NOTES FOR THE COURSE ON GEOMETRIC MEASURE THEORY

FIRST DRAFT

RAUL SERAPIONI

Contents

1. Measures, measurable functions, integration 3

2. Hausdorff Measures. Area and Coarea Formulas 16

Hausdorff Measures 16

Lipschitz functions 23

The Area and Coarea Formulas 27

3. Rectifiability 31

4. Appendix 33

References 38

Date: March 19, 2015.

1

I turn away in fright and horror from this lamentable plague of functions that do not have deriva-tives.C. Hermite, 1893

2

1. Measures, measurable functions, integration

The Lebesgue measure and the Lebesgue integral.

We recall, without proofs, the main steps in the construction of the Lebesgue1 measure and of theLebesgue integral. This sections is aimed to show the similarities of this classical construction withthe subsequent more general approach.

Denote by Q a closed cube in Rd with sides parallel to the coordinate axes. The volume of a closed

cube Q with sides of length ℓ ≥ 0 is defined as

|Q| = ℓd.

We define the Lebesgue outer measure m∗ of any set E ⊂ Rd as

m∗(E) := inf{∞∑

i=1

|Qi| : E ⊂⋃

i

Qi}.

The Lebesgue outer measure m∗ enjoys the following properties

Proposition 1.1. (1) m∗(Q) = |Q| for any closed and also open cube.(2) E1 ⊂ E2 =⇒ m∗(E1) ≤ m∗(E2).

(3) m∗(

∞⋃

i=1

Ei) ≤∞∑

i=1

m∗(Ei).

(4) m∗(E) = inf{m∗(O) : O open and O ⊃ E}.(5) dist (E1, E2) > 0 =⇒ m∗(E1 ∪E2) = m∗(E1) +m∗(E2)(6) if the Qi are almost disjoint cubes (i.e. cubes that may have in common only parts of the

boundary) m∗ is additive

m∗(

∞⋃

i=1

Qi) =

∞∑

i=1

m∗(Qi)

(7) also if Oi are disjoint open sets m∗ is additive

m∗(

∞⋃

i=1

Oi) =

∞∑

i=1

m∗(Oi)

In general it is false that m∗ is additive on generic disjoint sets. To get a general family of setss.t. the measure is additive when adding disjoint sets of this family we single out the Lebesguemeasurable sets defined as

Definition 1.2. (1) E ⊂ Rd is Lebesgue measurable (measurable for short) if for all ε > 0

there is an open O = Oε ⊃ E such that

m∗(O \E) ≤ ε.

(2) If E ⊂ Rd is measurable we define the Lebesgue measure of m(E) as

m(E) := m∗(E).

Lebesgue measurable sets enjoy the following properties

Proposition 1.3. (1) Open sets are measurable(2) m∗(E) = 0 =⇒ E is measurable.

1Henri Leon Lebesgue: Beauvais, Oise, France, June 28, 1875 – Paris, France, July 26, 1941

3

(3) Countable unions of measurable set are measurable, countable intersections of measurablesets are measurable, the complementary of a measurable set is measurable. In short: thefamily M of measurable sets is σ-algebra.

Observe that (1) and (3) yield that the σ-algebra generated by the opens sets, the so called theBorel σ-algebra, is contained in M. The measure m is additive over disjoint sets in M:

Theorem 1.4. (1) If Ei are disjoint and measurable then

m(

∞⋃

i=1

Ei) =

∞∑

i=1

m(Ei).

(2) If Ei are measurable and Ei ⊂ Ei+1 then

m(Ei) → m(

∞⋃

i=1

Ei) as i → ∞.

(3) If Ei are measurable, Ei ⊃ Ei+1 and m(E1) < ∞ then

m(Ei) → m(∞⋂

i=1

Ei) as i → ∞.

Lebesgue measurable sets can be well approximated from outside and inside by open and closedsets.

Theorem 1.5. Let E be measurable in Rd then for every ε > 0

(1) there is an open O ⊃ E s.t. m(O \ E) < ε.(2) there is a closed C ⊂ E s.t. m(E \ C) < ε.(3) if m(E) < ∞ there is a compact K ⊂ E s.t. m(E \K) < ε.

Corollary 1.6. (1) A set E ⊂ Rd is measurable if and only if it differs from a Gδ set or from

an Fσ set by a set of measure zero.(2) The σ-algebra M of measurable sets can be obtained by the σ-algebra of Borel sets adjoining

the sets of measure zero.

There exist non-measurable sets.

The Lebesgue measure is rotation and translation invariant and it is d-homogeneous with respectto homoteties of Rd.

Definition 1.7. A function f : Rd → R is measurable if

f−1((−∞, a])is measurable for all a ∈ R.

Proposition 1.8. (1) If f is measurable then f−1(O) or f−1(C) are measurable for any opensset O or any closed set C.

(2) Continuous functions are measurable. Moreover

f measurable and Φ continuous =⇒ Φ ◦ f is measurable.

(3) If fn are measurable functions then

supn

fn, infn

fn, lim infn

fn, lim supn

fnare measurable

and in particularf(x) := lim

n→∞fn(x) is measurable.

4

Measurable functions can be approximated by simple functions. A simple function φ is such that

φ(x) :=

∞∑

j=1

ajχEj(x)

where aj ∈ R and Ej is measurable with finite measure.

Theorem 1.9. Let f be a measurable function, then

(1) if f ≥ 0 then there is an increasing sequence of simple functions φk s.t.

f(x) = limk→∞

φk(x) for all x ∈ Rd.

(2) if f is real valued there is a sequence of simple functions φk s.t.

|φk(x)| ≤ |φk+1(x)| and f(x) = limk→∞

φk(x).

Measurable functions are ”almost” continuous and a sequence of measurable functions pointwiseconvergent is ”almost” uniformly convergent. This is the content of the following theorems, thatwill be proved later in a slightly more general context.

Theorem 1.10. [Lusin’s Theorem] Let A ⊂ Rd be measurable with m(A) < ∞. Let f : A ⊂

Rd → R be measurable. Then for any ε > 0 it is possible to find a compact set Kε ⊂ A such that

m(A \Kε) ≤ ε and the restriction f|Kε(of f to Kε), f|Kε

: Kε → R is continuous.

Theorem 1.11. [Egorov’s Theorem] Let A ⊂ Rd be measurable with m(A) < ∞. Let f1, f2, · · · :

A → R be measurable and limk→+∞ fk(x) = g(x) a.e. x ∈ A. Then, for any ε > 0, it is possible tofind a measurable set Bε ⊂ A such that m(A \Bε) ≤ ε and fk → g uniformly on Bε.

Some of these properties are sometimes summarized in the so called three Littlewood’s2 principles:

(1) Measurable sets are almost finite union of cubes (in another version: measurable sets arealmost open sets):

(2) Measurable functions are almost continuous functions.(3) Pointwise convergent sequences of measurable functions are almost uniform convergent se-

quences.

In all the instances the word ”almost” has the meaning of ”. . . but for a set of measure < ε. . . ”.Clearly (1) is related with Theorem 1.5, while (2) and (3) are related with Theorem 1.10 andTheorem 1.11.

Definition of the Lebesgue integral. The general notion is reached defining the integral before insmaller sets and then increasing the generality.

Step 1: Integral of measurable simple functions. If φ(x) :=∑∞

j=1 ajχEj(x) where each Ej is

measurable with finite measure, we define∫

Rd

φ(x) dx :=∞∑

j=1

ajm(Ej).

Notice that this is a good definition, i.e. it depends only on φ and not on the infinitedifferent ways of writing it as a linear combination of characteristic functions.

2John Edensor Littlewood: Rochester, England June 9, 1885 – Cambridge, England September 6, 1977

5

Step 2: Integral of measurable bounded functions supported on a measurable sets E of finitemeasure.By Theorem 1.9 there is a sequence of simple functions φk supported on E s.t. φk(x) → f(x)as k → ∞ for a.e.x ∈ E and s.t.

(1) limk→∞

∫

Rd

φk(x) dx exists

(2) if f(x) = 0 a.e. then limk→∞

∫

Rd

φk(x) dx = 0

and we define∫

Rd

f(x) dx = limk→∞

∫

Rd

φk(x) dx.

Notice that (2) yields that this is a good definition, i.e. the definition does not depend onthe approximating sequence φk.

Step 3: Integral of non negative functions. If f(x) ≥ 0 and measurable we define∫

Rd

f(x) dx = supg{∫

Rd

f(x) dx : 0 ≤ g(x) ≤ f(x)}

and the functions g are bounded, measurable and supported on sets of finite measure.The supremum cab finite or infinite. If the supremum is finite we say that f is Lebesgueintegrable.

Step 4: The general case: f is a measurable real valued function on Rd. Then we write

f = f+ − f−. If both f+ and f− are Lebesgue integrable we define∫

Rd

f(x) dx =

∫

Rd

f+(x) dx−∫

Rd

f−(x) dx.

Outer measures, measures and the σ-algebra of measurable sets.

Definition 1.12. Let X be a set. An outer measure µ on X is a function µ : P(X) → [0,∞] s.t.

(1) µ(∅) = 0(2) µ(A) ≤∑∞

k=1 µ(Ak) if A ⊂ ⋃k Ak.

Observe that from (2) it follows

µ(A) ≤ µ(B), whenever A ⊂ B.

Example 1.13. Let X be any set.

(1) The Dirac3 measure δx0Given a fixed point x0 ∈ X, the Dirac mass δx0

is the measure s.t.δx0

(E) = 1 if x0 ∈ E and δx0(E) = 0 otherwise.

(2) The Counting measure ♯ For any E ⊂ X, ♯(E) is the cardinality of E if it is finite and♯(E) = +∞ otherwise.

(3) Let X = Rd. Define the measure of a box B :=

∏di=1[ai, bi) as |B| := ∏d

i=1(bi − ai). Theouter d-dimensional Lebesgue measure m∗(E) of any E ⊂ R

n is defined as

m∗(E) := inf{∞∑

k=1

|Bk| : E ⊂∞⋃

k=1

Bk}.

3Paul Adrien Dirac: Bristol, England, August 8, 1902 – Tallahassee, Florida, October 20, 1984

6

(4) Let X = Rd. The outer d-dimensional Jordan4 measure J∗(E) of any E ⊂ R

n is defined as

(J)∗(E) := inf{N∑

k=1

|Bk| : E ⊂N⋃

k=1

Bk}.

