Formal definition of the pack Intermediate The geometric integration theory of Whitney, sharpened by Federer and Evans-Gariepy, measures the size of irregular sets and the volume of non-smooth images. Its tools are the s -dimensional Hausdorff measure H s , built by the Carathéodory construction from diameters of small covering sets; the resulting Hausdorff dimension; Lipschitz maps, whose almost-everywhere differentiability is Rademacher's theorem; the area formula, which computes the H m -measure of the image of an injective Lipschitz map f : R m → R n (m ≤ n ) by integrating the Jacobian; the coarea formula, its companion for m ≥ n that slices by level sets; and countably m -rectifiable sets, those covered up to H m -null error by countably many Lipschitz images of R m .
This pack collects ten exercises — three easy, four medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units. The exercises group loosely by topic: computing Hausdorff measure and dimension of explicit sets (easy); Lipschitz constants, Rademacher differentiability, and the area formula on graphs (medium); and the coarea formula and the structure of rectifiable sets (hard).
Conventions follow Evans-Gariepy: H s is normalized so that H n = L n (Lebesgue measure) on R n , with constant α ( s ) = π s /2 /Γ ( s /2 + 1 ) ; the Jacobian of a linear map L : R m → R n (m ≤ n ) is J L = det ( L ⊤ L ) .
Key theorem with full solution Intermediate Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).
Lead exercise. State the area formula for a Lipschitz map f : R m → R n with m ≤ n , and use it to derive the classical surface-area formula for the graph of a C 1 function u : Ω ⊂ R m → R .
Solution. The area formula states: for a Lipschitz map f : R m → R n with m ≤ n and a Lebesgue-measurable set A ⊂ R m ,
$$
\int_A \mathbf{J}f(x),d\mathcal{L}^m(x) = \int_{\mathbb{R}^n} #\bigl(A \cap f^{-1}(y)\bigr),d\mathcal{H}^m(y),
$$
where J f ( x ) = det ( D f ( x ) ⊤ D f ( x ) ) is the m -dimensional Jacobian (defined L m -a.e. by Rademacher's theorem), and # counts preimages. When f is injective on A the multiplicity is 1 , so the right side is H m ( f ( A )) — the m -area of the image.
Apply it to the graph map f : Ω → R m + 1 , f ( x ) = ( x , u ( x )) , which is injective with
$$
Df = \begin{pmatrix} I_m \ \nabla u^{\top} \end{pmatrix}, \qquad Df^{\top}Df = I_m + \nabla u,\nabla u^{\top}.
$$
The matrix I m + ∇ u ∇ u ⊤ is a rank-one perturbation of the identity; by the matrix-determinant lemma det ( I m + ∇ u ∇ u ⊤ ) = 1 + ∣∇ u ∣ 2 . Hence J f = 1 + ∣∇ u ∣ 2 , and the area formula gives
$$
\mathcal{H}^m(\mathrm{graph},u) = \int_{\Omega} \sqrt{1 + |\nabla u(x)|^2},d\mathcal{L}^m(x).
$$
This is exactly the elementary surface-area integral, now derived as a special case of the Lipschitz area formula. □
The area formula is the engine of the whole subject: it reduces the geometric measure of a parametrized surface to an integral of a Jacobian over the parameter domain, and it extends verbatim to merely Lipschitz f because Rademacher's theorem supplies D f almost everywhere.
Exercises Intermediate
Exercise 1 (easy, numeric). Hausdorff dimension of the middle-thirds Cantor set.
Compute the Hausdorff dimension of the standard middle-thirds Cantor set C ⊂ [ 0 , 1 ] .
Hint
C is self-similar under two maps each contracting by 1/3 . The similarity dimension s solves 2 ⋅ ( 1/3 ) s = 1 .
Answer
The Cantor set is the attractor of the iterated function system { x ↦ x /3 , x ↦ x /3 + 2/3 } : two similarities, each with ratio r = 1/3 , satisfying the open-set condition. The similarity dimension s solves N r s = 1 with N = 2 pieces:
$$
2\cdot (1/3)^s = 1 \implies 3^s = 2 \implies s = \frac{\log 2}{\log 3} \approx 0.6309.
$$
For an IFS satisfying the open-set condition, the Hausdorff dimension equals the similarity dimension, and 0 < H s ( C ) < ∞ at this exponent. So dim H C = log 2/ log 3 .
Exercise 2 (easy, numeric). Hausdorff measure scaling.
Let A ⊂ R n and let T ( x ) = λ x be a dilation by λ > 0 . Express H s ( T ( A )) in terms of H s ( A ) .
Hint
A cover of A by sets of diameter ≤ δ maps under T to a cover of T ( A ) by sets of diameter ≤ λ δ , and diam T ( U ) = λ diam U .
