1. Geometric Measure Theory, Professor Tatiana Toro.

Lecture 1: March 31, 2014

Consider E ⊂ R with 0 < m(E) < ∞, where m is Lebesgue measure. Then by thedifferentiation theorem, we have for m-a.e. x ∈ E,

limr→0

m((x− r, x+ r) ∩ E)

2r= 1

Besicovitch studied the following situations: When E ⊂ R2, andH1 E(A) := H1(E∩A),(where H1 is one-dimensional Hausdorff measure), what can one say about E if for H1-a.e.x ∈ E we have

limr→0

H1(E ∩B(x, r))

2r= 1

From 1928-1938, it was found that if the above holds, and 0 < H1(E) < ∞, thenE ⊂

⋃∞i=1 fi(R) ∪ E0, where H1(E0) = 0 and the fi are Lipschitz.

Recall 2X = P(X).

Definition 1. A mapping µ : 2X → [0,∞] is a measure, (outer), on X if: i) µ(∅) = 0, ii)µ(A) ≤

∑∞1 µ(Ai) when A ⊂ ∪∞1 Ai.

Remark 2. If A ⊂ B then µ(A) ≤ µ(B).

Definition 3. A ⊂ X is µ-measurable if µ(B) = µ(B ∩ A) + µ(B ∩ Ac) for all B ⊂ X.

Remark 4. If A is µ-measurable, the Ac is also.

Theorem 5. Let {Ak}∞1 be µ measurable, then:

(1) ∪∞1 Ai and ∩∞1 Ai are measurable.(2) µ(∪Ai) =

∑µ(Ai) if the Ai are pairwise disjoint.

(3) If Ak ⊂ Ak+1 then µ(∪Ai) = limµ(Ak).(4) If Ak+1 ⊂ Ak and µ(A1) <∞ then µ(∩Ak) = limµ(Ak).

Recall that A is a σ-algebra over X if A ⊂ 2X , both X and ∅ are in A, and A is closedunder complements and countable unions. Whence the µ-measurable sets form a σ-algebra.

Definition 6. Let (X, d) be a metric space. The Borel σ-algebra generated by the open setsof X is the smallest σ-algebra containing the open sets.

Definition 7. (1) A(n) (outer) measure µ on X is regular if for each A ⊂ X there is aµ-measurable set B ⊃ A with µ(A) = µ(B).

(2) A measure on (X, d) is called a Borel measure if all Borel sets are measurable.(3) A measure on (X, d) is Borel regular if it is Borel and for every A ⊂ X there is a

Borel set B ⊃ A with µ(A) = µ(B).(4) A measure µ on (X, d) is Radon if it is Borel regular and every compact set has finite

measure.

Theorem 8. Let µ be a regular measure on X, then if {Ak}∞1 ⊂ 2X and Ak ⊂ Ak+1, thenµ(∪Ak) = limµ(Ak).

Remark 9. Note that the Ak need not be measurable in this case.1

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Theorem 10. Let µ be a Borel regular measure on (X, d) and A a µ-measurable set withfinite measure. Then µ A is Radon.

Theorem 11. Let µ be Borel regular on (X, d). Suppose X = ∪∞1 Vi, Vi open with finitemeasure. Then:

• µ(A) = infA⊂U, U openµ(U), for each A ⊂ X.• µ(A) = supC⊂A, C closedµ(C) for A µ-measurable.

For example, d-dimensional Lebesgue measure.

Remark 12. : Assume A is µ measurable with finite measure and (1) holds in Theorem11. Then for any ε > 0 there is an open set U ⊃ A with µ(U) < µ(A) + ε. Thus µ(U) =µ(U ∩ A) + µ(U ∩ Ac) = µ(A) + µ(U \ A) < µ(A) + ε, so µ(U \ A) < ε.

We show that the statement (1) in the theorem implies (2).

