E=[0,1]\), but now define \(\mu(A)=\int_{A\cap[0,1]}2x\,dm(x)\). This is a finite Borel measure concentrated on \(E\), and its total mass is \(\mu(E)=\int_0^1 2x\,dx=[x^2]_0^1=1\). For a ball \(B(y,r)\), its intersection with \([0,1]\) is an interval, possibly empty, with endpoints \(a,b\) satisfying \(0\le a\le b\le1\) and \(b-a\le2r\). Its mass is \(b^2-a^2=(b-a)(a+b)\le2(b-a)\le4r\), since \(a+b\le2\). The empty intersection has mass zero and satisfies the same bound. Thus the mass distribution principle applies with \(s=1\), \(C=4\), and gives \(\mathcal{H}^1(E)\ge1/4\).
As a check, the calculation of total mass used the full density: \(\int_0^1 2x\,dx=1^2-0^2=1\). The ball estimate uses the largest possible value \(2x\le2\) on the support and the interval-length bound \(2r\), giving \(2\cdot2r=4r\); both quantities match the stated constant.
Turning Measure Bounds into Dimension Bounds
The Hausdorff dimension of \(E\) is \(\dim_H E=\inf\{t\ge0:\mathcal{H}^t(E)=0\}\), using the usual critical-exponent convention. The mass distribution principle is useful because a positive lower bound for \(\mathcal{H}^s(E)\) rules out vanishing Hausdorff measure at every smaller exponent.
Proof. The theorem gives \(\mathcal{H}^s(E)\ge\mu(E)/C>0\). Fix \(t\) with \(0\le t<s\). For any cover with diameters at most \(\delta\le1\), each diameter \(d_i\) satisfies \(d_i^t\ge d_i^s\): for \(d_i>0\) this follows from \(d_i^{s-t}\le1\), and for \(d_i=0\) both powers are zero when \(t>0\). At \(t=0\), the zero-diameter convention for Hausdorff contents counts nonempty covering sets with weight \(1\), so the same inequality holds. Taking infima gives \(\mathcal{H}^t_\delta(E)\ge\mathcal{H}^s_\delta(E)\); passing to the limits as \(\delta\downarrow0\) yields \(\mathcal{H}^t(E)\ge\mathcal{H}^s(E)>0\). Hence no exponent \(t<s\) belongs to the set of exponents at which the Hausdorff measure is zero, and the critical-exponent definition gives \(\dim_H E\ge s\). \(\square\)
In practice, the method is a three-step lower-bound argument: construct a nonzero finite measure concentrated on the set, establish a uniform ball-mass estimate at the desired exponent, and apply the theorem. The measure need not be normalized to total mass one; scaling it scales both \(\mu(E)\) and the admissible constant \(C\), so the ratio \(\mu(E)/C\) is the quantity that enters the conclusion.
A Common Covering Pitfall
A tempting but incorrect shortcut is to say that a set of diameter \(d\) lies in a ball of radius \(d/2\) centered at one of its points. For a concrete failure, take \(U=[0,1]\subset\mathbb{R}\), whose diameter is \(1\), and choose \(x=0\in U\). The point \(1\in U\) is not in \(B(0,1/2)=(-1/2,1/2)\). Applying that false containment to the uniform measure example above would replace the valid radius-\(d\) bound by a radius-\(d/2\) bound and incorrectly suggest a stronger Hausdorff lower bound. The correction is to use \(B(x,d+\eta)\) for \(\eta>0\), then choose \(\eta\) small enough to control \((d+\eta)^s\). This is precisely the enlargement handled in the proof.
There is also a useful diagnostic for the exponent: for \(s>0\), a measure with a positive atom cannot satisfy \(\mu(B(x,r))\le C r^s\) at that atom for every \(r>0\). Indeed, if \(\mu(\{x\})=a>0\), then \(a\le\mu(B(x,r))\le Cr^s\) for every \(r>0\), while \(Cr^s\to0\) as \(r\downarrow0\), a contradiction. Thus the small-ball condition genuinely constrains how mass may concentrate.
Check Your Understanding
In each question, keep track of the support condition and the requirement that the ball estimate hold for every positive radius.
- In the mass distribution principle, why is it important to allow a covering set of diameter zero, and where is the assumption \(s>0\) used?
- Suppose a finite Borel measure on a metric space is concentrated on \(E\) and satisfies \(\mu(B(x,r))\le 7r^{3/2}\). What lower bound follows for \(\mathcal{H}^{3/2}(E)\)?
- For Lebesgue area measure restricted to \([0,1]^2\), verify the ball estimate with \(C=4\) and \(s=2\) using a containing coordinate square.
- Explain why positive \(\mathcal{H}^s(E)\) implies \(\dim_H E\ge s\), using covers with diameters at most \(\delta\le1\).
- Give a specific point of \([0,1]\) showing that the interval does not fit inside a radius-\(1/2\) ball centered at \(0\), despite having diameter \(1\).