Answer
For any set U , diam T ( U ) = λ diam U . In the Carathéodory construction, the s -content of a cover { U i } contributes ∑ i α ( s ) ( diam U i /2 ) s ; replacing each U i by T ( U i ) multiplies every diameter by λ , hence multiplies each summand by λ s . The correspondence { U i } ↔ { T ( U i )} is a bijection between δ -covers of A and λ δ -covers of T ( A ) . Passing to the infimum and then δ → 0 ,
$$
\mathcal{H}^s(T(A)) = \lambda^s,\mathcal{H}^s(A).
$$
This is the homogeneity of Hausdorff measure under scaling — the reason H s is the right gauge for self-similar sets and the source of the dimension-defining exponent.
Exercise 3 (easy, proof). Lipschitz maps do not raise Hausdorff dimension.
Let f : R m → R n be Lipschitz with constant L . Show H s ( f ( A )) ≤ L s H s ( A ) for every A ⊂ R m , and conclude dim H f ( A ) ≤ dim H A .
Hint
diam f ( U ) ≤ L diam U . Push a δ -cover of A through f to get an L δ -cover of f ( A ) .
Answer
Lipschitz with constant L means ∣ f ( x ) − f ( y ) ∣ ≤ L ∣ x − y ∣ , so for any U , diam f ( U ) ≤ L diam U . Given a cover { U i } of A with diam U i ≤ δ , the sets { f ( U i )} cover f ( A ) with diam f ( U i ) ≤ L δ . Then
$$
\mathcal{H}^s_{L\delta}(f(A)) \le \sum_i \alpha(s)\Bigl(\tfrac{\mathrm{diam},f(U_i)}{2}\Bigr)^s \le L^s \sum_i \alpha(s)\Bigl(\tfrac{\mathrm{diam},U_i}{2}\Bigr)^s.
$$
Taking the infimum over δ -covers gives H L δ s ( f ( A )) ≤ L s H δ s ( A ) ; letting δ → 0 yields H s ( f ( A )) ≤ L s H s ( A ) .
If s > dim H A then H s ( A ) = 0 , so H s ( f ( A )) = 0 , hence dim H f ( A ) ≤ s . Taking the infimum over such s gives dim H f ( A ) ≤ dim H A . Lipschitz maps cannot increase Hausdorff dimension.
Exercise 4 (medium, proof). Rademacher in one variable.
State Rademacher's theorem and prove the one-dimensional case: a Lipschitz function f : R → R is differentiable L 1 -almost everywhere.
Hint
A Lipschitz function is absolutely continuous; invoke the Lebesgue differentiation theorem for the fundamental theorem of calculus, or use that monotone functions are differentiable a.e. after writing f as a difference of monotone functions.
Answer
Rademacher's theorem. A locally Lipschitz map f : R m → R n is differentiable at L m -almost every point.
One-dimensional proof. Let f : R → R be L -Lipschitz. Then f is absolutely continuous: for disjoint intervals ( a i , b i ) , ∑ i ∣ f ( b i ) − f ( a i ) ∣ ≤ L ∑ i ( b i − a i ) , so the variation is controlled by total length and the absolute-continuity ε -δ condition holds with δ = ε / L . By the fundamental theorem of calculus for absolutely continuous functions, f is differentiable L 1 -a.e., f ′ ∈ L ∞ with ∣ f ′ ∣ ≤ L a.e., and f ( b ) − f ( a ) = ∫ a b f ′ ( t ) d t .
(Equivalently: g ( x ) = Lx + f ( x ) and h ( x ) = Lx − f ( x ) are both nondecreasing because f is L -Lipschitz; monotone functions are differentiable a.e. by Lebesgue's theorem, so f = 2 L 1 ( g − h ) is differentiable a.e.) The higher-dimensional Rademacher theorem reduces partial-derivative existence to this one-variable statement along lines, then upgrades a.e. partials to full differentiability.
Exercise 5 (medium, numeric). Area of a paraboloid cap via the area formula.
Use the area formula to compute the H 2 -measure of the graph of u ( x , y ) = x 2 + y 2 over the disk { x 2 + y 2 ≤ 1 } .
Hint
For a graph, J f = 1 + ∣∇ u ∣ 2 . Here ∣∇ u ∣ 2 = 4 ( x 2 + y 2 ) ; switch to polar coordinates.