1) =⇒ 2). If A is µ-measurable, then so is Ac, and by following the remark, we have thatfor a given ε > 0, we have an open U ⊃ Ac with µ(U \ Ac) = µ(U ∩ A) < ε. Then we have

µ(A) = µ(A ∩ U c) + µ(A \ U c) = µ(U c) + µ(A \ U c)

which shows that µ(U c) ≤ µ(A) ≤ µ(U c) + ε. �

proof of (1) when X has finite measure. Assume µ(X) <∞ finite and let A be µ-measurable(Borel). Consider M = {A Borel : (1) holds}. Then X,∅ ∈ M, and M is closed undercountable unions and intersections. For any ε > 0, and for any k there is an open Uk ⊃Ak with µ(Uk \ Ak) < ε. Observe that we have the containment monotonicity property∪j+1

1 Uk ⊂ ∪j1Uk, and the same with ∩j1Ak. Then µ(∩∞1 Ak) = limj→∞ µ(∩jk=1Ak), so that

µ(∩jk=1Uk) ≤ ε+ µ(∩Ak) for large j. Consequently, we have that

µ(∩∞k=1Ak) ≤ µ(∩jk=1Uk) ≤ 2ε+ µ(∩∞k=1Ak)

Thus (1) is closed under countable intersections. Notice that the open sets are in M,and all the closed sets are too: Consider A = {A ∈M : X \ A ∈M}, and observe that anyclosed set is the countable intersection of open sets. If C is closed, it can be written as theintersection of Ui = ∪x∈CB(x, 1

2i). Since the Ui are in M, so is C.

To finish off the proof of the theorem when X may not have finite measure, apply theprevious result to µ Vj. �

Corollary 13. Let µ be a Radon measure on (X, d), a separable locally compact space, thenfor A µ-measurable, we have that (1) holds and

µ(A) = supK⊂A, K compact

µ(K).

Recall Caratheodory’s Theorem, which says that if µ is a measure on (X, d) such that forall A,B ⊂ X, µ(A ∪B) = µ(A) + µ(B) whenever d(A, B) > 0, then µ is Borel regular.

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2. Lecture 2. April 2nd, 2014

We now start discussing covering lemmas, beginning with the following

Theorem 14 (Vitali). Let F be a family of nondegenerate closed balls such that

sup{diam B : B ∈ F} = R < ∞.

Then there exists a pairwise disjoint subcollection G ⊂ F such that⋃B∈F

B ⊂⋃B∈G

B,

along with the property that for every B ∈ F there is an S ∈ G with B∩S 6= ∅, with B ⊂ S.

Here B is a ball with five times the radius of B with the same center.

Proof. Let Fj ={B ∈ F : 2−jR < diam B ≤ 2−j+1R

}. Now, set G1 as a maximal disjoint

subcollection of F1. Given G1, . . . ,Gj−1 set Gj as the maximal disjoint subcollection of{B ∈ Fj : B ∩B′ = ∅∀B′ ⊂

⋃j−1i=1 Gi

}. We show that letting G = ∪Gi has the desired prop-

erties. Clearly, we have F = ∪Fi, and if a particular B is in Fj, there is a D ∈ ∪j1Gi with

D ∩ B 6= ∅. Since diam B ≤ 2−j+1R ≤ 2 diam D, and as diam D ≥ 2−jR, we have B ⊂ D.

Hence we also have⋃B∈Fk

B ⊂⋃B∈G B, and the result follows. �

Definition 15. A collection F of (closed) nondegenerate balls coverA if A ⊂ ∪B∈FB. Wesay that F is a fine cover of A if F covers A, and for each x in A, we have

infB∈F{diam(B) : x ∈ B} = 0.

Corollary 16. Let F be the collection of nondegenerate closed balls. Let F be a fine coverof A, and assume that

sup {diam(B) : B ∈ F} = R <∞,then there is a pairwise disjoint subcollection of F , call it G, such that for any natural numberN ,

A\N⋃1

Bi ⊂⋃

B∈G\{Bi}N1

B.

