Hausdorff Measure, Lipschitz Maps, and Densities
Length and area can be defined for sets that have no smooth parametrization. Hausdorff measure does this by covering a set with small pieces and adding powers of their diameters. The exponent specifies the dimension being measured.
We develop this construction, examine its behaviour under Lipschitz maps, and prove almost-everywhere differentiability of those maps. Density estimates then describe how the measure of a set is distributed near a typical point. These are the tools needed for area, coarea, and rectifiability.
1. Hausdorff Measure and Lipschitz Functions
Lipschitz functions constitute a larger class of functions than the class of smooth functions (defined on \(\mathscr C^\infty\) manifolds) considered so far. Their uniform control on distances is enough for several of the measure-theoretic conclusions used below. We begin with defining the Hausdorff dimension of sets, and extend that to Hausdorff measures on \(\mathbb R^d\). For starters, we define the Hausdorff measure.
Definition 1. Put \(c_s=\omega_s2^{-s}\), where \(\omega_s=\pi^{s/2}/\Gamma(1+s/2)\). For \(0\leq s<\infty\) and \(\delta>0\), define
For \(s=0\), every nonempty covering set costs one, including a singleton; empty sets can be discarded. The empty cover gives zero to the empty set.
Definition 2. The \(s\)-dimensional Hausdorff outer measure is
The limit exists because imposing a smaller diameter bound can only increase the infimum. The variable tending to zero is the covering scale \(\delta\); the dimension \(s\) stays fixed.
Remark 3. The normalization gives \(\lambda_d(B_r(x))=\omega_dr^d\). We use the standard normalization theorem \(\mathcal H^d=\lambda_d\) on \(\mathbb R^d\), including outer measures. One direction follows from the isodiametric inequality \(\lambda_d^*(C)\leq c_d(\operatorname{diam}C)^d\) applied to every cover. The opposite direction is the Euclidean covering theorem; an exact statement and proof are in Simon, Chapter 1, §2. This normalization is an input to the linear area calculations below.
Proposition 4. \(\mathcal H^s\) is a metric outer measure, hence all Borel sets are Carathéodory measurable. It is Borel regular: every set \(A\) has a Borel superset \(B\) with \(\mathcal H^s(B)=\mathcal H^s(A)\). Its restriction to measurable sets is a complete measure.
Proof. Empty-set and monotonicity properties follow directly from covers. For countable subadditivity, choose a cover of each \(A_k\) within \(\varepsilon2^{-k}\) of its infimum at scale \(\delta\). Their union covers \(\bigcup_kA_k\), so
Let \(\varepsilon\downarrow0\) and \(\delta\downarrow0\). These are outer-measure assertions; \(\mathcal H_\delta^s\) need not be additive on Borel sets at a fixed positive scale.
If \(\operatorname{dist}(A,B)>\delta\), no covering set of diameter at most \(\delta\) meets both \(A\) and \(B\). Splitting a cover into the two subfamilies therefore gives
Letting \(\delta\downarrow0\) proves metric additivity; the metric outer-measure criterion then makes every Borel set measurable.
For Borel regularity assume \(\mathcal H^s(A)<\infty\). Closing each covering set preserves its diameter. Choose closed \(C_j^k\) covering \(A\), of diameter at most \(1/k\), whose total cost is at most \(\mathcal H^s_{1/k}(A)+1/k\). Set \(B=\bigcap_k\bigcup_jC_j^k\). Then \(A\subseteq B\subseteq\bigcup_jC_j^k\) for every \(k\), and
Take \(k\to\infty\) and use monotonicity for the reverse inequality. If \(\mathcal H^s(A)=\infty\), take \(B=\mathbb R^d\). Finally, \(s<d\) gives infinite measure to every nonempty open ball, so this measure is not locally finite on \(\mathbb R^d\) in that regime. ◻
Proposition 5. \(\mathcal H^0\) is counting measure; \(\mathcal H^1=\lambda_1\) on \(\mathbb R\); \(\mathcal H^s=0\) on \(\mathbb R^d\) for \(s>d\); and
for every affine isometry \(U\).
Proof. A sufficiently small scale separates any fixed finite collection of distinct points; thus the zero-dimensional cost is at least its cardinality. Singleton covers give the reverse inequality. On a line, replacing a set by its interval hull preserves diameter, yielding the usual interval-cover definition of Lebesgue outer measure. A bounded cube in \(\mathbb R^d\) can be covered by \(O(\delta^{-d})\) cubes of diameter at most \(\delta\), at total \(s\)-cost \(O(\delta^{s-d})\); exhaust space by bounded cubes. Scaling or applying an isometry to every cover, then applying the inverse map, proves the two equalities. ◻
Proposition 6. For a finite dimension parameter \(s\geq0\), if \(\mathcal H_\delta^s(A)=0\) at one positive scale, then \(\mathcal H^s(A)=0\).