(5) The restricted measure µ E. Given any fixed subset E ⊂ X, define (µ E)(A) := µ(A∩E).

Exercise 1.14. Prove that there is E ⊂ R such that m∗(E) < J∗(E).

Definition 1.15. (Caratheodory5 measurability) A ⊂ X is µ-measurable if

µ(B) = µ(A ∩B) + µ(Ac ∩B), for all B ⊂ X.

We denote as M the family of µ-measurable subsets of X.

Remark 1.16. (1) at least X and the empty set are always µ-measurable. Moreover it is possiblethat ∅ and X are the only measurable sets. As an example, let X = R and define µ(∅) = 0,µ(R) = 2 and µ(E) = 1 for any E ⊂ R and different from ∅ and R.

(2) To prove that A ⊂ X is µ-measurable it is enough to prove that

µ(B) ≥ µ(A ∩B) + µ(Ac ∩B), for all B ⊂ X

because the opposite inequality is always true.(3) If A is µ-measurable and if E is any subset of X then A is (µ E)-measurable.

Exercise 1.17. Prove that if µ(E) = 0 then E is µ-measurable.

Theorem 1.18. Let µ be a measure on X. Let M be the family of all µ-measurable subsets of X.Then

(1) M is a σ-algebra, i.e.(a) ∅ ∈ M;(b) if A ⊂ M then Ac ∈ M;(c) if A1, A2, · · · ∈ M then

⋃

iAi ∈ M.(2) additivity of the measure : if A1, A2, · · · ∈ M are pairwise disjointed, then

µ

(

⋃

i

Ai

)

=∞∑

i=1

µ(Ai).

(3) if A1, A2, · · · ∈ M and A1 ⊂ A2 ⊂ . . . , then µ

(

⋃

i

Ai

)

= limi→∞

µ(Ai).

(4) if A1, A2, · · · ∈ M, A1 ⊃ A2 ⊃ . . . and µ(A1) < ∞ then

µ

(

⋂

i

Ai

)

= limi→∞

µ(Ai).

Proof. Assume that A1, A2, . . . are µ-measurable subsets of X.

4Camille Jordan: Lyon, France, January 5, 1838 – Paris, France January 22, 19225Constantin Caratheodory: Berlin, German Empire, September 13, 1873 – Munich, West Germany, February 2,

1950

7

(i) Finite union and finite intersection of measurable sets are measurable.For any set B

µ(B) = µ(B ∩A1) + µ(B ∩Ac1)

= µ(B ∩A1) + µ((B ∩Ac1) ∩A2) + µ((B ∩Ac

1) ∩Ac2)

= µ(B ∩A1) + µ((B ∩Ac1) ∩A2) + µ((B ∩ (A1 ∪A2)

c)

observing that (B ∩A1) ∪ ((B ∩Ac1) ∩A2) = B ∩ (A1 ∪A2)

≥ µ(B ∩ (A1 ∪A2)) + µ(B ∩ (A1 ∪A2)c).

Hence A1∪A2 is µ-measurable and, consequently, any finite union of measurable sets is measurable.Because A1∩A2 = (Ac

1∪Ac2)

c and because A is measurable if and only if Ac is measurable, it followsthat the intersection of a finite number of measurable sets is measurable.

(ii) Proof of (2).Assume that the sets Ai are pairwise disjointed. Then

µ(

k⋃

i=1

Ai) = µ(

k⋃

i=1

Ai ∩Ak) + µ(

k⋃

i=1

Ai ∩Ack)

= µ(Ak) + µ(k−1⋃

i=1

Ai)

= µ(Ak) + µ(Ak−1) + · · ·+ µ(A1).

Thenk∑

i=1

µ(Ai) = µ(

k⋃

i=1

Ai) ≤ µ(

∞⋃

i=1

Ai),

hence∞∑

i=1

µ(Ai) ≤ µ(

∞⋃

i=1

Ai).

This concludes the proof because the opposite inequality is always true.

(iii) Proof of (3).Using that A1 ⊂ A2 ⊂ A3 ⊂ . . . , write each Ak as union of pairwise disjoint measurable sets

Ak = (Ak \ Ak−1) ∪ (Ak−1 \Ak−2) ∪ · · · ∪A1.

Hence from (2)

µ(Ak) = µ(A1) +

k∑

i=2

µ(Ai \Ai−1)

and, once more from (2),

limk→+∞

µ(Ak) = µ(A1) +

+∞∑

i=2

µ(Ai \ Ai−1) = µ(

+∞⋃

i=1

Ai).

(iv) Proof of (4).Observe that, without any assumption on the sets Ai,

∞⋂

i=1

Ai ⊂ Ak =⇒ µ(

∞⋂

i=1

Ai) ≤ µ(Ak), for all k ≥ 1,

8

hence

µ(∞⋂

i=1

Ai) ≤ limk→+∞

µ(Ak).

So that we have to prove the opposite inequality.Using that A1 ⊃ A2 ⊃ A3 ⊃ . . . , we have (A1 \ A2) ⊂ (A1 \ A3) ⊂ . . . and from (3) it follows

limk→+∞

µ(A1 \ Ak) = µ(

∞⋃

k=1

(A1 \ Ak)) ≥ µ(A1)− µ(

∞⋂

k=1

Ak),

because A1 = (A1 \(∩∞i=1Ai))∪ (∩∞

i=1Ai) = ∩∞i=1(A1 \Ai)∪ (∩∞

i=1Ai). Hence, using that µ(A1) < ∞,

µ(

∞⋂

k=1

Ak) ≥ µ(A1)− limk→+∞

µ(A1 \ Ak).

On the other side µ(Ak) = µ(A1)− µ(A1 \Ak), that yields

limk→+∞

µ(Ak) = µ(A1)− limk→+∞

µ(A1 \ Ak) ≤ µ(∞⋂

k=1

Ak)

and the proof is completed.

(v) M is a σ-algebra.From step (i) we know that any finite union of µ-measurable sets is µ-measurable. Hence, for allB with µ(B) < ∞,

µ(B) = µ(B ∩ (

k⋃

i=1

Ai)) + µ(B ∩ (

k⋃

i=1

Ai)c) = (µ B)(

k⋃

i=1

Ai) + (µ B)((

k⋃

i=1

Ai)c).

Moreover we know that µ-measurable sets are also µ B-measurable. Hence applying (3) and (4)

to the measure µ B and to the increasing and decreasing sequences of sets⋃k

i=1 Ai and (⋃k

i=1 Ai)c

we obtain

µ(B) = limk→+∞

(

(µ B)(

k⋃

i=1

Ai)) + (µ B)(

k⋃

i=1

Ai)c)

)

= limk→+∞

(µ B)(

k⋃

i=1

Ai)) + limk→+∞

(µ B)(

k⋃

i=1

Ai)c)

= (µ B)(

∞⋃

i=1

Ai)) + (µ B)(

∞⋃

i=1

Ai)c)

= µ(B ∩ (∞⋃

i=1

Ai)) + µ(B ∩ (∞⋃

i=1

Ai)c),

proving that⋃∞

i=1 Ai is µ-measurable. Finally, since⋂∞

i=1 Ai = (⋃∞

i=1Aci )

c we have also the µ-measurability of countable intersections of measurable sets. �

Definition 1.19. A measure µ on a set X is a regular measure if for all E ⊂ X there is a µ-measurable A ∈ M such that

E ⊂ A and µ(E) = µ(A).

Observe that there is no difference if in Definition 1.19 we assume that there is A ∈ M s.t. A ⊂ Eand µ(A) = µ(E). Another choice is: there is A ∈ M s.t. µ(A∆E) = 0.

9

Theorem 1.20. Le µ be a regular measure on X. If E1 ⊂ E2 ⊂ E3 ⊂ · · · ⊆ X, then

limi→+∞

µ(Ei) = µ(

∞⋃

i=1

Ei).

Observe that here, we do not assume that the Ei are µ-measurable sets.

Proof. First observe that Ek ⊂ ⋃∞i=1 Ei, for all k, hence

limk→+∞

µ(Ek) ≤ µ(

∞⋃

i=1

Ei),

and have to prove the inverse inequality.Since µ is regular, for each Ek there is Ak ∈ M s.t. Ek ⊂ Ak and µ(Ak) = µ(Ek). DefineBk :=

⋂

j≥k Aj . Then, using that E1 ⊂ E2 ⊂ . . . , it follows that Ek ⊂ Bk. Moreover, Bk isµ-measurable and

µ(Ek) = µ(Bk).

Indeed: Bk ⊂ Ak implies µ(Bk) ≤ µ(Ak). On the other side: Ek ⊂ Bk hence µ(Ak) = µ(Ek) ≤µ(Bk).Notice that, to avoid trivialities, we can assume that µ(E1) < ∞. Finally, using that Bk is adecreasing sequence, from (4) of Theor 1.18, we have

limk→+∞

µ(Ek) = limk→+∞

µ(Bk) = µ(∞⋂

i=1

Bi) ≥ µ(∞⋂

i=1

Ei).

�

Relations between measures and topology.

Definition 1.21. Le X be a topological space and µ a measure on X.

(1) The Borel6 sets in X are the elements of the smallest σ-algebra of subsets of X containingthe open sets.

(2) µ is a Borel measure if the Borel sets are µ-measurable.(3) µ is a Borel regular measure if it is a Borel measure and for every A ⊂ X there is a Borel

set B ⊃ A s.t. µ(A) = µ(B) .(4) µ is a Radon7 measure if it is a Borel regular measure and µ(K) < ∞ for every compact

K ⊂ X.

Observe that, on Rn, the restriction of a Borel regular measure to any measurable set of finite

measure produces a Radon measure.

Theorem 1.22. Let µ be a Borel regular measure on a topological space X and let A be a µ-measurable set with µ(A) < ∞. Then µ A is a Radon measure.

Proof. See [3] �

Theorem 1.23. Let µ be a Radon measure on a topological space X, then

6Emile Borel: Saint-Affrique, France, January 7, 1871 – Paris, France, February 3, 19567Johann Karl August Radon: Decin, Bohemia, Austria-Hungary, December 16, 1887 – Vienna, Austria, May 25,

1956

10

(1) for all A ⊂ Xµ(A) = inf{µ(V ) : A ⊂ V, V open}.

(2) for all µ-measurable subsets A ⊂ X

µ(A) = sup{µ(K) : K ⊂ A,Kcompact}.