Remark 17. Note that the assumption above that R be finite is actually redundant. In factsince the cover is fine one could replace the collection F with an appropriate subcollectionwhose balls have uniformly bounded diameter.

Proof of Corollary. Let x ∈ A \ ∪N1 Bi, where the Bi are in G. Then there is a B ∈ F suchthat x is in B and, by fineness of the cover, we can choose B so B ∩ Bi = ∅, i = 1, . . . , N.

By Vitali, then, there is a D in the collection G, distinct from the Bi, with B ⊂ D, andB ∩D 6= ∅. �

Theorem 18 (Besicovitch’s Covering Theorem). There is a constant Nn, depending onlyon n, with the property that if F is any collection of nondegenerate closed balls in Rn withdiameters uniformly bounded by a constant D <∞, i.e.

sup {diam(B) : B ∈ F} = D <∞,

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and A is the collection of the centers of the balls, then there exists G1, . . . ,GNn ⊂ F witheach Gi a countable collection of pairwise disjoint balls and

A ⊂Nn⋃1

⋃B∈Gi

B.

Remark 19. Although Besicovitch’s covering theorem may seem to be less useful than Vi-tali’s covering theorem, there are in fact different uses, (and required hypthoses!), of thesetwo theorems. Vitali’s theorem can be used in an arbitrary metric space, and it gives onedisjoint family of balls whose dilations cover the original set but are no longer disjoint. It isused in case we are dealing with a doubling measure. Besicovitch’s theorem can be used inRn, and gives us disjoint families with finite overlap of at most Nn. It is used in case we aredealing with a Radon measure.

Sketch of the proof of Besicovitch. If A is bounded, choose B1 = B(a1, r1) ∈ F , r1 ≥ 34D2,

and inductively choose Bj, j ≥ 2 as follows: SetAj = A \ ∪j−11 Bj. If Aj is empty, we stop atJ = j − 1, and A ⊂ ∪J1Bi. If not, find aj ∈ Aj such that

Bj = B(aj, rj), rj ≥3

4

1

2sup {r : B(a, r) ∈ F , a ∈ Aj} .

If Aj is not empty for all j, then J = +∞.• Claim 1: If j > i, then rj ≤ 4

3ri.

• Claim 2:{B(aj,

rj3

)}Ji=1

is a disjoint collection.• Claim 3: If J is +∞, then rj → 0.

• Claim 4: If A ⊂⋃Jj=1B(aj, rj).

Fix k ∈ N, set I = {j : 1 ≤ J ≤ k, Bj ∩Bk 6= ∅} and K = I ∩ {j : rj ≤ 3rk} .

• Claim 5: The cardinality of K is not more than 20n.• Claim 6: There is an Ln such that Card(I \ K) ≤ Ln. To show claim 6, one uses

geometry and trigonometry of the triangles between centers of balls.

�

Definition 20. A doubling measure µ is a Borel regular measure with a constant 0 < c <∞such that µ(B(x, 2r)) ≤ cµ(B(x, r)) for all x, r, and µ(B(x0, 1)) <∞ for some x0 ∈ X.Remark 21. This last condition guarantees that bounded sets have bounded measure.

Theorem 22. Let (X, d) be a metric space, and let µ be a doubling measure, F a finecover of A ⊂ X by nondegenerate closed balls. Then there is a countable subcollectionG = {Bi}∞1 ⊂ F of disjoint balls such that µ(A \ ∪∞1 Bi) = 0.