Proof. For \(s=0\), zero cost implies \(A=\varnothing\). For \(s>0\), a cover of total cost at most \(\varepsilon\) has each diameter at most \((\varepsilon/c_s)^{1/s}\). Hence the same cover works at that smaller scale. Given any fixed \(\eta>0\), let \(\varepsilon\downarrow0\) with \((\varepsilon/c_s)^{1/s}<\eta\); then \(\mathcal H_\eta^s(A)=0\). Take \(\eta\downarrow0\). ◻
Lemma 7. For \(0\leq s<t<\infty\), finite \(\mathcal H^s(A)\) implies \(\mathcal H^t(A)=0\); equivalently, positive \(\mathcal H^t(A)\) implies infinite \(\mathcal H^s(A)\).
Proof. Every scale-\(\delta\) cover gives
Send \(\delta\downarrow0\). The zero-dimensional convention makes the same comparison valid when \(s=0\). ◻
Definition 8. \(\dim_H A=\inf\{s\geq0:\mathcal H^s(A)=0\}\). Proposition 5 gives \(\dim_H A\leq d\) for subsets of \(\mathbb R^d\).
Proposition 9. If \(f:A\subseteq\mathbb R^d\to\mathbb R^q\) is \(L\)-Lipschitz and \(s>0\), then \(\mathcal H^s(f(A))\leq L^s\mathcal H^s(A)\). For \(s=0\), the assertion is simply that a map cannot increase cardinality.
Proof. For \(L>0\), intersect a scale-\(\delta\) cover with \(A\) and apply \(f\). Image diameters are at most \(L\) times the original diameters, so \(\mathcal H_{L\delta}^s(f(A))\leq L^s\mathcal H_\delta^s(A)\). Let \(\delta\downarrow0\). A constant map has image at most one point and has zero positive-dimensional Hausdorff measure. In particular Lipschitz maps carry \(\mathcal H^s\)-null sets to null sets. ◻
Proposition 10. For \(f\in L^1_{\mathrm{loc}}(\mathbb R^d)\) and \(0\leq s<d\), the exceptional set
has \(\mathcal H^s\) measure zero. The factor \(r^{-s}\) is part of the assertion.
Proof. At Lebesgue points of \(|f|\), the displayed quantity is \(\omega_dr^{d-s}\) times a convergent average, so it tends to zero. Work on a bounded ball and truncate \(f\) outside a larger ball; this reduces the covering argument to an integrable function. For \(\varepsilon>0\), the set \(E\) where the limsup exceeds \(\varepsilon\) is Lebesgue null. Choose an open \(U\supseteq E\) with \(\int_U|f|<\sigma\). For every \(x\in E\) there are arbitrarily small closed balls \(B_r(x)\subset U\) satisfying \(\int_{B_r(x)}|f|>\varepsilon r^s\).
Select disjoint such balls \(B_{r_i}(x_i)\) with \(r_i<\delta\) whose fivefold enlargements cover \(E\), by Lemma 22 below. Countability follows from disjoint positive-radius balls in Euclidean space. Then
Let \(\delta\downarrow0\), then \(\sigma\downarrow0\), and take the countable union over bounded balls and \(\varepsilon=1/j\). For \(s=0\) absolute continuity of the integral also shows directly that \(E\) is empty. The selected balls themselves need not cover \(E\); their enlargements do. ◻
The following proof supplies the derivative used in the area and coarea formulas.
Theorem 11 (Rademacher). Let \(f : \mathbb R^n \to \mathbb R^m\) be a locally Lipschitz function. Then, \(f\) is differentiable \(\lambda_n\)-a.e.
Proof. It suffices to prove the scalar case on an open ball \(B\) where \(f\) has a Lipschitz constant \(L\). A locally Lipschitz map admits a countable cover by such balls, and the vector-valued result follows by intersecting the full-measure sets for its finitely many components.
Fix a unit vector \(v\). The restriction of \(f\) to every line parallel to \(v\), on its intersection with \(B\), is Lipschitz and hence absolutely continuous. The one-dimensional differentiation theorem and Fubini imply that
exists for almost every \(x\in B\), with \(|D_vf|\le L\). The exceptional set is measurable: the upper and lower limits can be taken through rational nonzero \(t\), giving countable suprema and infima of continuous difference quotients. This is an almost-everywhere statement for each fixed direction, not a claim about all points.