Proof. See [3] �

Theorem 1.24. [Caratheodory criterion] Let (X, d) be a metric space and µ a measure on X.Then µ is a Borel measure if and only if

µ(A ∪B) = µ(A) + µ(B) whenever dist (A;B) > 0.

Proof. See [3] �

Measurable functions.

Definition 1.25. Let X be a set and µ a measure on X. Let Y be a topological space. A functionf : X → Y is said to be µ-measurable if f−1(U) is µ-measurable for each open U ⊂ Y .

In what follows Y will be almost always [0,∞], R, [−∞,∞], Rn or C.

Theorem 1.26. If f, g : X → R(C) are µ-measurable then

(1) λf , f + g, f · g, |f |, max(f, g), min(f, g) are all µ-measurable. The quotient f/g is µ-measurable provided it is defined as 0 when both f(x) = 0 and g(x) = 0.

(2) If the functions f1, f2, · · · : X → [−∞,∞] are µ-measurable then also infk fk, supk fk,lim infk fk, lim supk fk are µ-measurable.

Proof. See [3] �

Theorem 1.27. Let f : X → [0,∞] be µ-measurable. Then, for k = 1, 2, . . . , the sets Ek ⊂ Xdefined as

E1 := {x : f(x) ≥ 1} and, for k > 1, Ek := {x : f(x) ≥ 1

k+

k−1∑

j=1

1

jχEj

(x)}

are µ-measurable and

f(x) =∞∑

k=1

1

kχEk

(x).

Proof. See [3] �

Theorem 1.28. [Tietze’s8 Extension Theorem] Let f : Γ(⊂ Rn) → R

m be a continuous function

on a compact set Γ. Then there is f : Rn → Rm continuous and such that f|Γ = f .

Proof. See e.g. [13] page 246 or [3] page 15. �

8Heinrich Franz Tietze: Schleinz, Austria-Hungary, August 31, 1880 – Munich, West Germany, February 17, 1964

11

Theorem 1.29. [Lusin’s9 Theorem] Let µ be a Borel regular measure on Rn, let A ⊂ R

n be µ-measurable with µ(A) < ∞. Let f : A ⊂ R

n → Rm be µ-measurable. Then for any ε > 0 it is

possible to find a compact set Kε ⊂ A such that

(1) µ(A \Kε) ≤ ε and(2) the restriction f|Kε

(of f to Kε), f|Kε: Kε → R

m is continuous.

Notice that the Theorem does not say that f is continuous on Kε.

Proof. The idea is to construct a sequence of continuous functions converging to f uniformly on acompact set Kε.Fix i ∈ N and divide the codomain R

m in disjoint Borel sets {Bi,j}j∈N of diameter less than 1i

(choosing the Bi,j as half open boxes as in Example 1.13 is a good choice). Then let Ai,j :=A ∩ f−1 (Bi,j).Observe that, by the assumption µ(A) < ∞ and by Theor 1.22, (µ A) is a Radon measure. ByTheor 1.23(2), for any ε > 0 and any Ai,j there is a compact set Kε,i,j = Ki,j s.t.

Ki,j ⊂ Ai,j (µ A)(Ai,j \Ki,j) <ε

2i+j.

Then

µ(A \∞⋃

j=1

Ki,j) = (µ A)(A \∞⋃

j=1

Ki,j)

= (µ A)(

∞⋃

j=1

Ai,j \∞⋃

j=1

Ki,j)

= (µ A)(

∞⋃

j=1

(Ai,j \Ki,j)) <ε

2i.

Because N 7→ A \ ∪Nj=1Ki,j is a decreasing sequence of measurable sets, by Theor 1.18(4),

limN→+∞

µ(A \N⋃

j=1

Ki,j) = µ(A \∞⋃

j=1

Ki,j) <ε

2i

hence there is N(i) s.t.

µ(A \N(i)⋃

j=1

Ki,j) <ε

2i.

Observe that Ci :=⋃N(i)

j=1 Ki,j is a compact set. Now choose yi,j ∈ Bi,j and define gi : Ci → Rm as

gi(x) := yi,j for all x ∈ Ki,j , 1 ≤ i ≤ N(i).

Notice that gi is a continuous step function because the sets Ki,j (being compact and disjoint) havepositive distance from each other. Moreover

|f(x)− gi(x)|Rm <1

ifor all x ∈ Ci.

Define

Kε :=

∞⋂

i=1

Ci.

9Nikolai Nikolaevich Luzin: Irkutsk, Russian Empire, 9 December 1883 – Moscow, 28 January 1950

12

Then Kε is compact and A \Kε ⊂⋃∞

i=1(A \ Ci) hence

µ(A \Kε) ≤∞∑

i=1

µ(A \ Ci) < ε.

Since |f(x)− gi(x)|Rm < 1i for all x ∈ Kε ⊂ Ci we have that gi → f as i → +∞, uniformly on Kε

and consequently F|Kεis continuous on Kε. �

Theorem 1.30. [Egorov’s10 Theorem] Let µ be a Borel regular measure on Rn, let A ⊂ R

n be µ-measurable with µ(A) < ∞. Let f1, f2, · · · : Rn → R

m be µ-measurable and limk→+∞ fk(x) = g(x)for µ-a.e. x ∈ A. Then for any ε > 0 it is possible to find a µ-measurable set Bε ⊂ A such that

(1) µ(A \Bε) ≤ ε and(2) fk → g uniformly on Bε.

Proof. For i, j ∈ N define

Ci,j :=∞⋃

k=j

{

x : |fk(x)− g(x)|Rm ≥ ε/2i}

Because fk and g are measurable also |fk(x) − g(x)|Rm is measurable, hence each Ci,j is µ-measurable, moreover j 7→ Ci,j is decreasing for each i fixed.By Theor.1.18 (4), using the assumption µ(A) < ∞, we have

limj→∞

µ(A ∩Ci,j) = µ(∞⋂

j=1

(A ∩Ci,j))

= µ(A ∩∞⋂

j=1

Ci,j) = 0,

indeed, by assumption, |fk(x)− g(x)|Rm → 0 as k → +∞ for µ almost every x ∈ A. For any suchx and for each i, there is k s.t. |fk(x)− g(x)|Rm < 1/2i, hence x /∈ Ci,k.

Fix ε > 0. Since limj→∞ µ(A ∩Ci,j) = 0 there is N(i) = N(i, ε) ∈ N s.t.

µ(A ∩ Ci,N(i)) < ε/2i.

Let

Bε := A \ (∞⋃

i=1

Ci,N(i))

then

A \Bε =∞⋃

i=1

(A ∩ Ci,N(i))

and

µ(A \Bε) ≤∞∑

i=1

µ(A ∩ Ci,N(i)) < ε.

For each i, for all x ∈ Bε and for all k ≥ N(i) we have

|fk(x)− g(x)|Rm < 1/2i

that is fk converges to g uniformly on Bε. �

10Dmitri Fyodorovich Egorov: Moscow, Russian Empire, December 22, 1869 – Kazan, Soviet Union, September10, 1931

13

Covering Theorems.

Definition 1.31. Let F be a collection of sets Eα ⊂ Rd (usually balls, cubes, etc).

(1) A collection F is a cover of A ⊂ Rd if

A ⊂⋃

α

Eα.

(2) The collection F is fine at x ∈ Rd if

inf{diam(Eα) : x ∈ Eα} = 0.

(3) A collection F is a fine cover of A ⊂ Rd if it is a cover of A and if it is fine at each x ∈ A.

Theorem 1.32. [Vitali’s11 covering Theorem] Let F be a collection of closed balls Bα in Rd. We

assume only that sup{diam(Bα) : Bα ∈ F} < ∞. Then there is a sub collection G ⊂ F s.t.

(1) the balls Bα ∈ G are disjointed.

(2)⋃

Bα∈F

Bα ⊂⋃

Bα∈G

5Bα.

Proof. See [10] or [3]. �

Corollary 1.33. Let F be a collection of closed balls Bα in Rd. Let sup{diam(Bα) : Bα ∈ F} < ∞.

Let F be a fine cover of an open set O ⊂ Rd and let δ > 0. Then there is a countable disjoint

collection G ⊂ F of balls Bα contained in O with diam(Bα) < δ and s.t.

md(O \⋃

α∈G

Bα) = 0.

Proof. See [10] or [3]. �

Theorem 1.34. [Besicovitch12 covering Theorem] Let A ⊂ Rd. Let F be a collection of closed balls

Bα such that

• for each x ∈ A there is a Bα ∈ F with center in x;• sup{diam(Bα) : Bα ∈ F} < ∞.

Then there is an integer constant N = N(d) > 0, depending only on the dimension d of the space,and there are N subcollections G1, . . . ,GN ⊂ F such that

(1) each Gi, i = 1, . . . , N is a disjoint collection;

(2) A ⊂N⋃

i=1

⋃

Bα∈Gi

Bα.

Proof. See [3]. �

11Giuseppe Vitali: Ravenna, Italy, August 26, 1875 – Bologna, Italy, February 29, 193212Abram Samoilovitch Besicovitch: Berdyansk, Russian Empire, January 23, 1891 – Cambridge, UK, November

2, 1970

14

Corollary 1.35. Let µ be a Borel regular measure in Rd. Let A ⊂ R

d with µ(A) < ∞. Let F be acollection of closed balls B(x, r) such that for all x ∈ A there is r > 0 and B(x, r) ∈ F , moreover

for each x ∈ A inf{diam(B(x, r)) : B(x, r) ∈ F} = 0.

Then for all δ > 0 and for any open O ⊂ Rd there is a countable disjoint collection G ⊂ F of balls

Bα contained in O with diam(Bα) < δ and s.t.

µ(

(A ∩O) \⋃

Bα∈G

Bα

)

= 0.

Proof. See [3]. �

Riesz13 Representation Theorem.

Theorem 1.36. Let X be a locally compact Hausdorff space. Let L : Cc(X;Rk) → R be a linearfunctional such that

sup{L(f) : f ∈ Cc(Rd;Rk), sup

x∈Rd

|f(x)|Rk ≤ 1, spt(f) ⊂ K} < ∞

for any compact K ⊂ Rd. Then there are a Radon measure µ, called the variation measure associ-

ated with L, and a µ measurable function σ : X → Rk such that

(1) |σ(x)|Rk = 1 for µ a.e. x ∈ X;(2) for all f ∈ Cc(X;Rk),

L(f) =

∫

Xf(x) · σ(x) dµ(x).