Proof. Note that the doubling measure property gives µ(B) ≤ µ(8B) ≤ c3µ(B). We startwith the case when the measure of A has finite, (nonzero) measure. First, notice that thereis then an open set U ⊃ A with µ(U) < µ(A) + ε, and replacing F with the collectionF ′ := {B ∈ F : B ⊂ U} , we can replace X with U , i.e., we may assume that X has finitemeasure. By Vitali’s corollary there exists a G ⊂ F such that {Bi} ⊂ G and

A \N⋃1

Bi ⊂⋃

B∈G\BiN1

B

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Since µ(X) <∞ and G is a disjoint family, G is countable, since nondegenerate balls havepositive measure when the measure is doubling. Then

µ(A \N⋃1

Bi) ≤ µ(∞⋃N+1

Bj) ≤ c3∞∑N+1

µ(Bj)

since

µ(∞⋃1

Bi) =∞∑i=1

µ(Bi) ≤ µ(U) <∞,

letting N → ∞ gives the result in the case µ(X) < ∞. To extend to the case when X hasinfinite measure, we write X as a countable union of annuli by taking balls with radius goingto infinity. We do so in such a way that the boundary of the annuli has measure zero. Thatsuch a sequence of annuli exists follows from the doubling property, which in turn gives thateach bounded ball has finite measure. Thus there must never be an uncountable number ofspheres with positive measure, (spheres here are boundaries of balls, for simplicity consider

them to have the same center). We can now write Xa.e.= ∪∞1 Di, where the Di are the open

annuli. Now apply the above case to µ Di. �

The next theorem gives a similar result for Radon measures in Rn, which are not necessarilydoubling.

Theorem 23. Let µ be a Radon measure in Rn and A ⊂ Rn. If F is a fine collection ofnondegenerate closed balls with centers in the set A, then there is a countable subcollectionG = {Bi}∞1 ⊂ F with µ(A \ ∪Bi) = 0.

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3. Lecture 3: April 7, 2014

Last week we talked about two covering theorems. We finished by talking about thefollowing theorem:

Theorem 24. Let µ be a Radon measure in Rn and A ⊂ Rn. Define

F = {B(x, r) : x ∈ A}.

Suppose inf{r : B(x, r) ∈ F} = 0 for all x ∈ A (i.e. F is a fine cover of A). Thenthere exists a countable collection of disjoint balls G ⊂ F such that µ (A \

⋃iBi) = 0 where

{Bi}∞i=1 = G.

Remark: In the previous proof, we used a doubling measure, but we do not have this.

Proof. Since µ is a Radon measure and Rn is locally compact, µ(B(0, r)) <∞ for all r > 0(Remember balls are considered closed, unless otherwise indicated). Then µ(∂B(0, s)) > 0for only countably many s. This means there exists a sequence of ri increasing to ∞ suchthat µ(∂B(0, ri)) = 0. One sees that in measure

A =∞⋃i=1

(A ∩ int B(0, ri+1)) \B(0, ri).

Essentially, one is breaking the set A into sets which only live in an annulus. This procedureyields a decomposition of A (in measure) by sets which are disjoint and and contained indisjoint open sets. Hence,

µ(A) =∞∑i=1

µ((A ∩ int B(0, ri+1)) \B(0, ri)).

Note that because µ(B(0, r)) <∞, µ((A ∩ int B(0, r + 1)) \B(0, r)) ≤ µ(B(0, r + 1)) <∞.

Assume that 0 < µ(A) < ∞. Recall that in the Besicovitch’s covering lemma, there existsa N = Nn that depends only on n (i.e. where n is the dimension). This is called theBesicovitch number. Next, one see that there exists an open U such that A ⊂ U and

µ(U) < µ(A)(

1 + 14Nn

). Let

F = {B ∈ F : B ⊂ U and diam B < 1}.

By the assumption that inf{r : B(x, r) ∈ F} = 0 for all x ∈ A, F is a fine cover of A for F isa fine cover and sup{diam B : B ∈ F} < 1. Applying Besicovitch’s covering lemma, thereexists sub-collections G1,G2, . . . ,GNn ⊂ F such that Gi is a countable collection of pairwisedisjoint balls and

A ⊂Nn⋃i=1

( ⋃B∈Gi

B).

There exists an i∗ ∈ {1, . . . , Nn} such that

µ( ⋃B∈Gi∗

B)

= max{µ( ⋃B∈Gi

B)

: i ∈ {1, . . . , Nn}}.