Answer
The graph map f ( x , y ) = ( x , y , x 2 + y 2 ) has ∇ u = ( 2 x , 2 y ) , so ∣∇ u ∣ 2 = 4 ( x 2 + y 2 ) and J f = 1 + 4 ( x 2 + y 2 ) . By the area formula,
$$
\mathcal{H}^2(\mathrm{graph},u) = \int_{x^2+y^2\le 1} \sqrt{1 + 4(x^2 + y^2)},dx,dy.
$$
In polar coordinates (r ∈ [ 0 , 1 ] , area element r d r d θ ):
$$
= \int_0^{2\pi}!!\int_0^1 \sqrt{1 + 4r^2},r,dr,d\theta = 2\pi \cdot \frac{1}{12}(1 + 4r^2)^{3/2}\Big|_0^1 = \frac{\pi}{6}\bigl(5^{3/2} - 1\bigr).
$$
Numerically 5 3/2 = 5 5 ≈ 11.1803 , so the area is 6 π ( 11.1803 − 1 ) ≈ 6 π ( 10.1803 ) ≈ 5.330 .
Exercise 6 (medium, proof). Jacobian of a linear map.
For a linear map L : R m → R n with m ≤ n , show the area-formula Jacobian J L = det ( L ⊤ L ) equals the product of the singular values of L , and interpret it geometrically.
Hint
Use the singular value decomposition L = U Σ V ⊤ with Σ carrying singular values σ 1 , … , σ m ≥ 0 . Compute L ⊤ L .
Answer
Write the singular value decomposition L = U Σ V ⊤ , where U ∈ O ( n ) , V ∈ O ( m ) , and Σ ∈ R n × m has the singular values σ 1 , … , σ m ≥ 0 on its diagonal. Then
$$
L^{\top}L = V\Sigma^{\top}U^{\top}U\Sigma V^{\top} = V(\Sigma^{\top}\Sigma)V^{\top} = V,\mathrm{diag}(\sigma_1^2,\ldots,\sigma_m^2),V^{\top}.
$$
So det ( L ⊤ L ) = ∏ i = 1 m σ i 2 and J L = det ( L ⊤ L ) = ∏ i = 1 m σ i .
Geometrically, L maps the unit cube in R m to an m -parallelepiped in R n whose edges have lengths σ 1 , … , σ m along the orthonormal directions V e i ; its m -dimensional volume is the product of those lengths, ∏ σ i = J L . This is why J f ( x ) is the local volume-stretch factor of the area formula: it is the m -volume of the image of the infinitesimal unit cube under D f ( x ) .
Exercise 7 (medium, numeric). Coarea on a linear functional.
Apply the coarea formula to f : R 3 → R , f ( x ) = x 3 , on the unit ball B , to check that integrating the H 2 -measure of the level disks recovers the volume of B .
Hint
The coarea formula reads ∫ B J f d L 3 = ∫ R H 2 ( B ∩ f − 1 ( t )) d t , with J f = ∣∇ f ∣ .
Answer
Here ∇ f = ( 0 , 0 , 1 ) , so the coarea Jacobian is J f = ∣∇ f ∣ = 1 , and the left side is ∫ B 1 d L 3 = L 3 ( B ) = 3 4 π .
The level set f − 1 ( t ) = { x 3 = t } meets B in the disk of radius 1 − t 2 for ∣ t ∣ ≤ 1 , with H 2 -measure π ( 1 − t 2 ) . The right side is
$$
\int_{-1}^{1} \pi(1 - t^2),dt = \pi\Bigl[t - \tfrac{t^3}{3}\Bigr]_{-1}^{1} = \pi\Bigl(\tfrac{2}{3} + \tfrac{2}{3}\Bigr) = \frac{4\pi}{3}.
$$
Both sides equal 3 4 π : the coarea formula reproduces the volume by integrating the areas of horizontal slices. This is the Lipschitz generalization of Fubini's theorem along the level sets of f .
Exercise 8 (hard, proof). Coarea form of the layer-cake / Eilenberg inequality.
Let f : R m → R be Lipschitz with constant L . Prove the Eilenberg inequality
$$
\int_{\mathbb{R}} \mathcal{H}^{m-1}\bigl(A \cap f^{-1}(t)\bigr),dt \le L,\mathcal{L}^m(A)
$$
for measurable A ⊂ R m , as a corollary of the coarea formula.
Hint
The coarea formula gives equality with ∣∇ f ∣ in place of L inside the volume integral. Bound ∣∇ f ∣ ≤ L a.e. by Rademacher.
Answer
The coarea formula for a Lipschitz f : R m → R states, for measurable A ,
$$
\int_A |\nabla f(x)|,d\mathcal{L}^m(x) = \int_{\mathbb{R}} \mathcal{H}^{m-1}\bigl(A \cap f^{-1}(t)\bigr),dt,
$$
where ∇ f exists L m -a.e. by Rademacher's theorem. Since f is L -Lipschitz, ∣∇ f ( x ) ∣ ≤ L at every point of differentiability, hence L m -a.e. Therefore
$$
\int_{\mathbb{R}} \mathcal{H}^{m-1}\bigl(A \cap f^{-1}(t)\bigr),dt = \int_A |\nabla f|,d\mathcal{L}^m \le \int_A L,d\mathcal{L}^m = L,\mathcal{L}^m(A).
$$
This is the Eilenberg inequality: the total ( m − 1 ) -area of the level sets of a Lipschitz function inside A is controlled by the volume of A times the Lipschitz constant. It is the technical workhorse behind many size estimates in geometric measure theory, including the proof that a Lipschitz function maps null sets to "slice-thin" families of level sets.