Proof. See [3]. �

13Frigyes Riesz: Gyor, Austria-Hungary, January 22, 1880 – Budapest, Hungary, February 28, 1956

15

2. Hausdorff Measures. Area and Coarea Formulas

Hausdorff Measures. For 0 ≤ s < ∞ let α(s) :=πs/2

Γ( s2 + 1)where as usual Γ(t) :=

∫∞0 e−xxt−1 dx.

Observe that, if s is equal to a positive integer d then α(d) is the d dimensional Lebesgue measureof a unit ball in R

d.

Definition 2.1. [Hausdorff14 s-dimensional measures] Let 0 ≤ s < ∞ and let ζs : P(Rd) → R+ be

the evaluation function defined as ζs(∅) = 0 for 0 ≤ s and

ζs(E) :=

{

α(s)(

diam(E)2

)sif s > 0

ζ0(E) = 1 if s = 0for all E 6= ∅.

Then,

(1) for δ > 0 and for all A ⊂ Rd

Hsδ(A) := inf{

∞∑

j=1

ζs(Ej) : A ⊂∞⋃

j=1

Ej , diam(Ej) ≤ δ},

Hs∞(A) := inf{

∞∑

j=1

ζs(Ej) : A ⊂∞⋃

j=1

Ej}.

(2) for all A ⊂ Rd

Hs(A) := limδ→0+

Hsδ (A) = sup

δ>0Hs

δ (A).

Remark 2.2. It follows immediately from the definition that for all A ⊂ Rd

Hs(λA) = λsHs(A), for all λ > 0

Hs(L(A)) = λsHs(A), for all affine isometries L : Rd → Rd.

Remark 2.3. [Caratheodory’s construction] Observe that the preceding definition makes sense inany metric space. Moreover it can be seen as a particular instance of the following more generalCaratheodory’s construction.

On a metric space (X, d) let be given a collection F of subsets of X and an evaluation functionζ : F → R

+ with the following assumptions

(1) for all A ⊂ X and ∀δ > 0 there are F1, F2, · · · ∈ F s.t.

diam(Fj) ≤ δ, and A ⊂∞⋃

j=1

Fj ;

(2) ∀δ > 0 there is F ∈ F s.t. diamF ≤ δ and ζ(F ) ≤ δ.

Then for all A ⊂ X we define

Ψδ(A) := inf{∞∑

j=1

ζ(Fj) : Fj ∈ F , A ⊂∞⋃

j=1

Fj , diam(Fj) ≤ δ}.

andΨ(A) := lim

δ→0Ψδ(A) = sup

δ>0Ψδ(A).

14Felix Hausdorff: Breslau, Kingdom of Prussia, November 8, 1868 – Bonn, Germany, January 26, 1942

16

Example 1: in Rd it is possible to choose as F a smaller collection than ”all the subsets” of Rd.

Possible choices are: convex sets, spheres, cubes, etc. In particular when F is the collection of allballs in R

d the measures obtained are denoted as spherical Hausdorff measures.Example 2: the evaluation function ζ can be different from the ζs considered in Definition 2.1. Asan example, studying the trajectories of Brownian motions the function

ζ(t) := t2 log log t−1 was used in R3

ζ(t) := t2 log t−1 log log log t−1 was used in R2.

Theorem 2.4. For all s ≥ 0, Hs is a Borel regular measure in Rd

Proof. Hint of proof. For details see [3].Step 1 Prove that Hs

δ is subadditive, to get that Hs is a measure.Step 2 To prove that Hs is a Borel measure use Caratheodory’s criterion stated in Theorem 1.24.Step 3 To prove thatHs is Borel regular observe first that, for any set E ⊂ R

d, diam(E) = diam(E).Hence in the definition of Hs

δ we can as well use closed coverings, that is, for any A ⊂ Rd,

Hsδ(A) := inf{

∞∑

j=1

ζs(Ej) : Ej closed sets, A ⊂∞⋃

j=1

Ej , diam(Ej) ≤ δ}.

Let A be s.t. Hs(A) < ∞ hence Hsδ(A) ≤ Hs(A) < ∞.

For each integer k ≥ 1 let Ekj be a collection of closed sets with diam(Ej) < 1

k and A ⊂ ⋃

j Ekj ,

then∞∑

j=1

ζs(Ekj ) ≤ Hs

1

k

(A) +1

k.

Define Ek :=⋃

j Ekj and B :=

⋂

Ek. Observe that B is a Borel set because all the Ekj are closed.

Moreover Ek ⊃ A hence B ⊃ A.

Hs1

k

(B) ≤∞∑

j=1

ζs(Ekj ), because B ⊂ Ek

≤ Hs1

k

(A) +1

k.

Hence Hs1

k

(B) ≤ Hs(A) + 1k and, letting k → +∞, Hs(B) ≤ Hs(A). But A ⊂ B hence Hs(B) =

Hs(A). �

Proposition 2.5. For any A ⊂ Rd

(1) H0 is the counting measure.(2) if d = 1 then H1(A) = m1(A), where the 1-dimensional Lebesgue measure in R.(3) Hs(A) = 0 for all s > d.(4) if 0 ≤ s < t < ∞ then Hs(A) < ∞ =⇒ Ht(A) = 0;(5) if 0 ≤ s < t < ∞ then Ht(A) > 0 =⇒ Hs(A) = ∞.

Proof. For details see [3].17

Proof of (2). If I is an interval its length |I| = diam(I) = m1(I) = ζ1(I). Then, for all A ⊂ R andfor all δ > 0,

m1(A) = inf{∑

j

|Ij | : A ⊂⋃

j

Ij} because open sets are countable union of intervals

≤ inf{∑

j

|Ij | : A ⊂⋃

j

Ij, diam(Ij) ≤ δ}

= H1δ(A).

Letting δ → 0 it follows m1(A) ≤ H1(A). On the other hand, observing that |I| = ∑∞k=−∞

∣

∣I ∩[kδ, (k + 1)δ]

∣

∣,

m1(A) = inf{∑

j

|Ij | : A ⊂⋃

j

Ij}

= inf{∑

j

∞∑

k=−∞

∣

∣Ij ∩ [kδ, (k + 1)δ]∣

∣ : A ⊂⋃

j

Ij}

≥ H1δ(A).

Proof of (3). Let C be a cube of Rd with side ℓ > 0. For any integer k > 0, C is the disjoint unionof kd cubes Ck

j of side ℓk > 0.

Given δ > 0 let k be such that ℓk < δ. Then

Hsδ(C) ≤

∑

j

ζs(Ckj ) ≤ c(d, s)ℓd kd−s.

Hence, letting k → 0 it follows Hsδ(C) = 0 and also Hs(C) = 0. �

Given points (4) and (5) of Proposition 2.5 the following definition makes sense.

Definition 2.6. The Hausdorff dimension of a set A ⊂ Rd is defined as

Hdim(A) := inf{s ≥ 0 : Hs(A) = 0}.

Notice that if Hdim(A) > 0 then Hdim(A) = sup{t : Ht(A) = +∞}.

Theorem 2.7. Hd ≡ Ld in Rd for each positive integer d.

For d = 1 this result was stated in Proposition 2.5. For d > 1 the proof is less trivial andit depends on the isodiametric inequality (Theorem 2.9) that, in turn can be proved using thefollowing symmetrization procedure.

Definition 2.8. [Steiner15 symmetrization] Let σ ∈ Rd be a fixed unit vector. We denote as

• Lx0,σ := {x0 + tσ : t ∈ R} the line through x0 in direction σ;

• Pσ := {x ∈ Rd : x · σ = 0} the plane through the origin and orthogonal to σ.

For any A ⊂ Rd we define the Steiner symmetrization of A with respect to the plane Pσ as the set

Sσ(A) :=⋃

{x+ tσ : |t| ≤ 1

2H1(A ∩ Lx,σ)}

where the union is taken over all x ∈ Pσ such that A ∩ Lx,σ 6= ∅.15Jacob Steiner: Utzenstorf, Canton of Bern, March 18, 1796 – Bern, Switzerland, April 1, 1863

18

Using Steiner symmetrization it is possible to prove the following isodiametric inequality statingthat Euclidean balls in R

d are the sets of maximum d dimensional Lebesgue measure among allsets of a given diameter. Notice that the statement is not a trivial one: a set of a given diameteris not necessarily contained in the sphere of the same diameter.

Theorem 2.9. [Isodiametric Inequality] For all A ⊂ Rd

md(A) ≤ α(d)

(

diam(A)

2

)d

.

Proof. See [3]. �

Proof of Theorem 2.7. The proof of the inequality md ≤ Hd depends on the isodiametric inequality.The opposite inequality depends on Vitali’s covering theorem.

(Step 1) [md ≤ Hd] For δ > 0 let Ej be such that diam(Ej) ≤ δ and A ⊂ ⋃∞j=1Ej . Then, by the

isodiametric inequality,

md(A) ≤∞∑

j=1

md(Ej) ≤∞∑

j=1

α(d)

(

diam(A)

2

)d

≡∞∑

j=1

ζd(Ej).

Taking the infimum over all such coverings of A it follows

md(A) ≤ Hdδ(A), for all δ > 0

hence md(A) ≤ Hd(A).

(Step 2) [md ≥ Hd] Recalling that any open set is the disjoint union of half open cubes, it followsthat for all A ⊂ R

d and δ > 0

md(A) = inf{∞∑

j=1

md(Cj) : Cj cubes, A ⊂⋃

j

Cj ,diam(Cj) ≤ δ}.

Observe that for any cube C

ζd(C) ≡ α(d)

(

diam(C)

2

)d

= α(d)

(√n

2

)d

md(C),

hence

Hdδ(A) ≤ inf{

∞∑

j=1

ζd(Cj) : Cj cubes, A ⊂⋃

j

Cj,diam(Cj) ≤ δ}

≤ α(d)

(√n

2

)d

inf{∞∑

j=1

md(Cj) : Cj cubes, A ⊂⋃

j

Cj ,diam(Cj) ≤ δ}

= α(d)

(√n

2

)d

md(A).

This way we have proved that Hd ≪ md.

To get the general statement fix δ > 0, ε > 0 and choose disjoint cubes Cj such that

∞∑

j=1

md(Cj) ≤ md(A) + ε, A ⊂∞⋃

j=1

Cj, diam(Cj) ≤ δ.