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Hence,

µ(A) ≤Nn∑i=1

µ( ⋃B∈Gi

B)≤ Nnµ

( ⋃B∈Gi∗

B).

Because every B ∈ Gi∗ is contained in U and the balls in Gi∗ are disjoint,

(25) µ( ⋃B∈Gi∗

B)

=∑B∈Gi∗

µ(B) ≤ µ(U) ≤(1 +

1

4Nn

)µ(A).

By (25) and µ(A) <∞,∑

B∈Gi∗ µ(B) <∞; thus there exists a finite subcollection G ′i∗ ⊂ Gi∗such that

Nnµ

⋃B∈Gi∗

B

≤ 2Nnµ

⋃B∈G′

i∗

B

.

It follows that

µ(A) ≤ Nnµ

⋃B∈Gi∗

B

≤ 2Nnµ

⋃B∈G′

i∗

B

.

Define A1 = A \⋃B∈G′

i∗B. Since A and all the balls were contained in U ,

µ(A1) ≤ µ(U)− µ( ⋃B∈G′

i∗

B)≤ µ(U)− 1

2Nn

µ(A)

≤(1 +

1

4Nn

− 1

2Nn

)µ(A) =

(1− 1

4Nn

)µ(A).

Therefore,

µ(A1) ≤(1− 1

4Nn

)µ(A).

Let θ = 1 − 14Nn

so that µ(A1) ≤ θµ(A). Note that θ < 1 and so we have shrunk the set.

Moreover, one sees that A1 ⊂ U \⋃B∈G′

i∗B. Because G ′i∗ is finite, then the finite union

of closed sets is closed; hence, A1 is contained in the open set U \⋃B∈G′

i∗B and so there

exists an open U1 ⊃ A1 such that U1 ⊂ U \⋃B∈G′

i∗B and µ(U1) ≤

(1 + 1

4Nn

)µ(A). Set

F1 = {B ∈ F : B ⊂ U1} and note that F1 is a fine cover of A1. Assume that G ′i∗ = {Bj}M1j=1.

By repeating the procedure (i.e. applying Besicovitch and getting a bound on µ(A1)), onegets {Bj}M2

j=M1+1 ⊂ F1 such that

µ(A1 \

M2⋃j=M1+1

Bj

)≤ θµ(A1) ≤ θ2µ(A).

Because A \⋃M2

j=1Bj =(A \

⋃M1

j=1Bj

)∩(A1 \

⋃M2

j=M1+1Bj

), then

µ(A \

M2⋃j=1

Bj

)≤ µ

(A1 \

M2⋃j=M1+1

Bj

)≤ θ2µ(A).

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Continuing this procedure k times gives the existence of {Bj}Mkj=1 ⊂ F such that

µ(A \

Mk⋃j=1

Bk

)≤ θkµ(A).

Let k → ∞ and set G = {Bj}∞j=1 (Note θ < 1). Then this subcollection of F satisfies theresult. �

Hausdorff Measure and Densities

Definition 26. Let (X, d) be a metric space, m ∈ R+ (m is a non-negative real number),and δ > 0. For A ⊂ X, define

Hmδ (A) = inf

{∑i

ωm

(diam Ci

2

)m: A ⊂

∞⋃i=1

Ci and diam Ci ≤ δ

}.

Then the m-dimensional Hausdorff measure is given by

Hm(A) = limδ→0Hmδ (A).

Remark 27. • If m ∈ N, then ωm = Lm(B1(0)) where Lm(B1(0)) is Lebesgue measureof the unit ball in Rm.• The Hausdorff measure of a set always exists (it may be +∞) because Hm

δ (A) is adecreasing function of δ.• The Hausdorff measure is a regular Borel measure.

Definition 28. Let µ be a Borel regular measure in (X, d). Then for any A ⊂ X, define

(1) the upper density, θ∗,n(µ,A, x), to be

θ∗,n(µ,A, x) = lim supr→0

µ(A ∩B(x, r))

ωnrn.