Let \(g_i=D_{e_i}f\), choosing arbitrary values on their null exceptional sets, and put \(g=(g_1,\ldots,g_n)\). For \(\zeta\in C_c^\infty(B)\) and sufficiently small \(t\), translation of the integration variable gives
All integrations occur in a fixed compact subset of \(B\). Dominated convergence on the left, and smoothness of \(\zeta\) on the right, give
The last equality follows from the same identity in coordinate directions. Uniqueness of locally integrable distributions implies \(D_vf=v\cdot g\) almost everywhere.
Choose a countable dense set \(D\) in the unit sphere, containing the coordinate vectors. Intersect the full-measure sets just obtained for \(v\in D\). At every point \(x\) of the resulting set, all these directional derivatives exist and equal \(v\cdot g(x)\). We show that \(h\mapsto g(x)\cdot h\) is the full derivative.
For small nonzero \(t\), put
Uniformly over unit vectors \(v,w\), the Lipschitz bound gives \(|Q_t(v)-Q_t(w)|\le(L+|g(x)|)|v-w|\). Given \(\varepsilon>0\), choose a finite set \(v_1,\ldots,v_N\in D\) whose balls of radius \(\varepsilon/[2(L+|g(x)|+1)]\) cover the sphere. For sufficiently small \(|t|\), all \(|Q_t(v_j)|<\varepsilon/2\). The preceding Lipschitz estimate then gives \(\sup_{|v|=1}|Q_t(v)|<\varepsilon\). Setting \(t=|h|\) and \(v=h/|h|\) proves
This proves differentiability at every point of a full-measure set. The union of the null exceptional sets over the countable ball cover and the finitely many output components is still null. This completes the proof. ◻
The following linear estimates and decompositions prepare for the area and coarea formulas. They relate the Jacobian to the measure of images and fibres.
Lemma 12. For a linear map \(L:\mathbb R^n\to\mathbb R^m\) with \(n\leq m\),
This holds for every \(A\), with \(0\cdot\infty=0\) when the rank is deficient.
Proof. If \(L\) is injective, write \(L=OT\) with \(O\) an isometry onto its range and \(T=(L^*L)^{1/2}\) positive definite. Isometry invariance, \(\mathcal H^n=\lambda_n\) on an \(n\)-plane, and linear change of variables give the formula. If the rank is below \(n\), the image is contained in a lower-dimensional linear space, whose \(\mathcal H^n\) measure is zero by a cube covering. ◻
Lemma 13. Let \(f:\mathbb R^n\to\mathbb R^m\) be Lipschitz, \(n\leq m\), and let \(A\) be Lebesgue measurable. Then \(f(A)\) and the multiplicity \(N(f,A,y)=\mathcal H^0(A\cap f^{-1}(y))\) are \(\mathcal H^n\)-measurable, and
Proof. Write \(A\) as a countable union of compact sets together with a null remainder. The compact images are compact, and the remaining image is \(\mathcal H^n\)-null by Proposition 9. This proves image measurability, also for \(A\) intersected with any dyadic cube. Partition all of \(\mathbb R^n\) into half-open dyadic cubes of side \(2^{-k}\) and set
These measurable counts increase under refinement. Any finite collection of distinct preimages is separated by sufficiently small cubes, so \(N_k\uparrow N\), including infinite multiplicities. Finally,
Monotone convergence completes the proof. ◻
Lemma 14. Let \(f:\mathbb R^n\to\mathbb R^m\) be Lipschitz, \(n\leq m\), and \(t>1\). The differentiability set \(B=\{x:J_nf(x)>0\}\) has a countable Borel partition into sets \(E_k\) on which \(f\) is injective. There are positive definite symmetric \(T_k:\mathbb R^n\to\mathbb R^n\) such that
The maps and inverses here have their natural restricted domains and images.
Proof. Choose \(\eta>0\) with \(1+\eta\leq t\) and \((1-\eta)^{-1}\leq t\). Approximate \(Df(x)\) by a full-rank matrix \(A\) from a countable dense family. Put \(T=(A^*A)^{1/2}\). Group points for which
and for which the differentiability remainder, uniformly for \(0<|h|<1/j\), is at most \((\eta/2)\,\sigma_{\min}(T)|h|\). Subdivide each group into Borel pieces of diameter less than \(1/j\). Differentiability and the measurability of the derivative make these a countable Borel cover of \(B\); rational \(h\) suffice to express the uniform condition because \(f\) is continuous. Disjointize the cover.