19

By Corollary 1.33, for each Cj there is a disjoint sequence of balls Bk,j such that

md(Cj \⋃

k

Bk,j) = 0,⋃

k

Bk,j ⊂ Cj, diam(Bk,j) ≤ δ.

Hence

Hdδ(A) ≤

∞∑

j=1

Hdδ(Cj) =

∞∑

j=1

Hdδ

(

⋃

k

Bk,j

)

=∑

j

∑

k

Hdδ(Bk,j)

≤∑

j

∑

k

ζd(Bk,j) because diam(Bk,j) ≤ δ

=∑

j

∑

k

md(Bk,j) because the Bk,j are spheres

=∑

j

md(

⋃

k

Bk,j

)

=∑

j

md(Cj) ≤ md(A) + ε.

Hence, by the arbitrariaty of ε and δ, Hdδ ≤ md and finally Hd ≤ md �

20

Differentiation of measures. Density of a measure. Let µ and ν be Radon measures in Rd.

The upper and lower derivative of ν with respect to µ are defined as

Definition 2.10.

Dµν(x) :=

{

lim supr→0ν(B(x,r))µ(B(x,r)) if µ(B(x, r)) > 0 for all r > 0

+∞ if µ(B(x, r)) = 0 for some r > 0

Dµν(x) :=

{

lim infr→0ν(B(x,r))µ(B(x,r)) if µ(B(x, r)) > 0 for all r > 0

+∞ if µ(B(x, r)) = 0 for some r > 0

If both Dµν(x) and Dµν(x) exist, are finite and are equal then we define the (symmetric) derivativeDµν(x) as

Dµν(x) := Dµν(x) = Dµν(x).

Definition 2.11. The measure ν is said to be absolutely continuous with respect to the measureµ, written as

ν ≪ µ,

if ν(E) = 0 for any ⊂ Rd such that µ(E) = 0.

The measures µ and ν are said to be mutually singular, written as

ν⊥µ,

if there exists a Borel set B ⊂ Rd such that

ν(Rd \B) = µ(B) = 0.

Theorem 2.12. [Radon-Nikodym16 Theorem] Let µ and ν be Radon measures in Rd then

(1) the derivative Dµν(x) exists, is finite for µ-a.e. x ∈ Rd and is a µ-measurable function of

x;(2) if ν ≪ µ then

ν(E) =

∫

EDµν(x) dµ(x)

for any µ-measurable E ⊂ Rd.

Theorem 2.13. Let µ and ν be Radon measures in Rd then

(1) there are Radon measures νac and νs on Rd s.t.

νac ≪ µ νs⊥µ.

(2) For µ-a.e x ∈ Rd,

Dµν(x) = Dµνac(x), and Dµνs(x) = 0.

(3) For all Borel set E,

ν(E) =

∫

EDµνac(x) dµ(x) + νs(E).

16Otto Marcin Nikodym: Zabolotiv, Ucraina, August 13, 1887 – Utica, New York, May 4, 1974

21

Lebesgue points.

Theorem 2.14. [Lebesgue-Besicovitch] Let µ be a Radon measure in Rd and f ∈ L1

loc(Rd, µ) then

limr→0

1

µ(B(x, r))

∫

B(x,r)f(y) dµ(y) = f(x)

for µ a.e. x ∈ Rd.

Notice that balls centered in x are used in the averages.

Corollary 2.15. If E is a measurable then

limr→0

µ(B(x, r) ∩ E)

µ(B(x, r))=

{

1 for µ a.e. x ∈ E0 for µ a.e. x ∈ Ec.

That is, if E is µ measurable then µ a.e. x ∈ E is a point of density 1 for E and µ a.e. x ∈ Ec isa point of density 0 for E.

22

Lipschitz functions. The general definition of Lipschitz17 functions, making sense even for func-tions acting between metric spaces, is

Definition 2.16. Let (X, d) and (Y, δ) be metric spaces. The function f : X → Y is said to beLipschitz or L-Lipschitz if there is L ≥ 0 such that

(1) δ(f(x1), f(x2)) ≤ Ld(x1, x2),

for all x1, x2 ∈ X. The function f is said to be locally Lipschitz if for any compact subset of Xthere is L > 0 such that (1) holds for all x1, x2 in the compact.

A real valued L-Lipschitz function, defined on a subset E of a metric space (X, d), can always beextended to a L-Lipschitz function defined on the entire space X.

Theorem 2.17. Let (X, d) be a metric spaces and f : E(⊂ X) → R be L-Lipschitz in E. Then

there is f : X → R s.t. f|E = f and f is L-Lipschitz.

Proof. Define

(2) f(x) := infy∈E

(

f(y) + Ld(x, y))

.

(i): f is well defined for each x ∈ X. Indeed, for a fixed y0 ∈ E and for all y ∈ E ,f(y) + Ld(y, x) = f(y)− f(y0) + f(y0) + Ld(y, x)

≥ −Ld(y, y0) + f(y0) + Ld(y, x)

≥ f(y0)− Ld(y0, x).

(ii): f(y) = f(y), for y ∈ E . Indeed, taking y = x in (2) we see that f(y) ≤ f(y). On the other

side, f(y) ≤ f(y) + Ld(y, y) gives f(y) ≤ infy∈E(

f(y) + Ld(y, y))

= f(y).

(iii): f is L-Lipschitz because it is the infimum of a family of L-Lipschitz functions. Let us provethis fact. Assume that

g(x) = infα∈A

gα(x),

where, (gα)α is a family of L-Lipschitz real valued functions on X. Fix x, y ∈ X and ε > 0, thenchoose α = α(x) and β = β(y) such that

g(y) ≤ gβ(y) ≤ g(y) + ε, and g(x) ≤ gα(x) ≤ g(x) + ε,

thengα(x)− gβ(y)− ε ≤ g(x)− g(y) ≤ gα(x)− gβ(y) + ε.

Becausegα(x) ≥ gα(y)− Ld(x, y) ≥ gβ(y)− ε− Ld(x, y)

then−Ld(x, y)− 2ε ≤ g(x)− g(y) ≤ Ld(x, y) + 2ε.

�

The preceding theorem can be immediately extended to functions valued in finite dimensionalnormed vector spaces, provided we accept that the extended function has a larger Lipschitz constant.

Theorem 2.18. Let (X, d) be a metric spaces and f : E(⊂ X) → Rm be L-Lipschitz in E, then

there is f : X → Rm s.t. f|E = f and f is

√mL-Lipschitz.

Proof. Use Theorem 2.17 to extend each component of f . �

17Rudolf Otto Lipschitz: Konigsberg, May 14, 1832 – Bonn, October 3, 1903

23

Notice that a stronger result holds true for functions defined in a closed subset of Rd (and not ofgeneral metric spaces).

Theorem 2.19. [Kirszbraun18 Theorem] Let f : E(⊂ Rd) → R

m be L-Lipschitz in a closed set E,then there is f : Rd → R

m s.t. f|E = f and f is L-Lipschitz.

The proof of Kirszbraun’s theorem depends crucially on very euclidean properties of the domainspace R

d, hence its proof cannot be extended to general metric spaces.

If f : R → R is Lipschitz then it is absolutely continuous hence it is differentiable m1-a.e. in R. Itis a consequence of Fubini’s Theorem that also Lipschitz functions defined on Euclidean spaces Rd,d > 1, are differentiable md-a.e. This is the content of Rademacher19 theorem.

Theorem 2.20. [Rademacher’s Theorem] Let f : Rd → Rm be locally Lipschitz. Then f is differ-

entiable md-a.e. That is, ∇f(x) :=( ∂f

∂x1(x), . . . ,

∂f

∂xd(x))

exists for md-a.e. x and

limy→x

|f(y)− f(x)−∇f(x) · (y − x)||y − x| = 0.

Moreover ∇f is a measurable function.

Proof. The proof is classical. We are following here the one in [3] or in [14].

Observe that, considering separately each component of f , we can assume that m = 1.

Let v ∈ Rd with |v| = 1 and Πv = {y ∈ R

d : y · v = 0} be the orthogonal subspace to v. Then, forall y ∈ Πv the function ϕy,v : R → R defined as

t 7→ ϕy,v(t) := f(x+ tv)

is Lipschitz in R hence differentiable for m1-a.e. t ∈ R. Moreover, in each point x := y + tv suchthat the derivative of ϕy,v exists in t, we have

Dvf(x) =d

dtϕy,v(t).

Claim 1: the set Bv := {x : Dvf(x) does not exists} is a Borel set and hence it is d-measurable.

Indeed, the function Dvf defined as

Dvf(x) := lim supt→0

f(x+ tv)− f(x)

t= lim

k→∞sup

− 1

k<t< 1

k; t6=0

f(x+ tv)− f(x)

t

is a Borel function, because sup− 1

k<t< 1

k; t6=0

f(x+tv)−f(x)t is hemicontinuous, being the supremum of

bounded continuous functions, hence it is Borel and also Dvf is Borel being the limit of a sequenceof Borel functions.

With the same argument also Dvf is Borel. Hence

Bv := {x : Dvf(x) does not exists} = {x : Dvf(x) < Dvf(x)}is a Borel set and Dvf is a Borel function.

18Moizesz David Kirszbraun: Polish mathematician, 1903 or 1904 – Warsaw (?), 194219Hans Adolph Rademacher: Wandsbeck, German Empire, April 3, 1892 – Haverford, Pennsylvania, February 7,

1969

24

Being Bv a md-measurable set and because the m1-measure of all its one dimensional sections indirection v vanish, we can conclude, using Fubini’s theorem, that

(3) md(Bv) = 0.

Let v1, v2, . . . be a sequence of unit vectors that is dense in the unit sphere of Rd (and assume alsothat the unit vectors e1, . . . , ed are part of the sequence). Then md(Bvi) = 0, for i = 1, 2, . . . , hence

md(

⋃

i

Bvi

)

= 0

and there is a set, precisely(⋃

i Bvi

)c, of full measure in R

d such that for each point x in this set

∇f(x) :=( ∂f

∂x1(x), . . . ,

∂f

∂xd(x))

and Dvif(x)

exist and are Borel functions.

Claim 2: ∇f and Dvif are weak derivatives in the sense of distributions.

Let φ be a C∞ test function in Rd and let v be any unit vector in the preceding chosen sequence.

Then, by a simple change of variables, for all t 6= 0,∫

Rd

f(x+ tv)− f(x)

tφ(x) dx = −

∫

Rd

φ(x)− φ(x− tv)

tf(x) dx.