Exercise 9 (hard, proof). A Lipschitz graph is rectifiable.
Show that the graph Γ = {( x , u ( x )) : x ∈ R m } of a Lipschitz function u : R m → R n is a countably m -rectifiable set, and that H m ( Γ ∩ K ) < ∞ for every compact K .
Hint
Rectifiable means covered up to H m -null error by countably many Lipschitz images of R m . The graph map f ( x ) = ( x , u ( x )) is itself such an image.
Answer
A set E ⊂ R m + n is countably m -rectifiable if there are Lipschitz maps f i : R m → R m + n with H m ( E ∖ ⋃ i f i ( R m ) ) = 0 .
The graph map f ( x ) = ( x , u ( x )) is Lipschitz: ∣ f ( x ) − f ( y ) ∣ 2 = ∣ x − y ∣ 2 + ∣ u ( x ) − u ( y ) ∣ 2 ≤ ( 1 + L 2 ) ∣ x − y ∣ 2 , so f has Lipschitz constant 1 + L 2 where L = Lip ( u ) . Its image is exactly Γ . Thus a single Lipschitz image covers Γ with zero error, so Γ is countably m -rectifiable.
Finiteness: for compact K ⊂ R m + n , Γ ∩ K ⊂ f ( B ) for a bounded set B ⊂ R m (the projection of K ). By the Lipschitz size estimate (Exercise 3), H m ( f ( B )) ≤ ( 1 + L 2 ) m H m ( B ) = ( 1 + L 2 ) m /2 L m ( B ) < ∞ . Hence H m ( Γ ∩ K ) < ∞ . Equivalently, the area formula (lead exercise) gives the exact value ∫ B J f d L m , with J f = det ( I + D u ⊤ D u ) bounded by ( 1 + L 2 ) m /2 .
Exercise 10 (hard, proof). Approximate tangent plane of a rectifiable set.
State what it means for a countably m -rectifiable set E with H m └ E locally finite to have an approximate tangent plane H m -a.e., and explain why this fails for a purely unrectifiable set such as a positive-H 1 -measure subset of the plane that projects to null on every line.
Hint
An approximate tangent plane at x is an m -plane P such that the rescaled measures ( η x , λ ) # ( H m └ E ) converge to H m └ P as λ → 0 , where η x , λ ( y ) = ( y − x ) / λ . Purely unrectifiable sets have no blow-up limit concentrated on a plane.
Answer
For a countably m -rectifiable E with θ = H m └ E locally finite, the rectifiability structure theorem gives: at H m -a.e. x ∈ E there is an m -plane T x E (the approximate tangent plane) such that the blow-ups
$$
(\eta_{x,\lambda})#,\theta ;\xrightarrow[\lambda \to 0]{}; \mathcal{H}^m \llcorner T_x E \quad \text{weakly},
$$
where $\eta {x,\lambda}(y) = (y-x)/\lambda. S in ce Ei s , u pt o n u l l er r or , a co u n t ab l e u ni o n o f L i p sc hi t z g r a p h s , an d e a c h L i p sc hi t z g r a p hha s a t r u e t an g e n tpl an e a . e . ( b y R a d e ma c h e r ′ s d i f f er e n t iabi l i t y o f t h e d e f inin g f u n c t i o n ) , t h e t an g e n tpl an es p a t c h t o g e t h er t o g i v e T_x E$ almost everywhere.
For a purely 1 -unrectifiable set E ⊂ R 2 with 0 < H 1 ( E ) < ∞ — for instance a "four-corner" Cantor dust whose orthogonal projections onto L 1 -almost every line are null (Besicovitch) — no such tangent line exists at H 1 -a.e. point: the blow-up measures do not concentrate on any line, because the set looks the same (totally disconnected, projecting to null) at every scale. The Besicovitch-Federer projection theorem makes this precise: E is purely unrectifiable iff its projection to H 1 -almost every line is L 1 -null, the exact opposite of the rectifiable case, where projections have positive measure in most directions. Thus the existence of approximate tangent planes a.e. characterizes rectifiability.
Exercise pack — Whitney Geometric Integration Theory / Federer Ch. 2-3 supplement: Hausdorff measure and dimension, Lipschitz functions and Rademacher's theorem, the area and coarea formulas, and rectifiable sets. Ten exercises, three easy / four medium / three hard.