For \(x,y\) in one piece the two errors give
Since \(|Av|=|Tv|\), the image distance lies between \((1-\eta)|T(y-x)|\) and \((1+\eta)|T(y-x)|\). This proves injectivity and the two Lipschitz bounds. Every singular value of \(Df(x)T^{-1}\) lies between \(1-\eta\) and \(1+\eta\), so its \(n\)-Jacobian is between their \(n\)th powers. Multiplication by \(\det T\) proves the last bounds. ◻
Lemma 15. For linear \(L:\mathbb R^n\to\mathbb R^m\), \(n\geq m\), and Lebesgue measurable \(A\),
Slice measurability and the integral are understood for almost every \(y\), with completed measures.
Proof. If \(L\) has full rank, orthogonal changes of coordinates put it into the form \(L(z,w)=Dz\), with \(D\) invertible, \(z\in\mathbb R^m\) and \(w\in\mathbb R^{n-m}\). Fibres are isometric copies of the \(w\)-space. Fubini followed by \(y=Dz\) gives the formula and almost-everywhere measurability. A null change in \(A\) changes almost no slice. If the rank is below \(m\), the image lies in a Lebesgue-null subspace; all fibres outside it are empty, and both sides are zero. ◻
Lemma 16. The covering inequality needed for coarea is the following: if \(E\subset\mathbb R^N\) is \(\mathcal H^q\)-measurable with finite measure and \(f:\mathbb R^N\to\mathbb R^m\) is \(L\)-Lipschitz, with integers \(q\geq m\geq1\), then
When the right side is finite, the fibres are measurable for almost every \(y\), and their measure has a measurable representative. In particular this applies to \(q=N=n\). No measurability of \(f(E)\) is asserted in this dimension regime.
Covering argument. At scale \(\delta\), cover \(E\) by closed bounded sets \(C_i\) of near-minimal total \(q\)-cost. For each \(y\), their intersections with \(f^{-1}(y)\) cover the slice. Thus its \((q-m)\)-dimensional content is at most
The images are compact, and the isodiametric inequality bounds their Lebesgue measures by \(c_mL^m(\operatorname{diam}C_i)^m\). Use upper integrals until slice measurability is established, let \(\delta\downarrow0\) using Fatou, and let the covering costs decrease to \(\mathcal H^q(E)\). The constant is \(c_{q-m}c_m/c_q=\omega_{q-m}\omega_m/\omega_q\). For null \(E\) this argument gives zero outer measure on almost every fibre, even if \(E\) is not Borel. For finite positive measure choose a Borel hull differing by a null set; the measurable-section theorem for Hausdorff measures supplies measurability for the Borel part. These standard section-measurability details are included in the coarea framework in Simon, Chapter 2, Theorems 1.8–1.9, pp.44–46. The result extends to sigma-finite \(E\) by a disjoint finite-measure decomposition and Tonelli.
The missing image-measurability assertion cannot be restored: a nonmeasurable subset \(V\subset\mathbb R\) gives a Lebesgue measurable null subset \(V\times\{0\}\) of \(\mathbb R^2\), whose coordinate projection is \(V\).
Lemma 17. For a Lipschitz map \(h:\mathbb R^n\to\mathbb R^n\) and \(t>1\), the set where \(Dh\) exists and is invertible has a countable Borel partition \(D_k\). For invertible linear maps \(S_k\), the map \(F_k=S_k^{-1}\circ h|_{D_k}\) and its inverse are \(t\)-Lipschitz; moreover every singular value of \(S_k^{-1}Dh(x)\) is in \([t^{-1},t]\) for \(x\in D_k\).
Proof. Use the construction of Lemma 14 with a countable dense family of invertible square matrices \(S\), but apply the error bound after multiplication by \(S^{-1}\). On sufficiently small pieces,
The same choices of \(\eta\) give the claimed bounds. No symmetry of \(S\) is needed. A dimension-reducing map \(f:\mathbb R^n\to\mathbb R^m\) with \(n>m\) cannot itself have the asserted local inverse; coarea applies this lemma to the square augmented map \(h=(f,P)\) instead. ◻
Remark 18. The two Jacobians use different Gram matrices: \(J_nL=\sqrt{\det(L^*L)}\) for \(n\leq m\), and \(J_mL=\sqrt{\det(LL^*)}\) for \(n\geq m\). A rectangular map itself has no determinant.
2. Densities
Let \((X,d)\) be a metric space, \(\mu\) be an outer measure, and \(A \subset X\) is a subset (\(\mu\)-measurable) with \(x \in X\).