When t → 0 the integrands converge pointwise on the left hand side to Dvf(x)φ(x) and on the right

hand side to Dvφ(x)f(x). By dominated convergence theorem (remember that |f(x+tv)−f(x)t φ(x)| ≤

L|φ(x)|)∫

Rd

Dvf(x)φ(x) dx = −∫

Rd

Dvφ(x) f(x) dx

= −∫

Rd

d∑

j=1

vj∂φ

∂xj(x) f(x) dx because φ ∈ C∞

= −d∑

j=1

vj∫

Rd

∂φ

∂xj(x) f(x) dx

=

d∑

j=1

vj∫

Rd

∂f

∂xj(x)φ(x) dx using the same argument as above.

Hence finally∫

Rd

Dvf(x)φ(x) dx =

∫

Rd

d∑

j=1

vj∂f

∂xj(x)φ(x) dx

for all test functions φ. This yields that Dvf(x) =∑d

j=1 vj ∂f∂xj

(x) for md-a.e. x ∈ Rd.

Putting all together, there are a Borel set A with

md(Rd \ A) = 0

and a dense sequence of unit vectors v1, v2, . . . such that for all x ∈ A, Dvif(x) and ∇f(x) existand

(4) Dvif(x) = vi · ∇f(x), for all v1, v2, . . . .

Claim 3: f is differentiable in A.25

Let x ∈ A. We have to prove that for all ε > 0 there is δ > 0 such that∣

∣f(y)− f(x)− (y − x) · ∇f(x)∣

∣

|y − x| < ε for all |y − x| < δ.

Let us introduce the following notation: for x ∈ A, for v with |v| = 1 and for t ∈ R let

Q(x, v, t) :=f(x+ tv)− f(x)

t− v · ∇f(x)

Observe that if |v| = |w| = 1, then

∣

∣Q(x, v, t) −Q(x,w, t)∣

∣ =∣

∣

f(x+ tv)− f(x+ tw)

t− (v − w) · ∇f(x)

∣

∣

≤ L |v − w|+ |v − w| |∇f(x)|≤ L(1 +

√d)|v − w|.

(5)

Now, for ε > 0 choose v1, . . . , vN inside the sequence in (4) such that

(6) minj=1,...,N

|v − vj | ≤ ε for all v with |v| = 1.

Moreover observe that by definition for each vi, limt→0 Q(x, vi, t) = 0, hence, for ε > 0 there isδ > 0 s.t.

(7) maxj=1,...,N

|Q(x, vj , t)| ≤ ε for 0 < |t| ≤ δ.

Finally, for y 6= x, let v := y−x|y−x| and t := |y − x|. Then, from (6), we can choose j, 1 ≤ j ≤ N ,

such that |v − vj| < ε, hence if |y − x| < δ∣

∣f(y)− f(x)− (y − x) · ∇f(x)∣

∣

|y − x| ≡ |Q(x, v, t)|

≤ |Q(x, v, t) −Q(x, vj , t)|+ |Q(x, vj , t)|≤ L(1 +

√d)ε+ ε from (5) and (7).

�

The following theorem gives a necessary and sufficient condition for the existence of a C1 extensionof a given function from a compact subset of Rd. For a proof see [12].

Theorem 2.21. [Whitney20 extension theorem] Let A be a closed set in Rd. Let f : A → R and

h : A → Rd be continuous functions. Define

R(x, y) :=f(x)− f(y)− h(y) · (x− y)

|x− y| , for all x, y ∈ A, x 6= y.

Then, if for all compact sets C ⊂ A

limr→0+

sup0<|x−y|<r,x,y∈C

|R(x, y)| = 0,

there is f ∈ C1(Rd) such that

f|A = f (∇f)|A = h.

Whitney’s theorem together with a.e. differentiability of Lipschitz functions yield that any Lipschitzfunction coincides, outside of a set of measure ε > 0, with a C1 function.

20Hassler Whitney: New York City, New York, March 23, 1907 – Princeton, New Jersey, May 10, 1989

26

Theorem 2.22. Let f : Rd → R be Lipschitz. Then for each ε > 0 there is a g ∈ C1(Rd) such that

md({x : f(x) 6= g(x)}) ∪md({x : ∇f(x) 6= ∇g(x)}) < ε.

Proof. (Hint) Because f is Lipschitz f is differentiable almost everywhere, i.e. there is an excep-tional set E ⊂ R

d, with md(E) = 0, such that, for all x ∈ Rd\E, ∇f(x) exists and f is differentiable

in x.

By Lusin’s theorem, for any ε > 0 there is a closed set A ⊂ (Rn \ E) such that Ld(Rd \ A) < ε/2and ∇f is continuous in A.

Because f is differentiable in A, for x, y ∈ A,

(8) limx→y

f(x)− f(y)−∇f(y) · (x− y)

|x− y| = 0.

By Egoroff theorem, there is B ⊂ A such that Ld(Rd \ B) < ε where the convergence in (8) isuniform. This way we have proved that the assumptions of Whitney’s extension theorem (Theorem2.21) hold in B hence there is a C1 function g coincident with f in B. �

The Area and Coarea Formulas. Before stating the two main theorems (Theorem 2.24 andTheorem 2.29) we recall a few linear algebra definitions and results.

Proposition 2.23. [Polar Decomposition] Let L : Rd → Rk be a linear function. Then

(1) If d ≤ k there are a symmetric linear function S : Rd → Rd and a orthogonal linear function

O : Rd → Rk such that

L = O · S.We define the d-dimensional Jacobian JdL of L as JdL := |detS|. Then it holds

JdL =√

det(LT · L).

(2) If d ≥ k there are a symmetric linear function S : Rk → Rk and a orthogonal linear function

O : Rk → Rd such that

L = S · OT .

We define the k-dimensional Jacobian JkL of L as JdL := |detS|. Then it holds

JkL =√

det(L · LT ).

More generally we use the following notations: if f : Rd → Rk is differentiable in a ∈ R

d we define

(1) If d ≤ k, Jdf(a) :=√

det(

(Df(a))TDf(a))

;

(2) If d ≥ k, Jkf(a) :=√

det(

Df(a)(Df(a))T)

.

Clearly, if d = k it follows Jdf(a) =∣

∣ det(

Df(a))∣

∣.

27

The Area Formula.

Theorem 2.24. [Area Formula] Let d ≤ k and let f : Rd → Rk be a Lipschitz function. Then

(9)

∫

AJdf(x) dm

d(x) =

∫

Rk

H0(A ∩ f−1({y})) dHd(y)

for each Lebesgue measurable subset A ⊂ Rd.

Notice that if y /∈ f(A) then H0(A ∩ f−1({y})) = 0 hence (9) can be equivalently written as

(10)

∫

AJdf(x) dm

d(x) =

∫

f(A)H0(A ∩ f−1({y})) dHd(y).

Moreover, if f is injective then (9) reduces to∫

AJdf(x) dm

d(x) =

∫

f(A)dHd(y) = Hd(f(A)).

More generally, from Theorem 2.24 the following Change of Variables Theorem follows.

Theorem 2.25. [Change of Variables] Let d ≤ k and let f : Rd → Rk be a Lipschitz function. Let

g : Rd → R be md-integrable. Then

(11)

∫

Ag(x)Jdf(x) dm

d(x) =

∫

Rk

(

∑

x∈f−1({y})

g(x))

dHd(y)

for each Lebesgue measurable subset A ⊂ Rd.

Proof. Assume g ≥ 0. Then from Theorem 1.27, there are measurable subsets Ai ⊂ A such thatg(x) =

∑∞i=1

1iχAi

(x) for md-a.e. x ∈ A. Moreover, we define gN :=∑N

i=11iχAi

. Then gN ր g asN → +∞. Hence

∫

Ag Jdf dmd = lim

N→+∞

∫

Ag Jdf dmd, by monotone convergence,

= limN→+∞

N∑

i=1

1

i

∫

Ai

Jdf dmd

= limN→+∞

N∑

i=1

1

i

∫

f(Ai)H0(

Ai ∩ f−1({y}))

dHd(y), by (10).

On the other side, once more by monotone convergence,∫

f(A)

(

∑

x∈A∩f−1({y})

g(x))

dHd(y)

= limN→+∞

∫

f(A)

(

∑

x∈A∩f−1({y})

gN (x))

dHd(y)

= limN→+∞

∫

f(A)

(

∑

x∈A∩f−1({y})

N∑

i=1

1

iχAi

(x))

dHd(y)

= limN→+∞

N∑

i=1

1

i

∫

f(Ai)

(

∑

x∈A∩f−1({y})

χAi(x))

dHd(y)

= limN→+∞

N∑

i=1

1

i

∫

f(Ai)H0(

Ai ∩ f−1({y}))

dHd(y).

28

Hence (11) is proved for g ≥ 0. For a general g, let g = g+ − g−. �

Notice that if d = k and f is injective then

∑

x∈f−1({y})

g(x) =

{

g(f−1(y)) if y ≡ f(x) ∈ f(A)0 if y /∈ f(A),

moreover, Jdf(x) = |detDf(x)| hence (11) takes the form∫

Ag(x) |detDf(x)| dmd(x) =

∫

f(A)g(f−1(y)) dmd(y).

Let us prove now Theorem 2.24. Its proof of goes through the following steps

(Step 1) Formula (9) holds for linear functions. Precisely, in Lemma 2.26 it is proved that if T :Rd → R

k is a linear function, then for all A ⊂ Rd md-measurable,

md(T (A)) = Hd(T (A)) = JdT md(A).

(Step 2) If f : A(⊂ Rd) → R

k is ‘well approximated’ on the set A by a linear function T : Rd → Rk

(see the assumptions of Lemma 2.27) then

(1− ε)JdT ·md(A) ≤ Hd(f(A)) ≤ (1 + ε)JdT ·md(A)

with ε explicitly dependent on the approximation.(Step 3) If f : A(⊂ R

d) → Rk is of class C1 then A can be splitted in the union of small subsets

Aj such that in each one of them f is ‘well approximated’, in the sense of (Step 2), by thedifferential Df(aj). Hence Area Formula for C1 functions follows.

(Step 4) If f : A(⊂ Rd) → R

k is Lipschitz then it can be approximated by C1 functions in the senseof Theorem 2.22. Hence Area Formula follows in its generality and the proof is completed.