Note that this could be +∞.(2) the lower density, θn∗ (µ,A, x), to be

θn∗ (µ,A, x) = lim infr→0

µ(A ∩B(x, r))

ωnrn

where B(x, r) is the closed ball.

If θ∗,n(µ,A, x) = θn∗ (µ,A, x), then define the density, θn(µ,A, a), to be θn(µ,A, x) =θn∗ (µ,A, x).

Remark 29. If A = X, then we drop the set in the notation (i.e. θ∗,n(µ, x) and θn∗ (µ, x)).

Some questions:

(1) What do the upper and lower density tell you about the measure µ.

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Example (The Four Corner Cantor Set): Let C denote the four corner Cantor set. Setµ = H1 C. Then one can show that for µ a.e. x ∈ C,

0 < θ1∗(µ C, x) < θ∗,1(µ C, x)︸ ︷︷ ︸never the same

<∞.

However, there exists a C0 > 1 such that

1

C0

≤ µ(B(x, r) ∩ C)2r

≤ C0.

Theorem 30. (Marstrand) If θs(µ A, x) exists, positive, and finite for µ(A) > 0, thens ∈ N.

Remark 31. • There is only 1 interesting n.• Each θn,∗, θn∗ are measurable functions.

Theorem 32. Let µ be a Borel regular on (X, d) and t ≥ 0.

(1) If A1 ⊆ A2 ⊆ X and θ∗,n(µ,A2, x) ≥ t for all x ∈ A1, then tHn(A1) ≤ µ(A2).(2) If A ⊂ X and θ∗,n(µ,A, x) ≤ t for all x ∈ A, then µ(A) ≤ 2ntHn(A).

Important Remarks:

• We are not assuming that A, A1, or A2 are measurable.• If A1 = A2 = A, then from (1) if θ∗,n(µ,A, x) ≥ t for all x ∈ A, then tHn(A) ≤ µ(A).• All differentiation theorems require a covering lemma (including this one).• All balls, B(x, r), are closed unless otherwise indicated.

Proof. (Proof of Theorem 32, Part (1))We can assume that µ(A2) < ∞ and t > 0 for otherwise the result is trivial. We can alsoassume the strict inequality θ∗,n(µ,A2, x) > t since to obtain the conclusion of (1) for t equalto a given t1 > 0 it suffices to prove it for each t < t1. By the definition of lim sup, we havethat

infδ>0

supr<δ

µ(A2 ∩B(x, r))

ωnrn> t.

For every δ0 > 0 and x ∈ Rn, there exists rk = rk,x < δ0 such that µ(A2 ∩B(x, rk)) > tωnrnk .

Fix δ > 0. Consider the set

Fδ =

{B(x, r) : x ∈ A1, r <

δ

2, and µ(A2 ∩B(x, r)) > tωnr

n

}.

Using the definition of a lim sup, it follows that Fδ is a fine cover of A1 (i.e. the diameters ofthe balls are bounded by δ

2and for every x ∈ A1 and ε > 0 there exists a B ∈ Fδ such that

diam B < ε and x ∈ B). Applying the corollary to the Vitali covering lemma, there existsa disjoint subcollection G ⊂ Fδ (where {Bi} = G) such that for all M ∈ N,

(33) A1 \M⋃i=1

Bi ⊂⋃

B∈G\{Bi}Mi=1

B,

where B = B(x, 5r). In fact, this disjoint subcollection G can be assumed to be countable. Infact since µ is Radon and µ(A2) <∞, there exists U open such that A2 ⊂ U and µ(U) <∞.Choose balls B ∈ G such that B ⊂ U . Using the fact that G is a disjoint subcollection and

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B ⊂ U where µ(U) <∞, it follows that G is actually countable.) Hence, denote G = {Bi}∞i=1.Fix a M ∈ N. Using the countability and (33),

A1 ⊂

(M⋃i=1

Bi

)∪

(∞⋃

i=M+1

Bi

),

so A1 is covered by these two sets. Moreover since G ⊂ Fδ, µ(A2 ∩Bi) > tωn(radius Bi) andradius Bi <

δ2. Denote ri = radius Bi. Using the definition of Hn

δ ,

(34) Hn5δ(A1) ≤

M∑i=1

ωnrni +

∞∑i=M+1

ωn5nrni .