Definition 19. For a Borel regular measure \(\mu\) on a metric space and a measurable set \(A\), define
Here \(n\geq1\) and balls have their specified centres and radii.
Remark 20. If the restriction \(\mu\mathbin{\restriction}A\) is finite on a slightly larger ball around \(x\), then \(z\mapsto\mu(A\cap\overline B_r(z))\) is upper semicontinuous at \(x\). Indeed, for \(z\) near \(x\) it is at most \(\mu(A\cap\overline B_{r+\varepsilon}(x))\); continuity from above as \(\varepsilon\downarrow0\) proves the assertion. Local finiteness supplies this hypothesis in Euclidean space. The function need not be bounded on the whole space.
Theorem 21. Let \(\mu\) be a Borel regular measure, let \(A_2\) be measurable, and \(A_1\subseteq A_2\). If \(\Theta^{*n}(\mu,A_2,x)\geq t\) for every \(x\in A_1\), then
For \(t=0\) the assertion means only nonnegativity.
Lemma 22. A family of closed balls with positive radii bounded above by \(R<\infty\) has a pairwise disjoint subfamily such that every original ball meets a selected ball of radius at least half its radius, and is contained in that selected ball’s fivefold enlargement. In a separable metric space the selected family is countable.
Proof. Group radii into \((R2^{-j-1},R2^{-j}]\), \(j=0,1,\ldots\). At each level choose a maximal disjoint subfamily among balls disjoint from all earlier selections. Maximality follows from Zorn’s lemma, since unions of chains of disjoint families remain disjoint. Every rejected ball meets a selection at its own or an earlier level, whose radius \(r'\) satisfies \(r\leq2r'\). For any \(z\) in the rejected ball, the intersecting-ball triangle inequality gives \(d(z,x')\leq2r+r'\leq5r'\). Separability makes a disjoint family of positive-radius balls countable, by assigning distinct points of a fixed countable dense set in their interiors. In a general metric space the family may be uncountable, but positive measure inside each selected ball and finite total measure also give countability. ◻
Proof of Theorem 21. We may assume \(t>0\) and \(\mu(A_2)<\infty\). Fix \(0<\tau<t\) and \(\delta>0\). Take every closed ball centred at \(x\in A_1\), of radius \(0<r<\delta/10\), for which \(\mu(A_2\cap B_r(x))>\tau\omega_nr^n\). There are arbitrarily small such balls at every \(x\). Lemma 22 gives disjoint selections \(B_i=B_{r_i}(x_i)\), which are countable since each has positive \(\mu\mathbin{\restriction}A_2\) measure. They satisfy
For every \(N\),
To justify this stronger tail assertion, a point outside the first finitely many closed balls has a positive distance from their union. Choose a sufficiently small eligible ball centred there, disjoint from those balls. The selection property forces it to meet a selected ball with index greater than \(N\), whose enlargement contains the point.
The first \(N\) original balls and the enlarged tail therefore cover \(A_1\), all at diameter at most \(\delta\). Hence
The series is finite; let \(N\to\infty\), then \(\delta\downarrow0\), then \(\tau\uparrow t\). This proves the sharp constant one. Merely using the enlarged balls throughout would give the weaker constant \(5^n\). ◻
Theorem 23. If \(\mu\) is Borel regular on a metric space and \(A\) is measurable with \(\mu(A)<\infty\), then \(\Theta^{*n}(\mu,A,x)=0\) for \(\mathcal H^n\)-almost every \(x\notin A\).
Proof. Put \(\nu=\mu\mathbin{\restriction}A\), a finite Borel regular measure. Its metric regularity gives closed \(C_j\subset A\) with \(\mu(A\setminus C_j)\to0\) (in Euclidean space this is the usual compact approximation). Set \(S_t=\{x\notin A:\Theta^{*n}(\mu,A,x)\geq t\}\). For \(x\in S_t\), sufficiently small balls miss \(C_j\), so
Apply Theorem 21 to \(\nu\), \(A_1=S_t\), \(A_2=X\setminus C_j\). This gives \(t\mathcal H^n(S_t)\leq\mu(A\setminus C_j)\). Let \(j\to\infty\), then take the union over \(t=1/k\). For locally finite restrictions, the same conclusion holds locally by a countable exhaustion. ◻
The density comparison and regularity inputs are also stated in Simon, Chapter 1, Theorems 3.3 and 3.6; the selection and tail-cover steps above give the argument used here.
Reading: Carmo (1992), Riemannian geometry.