Lemma 2.26. Let d ≤ k and let T : Rd → Rk be a linear function. Then for all A ⊂ R

d Lebesguemeasurable

(12) md(T (A)) = Hd(T (A)) = JdT md(A)

Lemma 2.27. [Main Estimate] Let d ≤ k and A ⊂ Rd be Lebesgue measurable. Let f : A → R

k

and T : Rd → Rk be a linear function. Assume that there is ε > 0 s.t.

(1) Df(a) exists for all a ∈ A(2) ‖Df(a)− T‖ < ε for all a ∈ A(3)

∣

∣f(y)− f(a)− (y − a) ·Df(a)∣

∣ < ε |y − a| for all y, a ∈ A

(4) The orthogonal projection from f(A) to T (Rd) is 1− 1.

Then

(13) (1− c1ε)dJdT ·md(A) ≤ Hd(f(A)) ≤ (1 + c2ε)

dJdT ·md(A)

Notice that we obtain also the following version of Sard’s theorem

Theorem 2.28. [Sard21 Theorem] If f : Rm → Rn is a C1 function and A = {x : Jmf(x) = 0},

then Hm[f(A)] = 0.

21Arthur Sard: New York City, New York, July 28, 1909 – Basel, Switzerland, August 31, 1980

29

The Coarea Formula.

Theorem 2.29. [Coarea Formula] If f : Rd → Rk is a Lipschitz function and d ≥ k, then

∫

AJkf(x)dm

d(x) =

∫

Rk

Hd−k(A ∩ f−1({y}))dmk(y)

for each Lebesgue measurable set A ⊂ Rd. Here

Jkf(x) =√

det[(Df(x))(Df(x))T ]

Remark 2.30. It is a generalization of Fubini’s Theorem.

30

3. Rectifiability

Definitions. Rectifiable sets provide a useful generalization of C1 submanifolds. They can be defined- and characterized - in many different ways: using Lipschitz maps or C1 submanifolds or measuretheoretic tangency properties. In the first definition we require that, but for a negligeable subset,S be the countable union of Lipschitz images of measurable subsets of Rd.

Definition 3.1. Let 1 ≤ m ≤ d be integers. A set S ⊂ Rd is countably m-rectifiable if

S = S0 ∪(

∪∞j=1fj(Aj)

)

,

where Hm(S0) = 0, Aj ⊂ Rm are Hm-measurable and fj : Aj → R

n are Lipschitz functions forj = 1, 2, . . . .

Since Lipschitz functions defined on subsets of Rm can be extended to all of Rm, keeping a controlof the Lipschitz constant, (see Theorem 2.18) then we can equivalently say that S is countablym-rectifiable if it is Hm-measurable and

S ⊂ S0 ∪(

∪∞j=1fj(R

m))

,

where Hm(S0) = 0 and fj : Rm → R

n are Lipschitz functions for j = 1, 2, . . . .

Proposition 3.2. Let 1 ≤ m ≤ n, then S ⊂ Rn is countably m-rectifiable if and only if S is

Hm-measurable and there exist m-dimensional, embedded C1 submanifolds Mj , for j = 1, 2, . . . ,such that

S ⊂ S0 ∪(

∪∞j=1Mj

)

,

where Hm(S0) = 0.Aggiungere

Proof. �

We can also require that, possibly enlarging the negligeable subset S0, that the parts of S leavingon different submanifolds are pairwise disjointed.

Proposition 3.3. Let 1 ≤ m ≤ n, then S ⊂ Rn is countably m-rectifiable if and only if S is

Hm-measurable and there exist m-dimensional, embedded C1 submanifolds Mj and Hm-measurablesubsets Sj ⊂ Mj , for j = 1, 2, . . . , such that

S = S0 ∪(

∪∞j=1Sj

)

,

where Hm(S0) = 0 and Si ∩ Sj = ∅, for 1 ≤ i 6= j.Aggiungere

Proof. �

Rectifiable sets have an approximate tangent plane Hm-almost everywhere, see Definition 3.4. Thisfact is the content of Theorem 3.5.For x ∈ R

n and λ > 0, let δλ,x : Rn → Rn be the family of dilations defined as

δλ,x(y) :=1

λ(y − x).

Definition 3.4. Let S be a countably m-rectifiable set in Rn. An m-dimensional linear subspace

V of Rn is the approximate tangent plane to S at x ∈ S, denoted as TxS, if

Hm (δλ,xS) ⇀ Hm V, for λ → 0,

that is

limλ→0

∫

δλ,xSg dHm =

∫

Vg dHm, for all g ∈ C0(R

n,R).

31

Notice that if S is a C1 embedded submanifold, then approximate tangent spaces exist everywhereand they coincide with the usual (i.e. given by differentiation) tangent spaces.

Theorem 3.5. Let S be a countably m-rectifiable set in Rn such that Hm(S∩W ) < ∞ for all open

W ⋐ Rn. Then TxS exists for Hm-almost every x ∈ S.

Proof. Write S as in Proposition 3.3. For any j ≥ 1, using the fact that the Si are pairwisedisjointed for i ≥ 1, we have, for Hm almost every x ∈ Sj ,

(i) Θ∗m (Hm (S \ Sj), x) = 0,

and

(ii) limr→0

Hm(Sj ∩B(x, r))

Hm(Mj ∩B(x, r))= 1.

Since Mj is a C1 submanifold, its tangent m-plane exists, hence, by (ii), also TxSj exists andcoincides with the tangent m-plane. Finally, because of (i), TxSj = TxS for for Hm almost everyx ∈ Sj. �

If f : S → Rk is C1, using once more the decomposition of S as in Proposition 3.3, it follows that

it is possible to define the restricted k Jacobian JSk f (and, if k = 1, the restricted gradient ∇Sf).

Finally, Area and Coarea formulas can be restated for rectifiable sets as follows

Theorem 3.6. Let m ≤ n, m ≤ p and f : Rp → Rn be a Lipschitz function. Let S ⊂ R

p be Hm-measurable, m-rectifiable with Hm-measure locally finite. Then JS

mf exists Hm almost everywherein S and, for any Hm-measurable g : S → R,

(14)

∫

Sg(x)JS

mf(x) dHm(x) =

∫

f(S)g(y)H0(f−1(y)) dHm(y).

Theorem 3.7. Let k ≤ m ≤ p and f : Rp → Rk be a Lipschitz function. Let S ⊂ R

p be Hm-measurable, m-rectifiable with Hm-measure locally finite. Then JS

k f exists Hm almost everywherein S and, for any Hm-measurable g : S → R,

(15)

∫

Sg(x)JS

k f(x) dHm(x) =

∫

Rk

(

∫

S∩f−1(y)g(z) dHm−k(z)

)

dHk(y).

In particular, if k = 1, we have ∇Sf instead of JSk f and the preceding formula becomes

(16)

∫

Sg(x)∇Sf(x) dHm(x) =

∫ +∞

−∞

(

∫

S∩f−1(y)g(z) dHm−1(z)

)

dy.

Tangent measures and Rectifiability criterions.

Marstrand’s Theorem.

Theorem 3.8. [Marstrand’ Theorem] Let µ be a measure on Rn, α a non negative real number,

and E a Borel set with µ(E) > 0. Assume

(17) 0 < θα∗ (µ, x) = θα∗(µ, x) < ∞ for µ− a.e. x ∈ E.

Then α is an integer.

32

4. Appendix

A proof of Kirszbraun Theorem.

Theorem 4.1. [Kirszbraun Theorem] Let f : E(⊂ Rn) → R

m be L-Lipschitz in a closed set E, thenthere is f : Rn → R

m s.t. f|E = f and f is L-Lipschitz.

Lemma 4.2. [Main Lemma] Let F := {x1, x2, . . . , xN} ⊂ Rm and x0 /∈ F . If g : F → R

n is1-Lipschitz then there is g : F ∪ {x0} such that

g|F = g and g is 1-Lipschitz.

The Lemma can be stated in equivalent more geometrical formulations. The equivalence amongthem is an exercise.

Main Lemma: Geometric version 1 Let be given x1, x2, . . . , xN ∈ Rn and y1, y2, . . . , yN ∈ R

m

such that

|yi − yj|n ≤ |xi − xj|m for 1 ≤ i, j ≤ N .

Let be given x0 ∈ Rn such that ri := |x0 − xi|m > 0 for 1 ≤ i ≤ N .

Then there is y0 ∈ Rm such that

|yi − y0|n ≤ ri = |xi − x0|m for 1 ≤ i ≤ N .

Main Lemma: Geometric version 2 Let be given N points x1, x2, . . . , xN ∈ Rn, N point

y1, y2, . . . , yN ∈ Rm and positive numbers r1, . . . rN . If, for 1 ≤ i, j ≤ N

|yi − yj|n ≤ |xi − xj |m and

N⋂

i=1

B(xi, ri) 6= ∅

then alsoN⋂

i=1

B(yi, ri) 6= ∅.

Proof. (Main Lemma: Geometric version 1)

Step 1: Construction of the candidate point y0: For all t > 0 let

Yt :=

N⋂

j=1

B(yj, trj) = {y ∈ Rm : |y − yj|m ≤ trj}

and let c := inf{t : Yt 6= ∅}. Then c > 0 and Yc 6= ∅. Let b be any point in Yc.Observe that

|b− yj|m ≤ c |x0 − xj |n = c rj for 1 ≤ j ≤ N .

hence the thesis follows if we can prove that c ≤ 1.

Step 2: Since c = inf{t : Yt 6= ∅}, for all c′ < c⋂

j

B(yj, c′rj) = ∅

and there is a subset of the 2N points xj , yj s.t.

|b− yj|m = c |x0 − xj|n for 1 ≤ j ≤ N ′ ≤ N .33

Notice that we limit ourselves to consider only these points such that and we keep callingthese points xj and yj.

Step 3: Without loss of generality we can assume x0 = 0 and b = 0. Hence we are reducedto the following situation:we are given points x1, x2, . . . , xN ∈ R

n, y1, y2, . . . , yN ∈ Rm such that

(1) |yi − yj|m ≤ |xi − xj |n for 1 ≤ i, j ≤ N

(2) |yj|m = c |xj |n = crj hence in particular 0 ∈⋂

j

B(yj, crj)

(3)⋂

j

B(yj, c′rj) = ∅ for all c′ < c

and we have to prove that in this situation c ≤ 1. To do this we argue by contradiction.We prove that, if c > 1 then all the yj are contained in an open halfspace. In this case thepoints yj can be moved so to keep (1) true and getting (3) with a smaller c.

Step 4: If c > 1 then the projections on the unit ball of xj and yj satisfy (1) of Step 3 witha strict inequality

∣

∣

∣

∣

yi|yi|m

− yj|yj|m

∣

∣

∣

∣

m

<

∣

∣

∣

∣

xi|xi|n

− xj|xj |n

∣

∣

∣

∣

n

for 1 ≤ i, j ≤ N

Step 5: The pointsyi

|yi|mare contained in an open hemisphere, hence yi are contained in

open halfspace.

Step 6: Moving the yj we get the contradiction.

�

Lemma 4.3. Let be given y1, . . . , yN ∈ Rm and positive numbers r1, . . . , rN . For all t > 0 let

Yt :=N⋂

j=1

B(yj, trj) = {y ∈ Rm : |y − yj|m ≤ trj}

and let c := inf{t : Yt 6= ∅}. Then c > 0 and Yc contains a single point b where

b =

H∑

i=1

λiyji, with λi > 0 and∑

i

λi = 1

and yj1, . . . , yjH are those elements in y1, . . . , yN such that

|b− yji|m = crji for i = 1, . . . ,H

Proof. c = inf{t : Yt 6= ∅} exists because Yt 6= ∅ for t large enough. Moreover, Yc is not emptybecause each Yt is compact and Yc =

⋂

t>c Yt.

Assume, by contradiction, that there are z, w ∈ Yc with z 6= w. Then it follows that z+w2 ∈ Yc′ for

a c′ < c. Indeed, from the strict convexity of the norm,∣

∣

∣

∣

z + w

2

∣

∣

∣

∣

m

<1

2|z − yj |m +

1

2|w − yj|m ≤ crj

for j = 1, . . . , N . Hence there is c′ < c such that∣

∣

z+w2

∣

∣

m≤ c′rj so that Yc′ 6= ∅. A contradiction.

Finally, once more by contradiction, assume that b is not in the convex hull of yj1 , . . . , yjH . Thenthere is an hyperplane Π through b and leaving yj1, . . . , yjH strictly on one side of Π. If this is thecase, it is possible to move b towards a new point b′ inside the half space containing yj1 , . . . , yjH so

34

that we strictly decrease the distances |b′ − yji|m < |b − yji |m. Hence, once more, there is c′ < csuch that Yc′ 6= ∅. A contradiction. �

Notice that the proof of this Lemma holds for any norm in Rm that is strictly convex and has the

following property

if y ∈ Rm and 〈y, em〉 > 0 then

d

dt|y + tem||t=0

> 0

Lemma 4.4. Let x1, . . . , xN ∈ Rn and y1, . . . , yN ∈ R

m be such that

|yi − yj|m ≤ |xi − xj |n for all 1 ≤ i, j ≤ N .

Let x0 ∈ Rn with x0 6= xj and denote rj := |x0 − xj |n for j = 1, . . . , N . As in Lemma 4.3 let

c := inf{t :N⋂

j=1

B(yj, trj) 6= ∅}.

Then c ≤ 1.

Proof. In Lemma 4.3 we proved that c > 0, that there is b ∈ Rm such that

{b} =

N⋂

j=1

B(yj, crj)

and that b is in the convex hull of the points yji such that |b− yji |m = crji . �

Rademacher Theorem. A different proof of Theorem 2.20.

Theorem 4.5 (Rademacher’s Theorem). Let f : Rd → Rm be locally Lipschitz. Then f is dif-

ferentiable Ld-almost everywhere. That is, ∇f(x) :=

(

∂f

∂x1(x), . . . ,

∂f

∂xd(x)

)

exists for Ld-a.e. x

and

limy→x

|f(y)− f(x)−∇f(x) · (y − x)||y − x| = 0.

Moreover ∇f is a measurable function.

Proof. The proof is classical. We are following here the one in [6]. We assume m = 1 and we workby induction on the dimension d.

If d = 1 Lipschitz functions are absolutely continuos hence they are differentiable m1-a.e.

When d > 1 then f is absolutely continuos in each single variable. hence partial derivatives existm1-a.e. on each line parallel to the coordinate axes. After proving – as in the proof of Theorem2.20 – the md-measurability of the set where all partial derivatives exist, by Fubini’s theorem weconclude that at md-a.e. point of Rd the partial derivatives exist (and are measurable functions).We have to show that these derivatives actually represent the differential of f , at least in all thepoints where they are approximately continuous.

Write Rd as Rd−1 × R and points of Rd as p = (x, y). We consider a point p0 = (x0, y0) where the

following two conditions hold:

(i): f is differentiable as a function of the first d− 1 variables.

(ii): all the d partial derivatives of f exist and are approximately continuous in (x0, y0).35

Without loss of generality we assume that p0 = (0, 0) and (possibly adding a linear function) thatall the partial derivatives vanish at p0. Moreover we denote by L > 0 the Lipschitz constant of fin a neighborhood of (0, 0).

Fix ε > 0 small. By (i), we can choose r0 > 0 such that

|f(x, 0)| ≤ ε, for all x with |x| < r0.

By (ii) we know that the d-dimensional density at (0, 0) of the set

{

(x′, y′) :

∣

∣

∣

∣

∂f

∂y(x′, y′)

∣

∣

∣

∣

> ε}

is zero. Hence, possibly choosing a smaller r0, we may assume that

(18) md({

(x′, y′) :

∣

∣

∣

∣

∂f

∂y(x′, y′)

∣

∣

∣

∣

> ε, |x′| < 2r, |y′| < 2r})

≤ 1

2α(d− 1)εdrd,

for all 0 < r < r0.

Now consider (x, y) 6= (0, 0) with max{|x|, |y|} < r < r0; to prove the differentiability of f in (0, 0)we prove that there is c = c(L, d) > 0 such that |f(x, y)| ≤ cεr.

Indeed, assume for the moment the existence of x′ ∈ Rd−1, with |x− x′| < εr, such that

(19) m1({

y′ :

∣

∣

∣

∣

∂f

∂y(x′, y′)

∣

∣

∣

∣

> ε, |y′| < 2r})

< εr.

Such an x′ given, we have

|f(x′, y)− f(x′, 0)| =∣

∣

∫ y

0

∂f

∂y(x′, t) dt

∣

∣ ≤ ε|y|+ Lεr < (L+ 1)εr.

Notice that∣

∣

∣

∂f∂y (x

′, y′)∣

∣

∣< L for m1-a.e. y′.

Finally because f is L-Lipschitz, we have

|f(x, y)− f(x′, y)| ≤ L|x− x′| < Lεr,

|f(x, 0)− f(x′, 0)| ≤ L|x− x′| < Lεr;

and, because f is differentiable in (0, 0) with respect to the first d− 1 variables,

|f(x, 0)| ≤ ε|x| < εr.

Putting together these inequalities we conclude that

|f(x, y)| ≤ (3L+ 2)εr.

This concludes the proof but for the fact that we have to prove (19). Assume, by contradiction,that for all x′ ∈ R

d−1 with |x− x′| < εr we have

m1({

(x′, y′) :

∣

∣

∣

∣

∂f

∂y(x′, y′)

∣

∣

∣

∣

> ε, |y′| < 2r})

≥ εr.

Then

md({

(x′, y′) :

∣

∣

∣

∣

∂f

∂y(x′, y′)

∣

∣

∣

∣

> ε, |x′| < 2r, |y′| < 2r})

≥ md({

(x′, y′) :

∣

∣

∣

∣

∂f

∂y(x′, y′)

∣

∣

∣

∣

> ε, |x− x′| < εr, |y′| < 2r})

≥ εrmd−1({

x′ ∈ Rd−1 : |x− x′| < r

})

≥ α(d − 1)εd rd.

This contradicts (18) and the proof is completed. �

36

Abram Samoilovitch Besicovitch: Berdyansk, Russian Empire, January 23, 1891 – Cam-bridge, UK, November 2, 1970

Emile Borel: Saint-Affrique, France, January 7, 1871 – Paris, France, February 3, 1956Constantin Caratheodory: Berlin, German Empire, September 13, 1873 – Munich, West

Germany, February 2, 1950Paul Adrien Dirac: Bristol, England, August 8, 1902 – Tallahassee, Florida, October 20,

1984Dmitri Fyodorovich Egorov: Moscow, Russian Empire, December 22, 1869 – Kazan, So-

viet Union, September 10, 1931Felix Hausdorff: Breslau, Kingdom of Prussia, November 8, 1968 – Bonn, Germany, January

26, 1942Camille Jordan: Lyon, France, January 5, 1838 – Paris, France, January 22, 1922Moizesz David Kirszbraun: Polish mathematician, 1903 or 1904 – Warsaw (?), 1942Henri Leon Lebesgue: Beauvais, Oise, France, June 28, 1875 – Paris, France, July 26, 1941Rudolf Otto Lipschitz: Konigsberg, May 14, 1832 – Bonn, October 3, 1903John Edensor Littlewood: Rochester, England, June 9, 1885 – Cambridge, England, Sep-

tember 6, 1977Nikolai Nikolaevich Luzin: Irkutsk, Russian Empire, December 9, 1883 – Moscow, Soviet

Union, January 28, 1950Otto Marcin Nikodym: Zabolotiv, Ucraina, August 13, 1887 – Utica, New York, May 4,

1974Hans Adolph Rademacher: Wandsbeck, German Empire, April 3, 1892 – Haverford, Penn-

sylvania, February 7, 1969Johann Karl August Radon: Decin, Bohemia, Austria-Hungary, December 16, 1887 – Vi-

enna, Austria, May 25, 1956Frigyes Riesz: Gyor, Austria-Hungary, January 22, 1880 – Budapest, Hungary, February

28, 1956Arthur Sard: New York City, New York, July 28, 1909 – Basel, Switzerland, August 31,

1980Jacob Steiner: Utzenstorf, Canton of Bern, March 18, 1796 – Bern, Switzerland, April 1,

1863Heinrich Franz Tietze: Schleinz, Austria-Hungary, August 31, 1880 – Munich, West Ger-

many, February 17, 1964Giuseppe Vitali: Ravenna, Italy, August 26, 1875 – Bologna, Italy, February 29, 1932Hassler Whitney: New York City, New York, March 23, 1907 – Princeton, New Jersey, May

10, 1989

37

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