Since Bi ∈ Fδ, then

M∑i=1

ωnrni <

1

t

∞∑i=1

µ(A2 ∩Bi) =1

t

∞∑i=1

(µ A2)(Bi) =1

t(µ A2)

(∞⋃i=1

Bj

)≤ 1

tµ(A2) <∞.

Let M →∞ in (34). Then we have

Hn5δ(A1) ≤

∞∑i=1

ωnrni <

1

tµ(A2).

Next, let δ → 0. Then Hn(A1) ≤ 1tµ(A2). �

Proof. (Proof of Theorem 32, Part (2))Like in the proof of Theorem 32 Part (1), assume θ∗,n(µ,A, x) < t for all x ∈ A (i.e. otherwisepass to the limit ). By the definition of lim sup, we have that

infδ>0

supr<δ

µ(A2 ∩B(x, r))

ωnrn< t.

Hence, there exists a δx such that for all r < δx, µ(A ∩B(x, r)) < tωnrn. Define

Ak = {x ∈ A : µ(A ∩B(x, r)) < tωnrn for all 0 < r < 1

k}.

Note that Ak ⊂ Ak+1 and A =⋃∞k=1Ak. Using that µ is a Borel regular measure, and

Theorem 8 we have that

µ(A) = limk→∞

µ(Ak) (remember A may not be measurable).

Let δ ∈ (0, 14k

) and let {Ci}∞i=1 be a countable cover for Ak (i.e. Ak ⊂⋃Ci) such that

diam Ci ≤ δ. For xi ∈ Ak ∩ Ci, Ci ⊂ B(xi, diam Ci) where diam Ci <14k

. By the definitionof Ak,

µ(Ci ∩ Ak) ≤ µ(B(xi, diam Ci) ∩ Ak) < tωn(diam Ci)n ≤ tωn2n

(diam Ci

2

)n.

Hence,

µ(Ak) ≤∞∑i=1

µ(Ci ∩ Ak) ≤∞∑i=1

tωn2n(

diam Ci2

)n.

Taking the inf over all such covers {Ci}, we have by definition ofHnδ that µ(Ak) ≤ t2nHn

δ (Ak).Let δ → 0 and conclude that µ(Ak) ≤ t2nHn(Ak). Since Hn is a regular Borel measure,limk→∞Hn(Ak) = Hn(A). By letting k →∞, µ(A) ≤ t2nHn(A). �

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Applications of the Theorem:

Theorem 35. Suppose µ is Borel regular measure on (X, d). Let A ⊂ X such that µ(A) <∞, A is µ-measurable, and

θ∗,n(µ,A, x) = 0 µ a.e. x ∈ A.If Hn(A) <∞, then

θ∗,n(Hn A, x) = 0 for Hn a.e. x ∈ X \ A.

Theorem 36. Suppose A is a subset of X.

(1) If Hn(A) <∞, then θ∗,n(Hn A, x) ≤ 1 for Hn a.e. x ∈ A.(2) If Hn

δ (A) <∞ for each δ > 0, then θ∗,n(Hn∞ A, x) ≥ 1

2nfor Hn a.e. x ∈ A.

Complete proofs can be found in Lectures on Geometric Measure Theory by Leon Simon;theorems 3.5 and 3.6 (pg. 14-17).

Corollary 37. If 0 < Hn(A) <∞ and A a Hn-measurable subset of X, then

1

2n≤ θ∗,n(Hn A, x) ≤ 1.

Proof. This follows directly from Theorem 36 since Hn ≥ Hnδ ≥ Hn

∞. �

12

4. Lecture 4: April 9, 2014

Definition 38. Let x ∈ Rn and µ and ν Radon measures in Rn. Denote

spt µ = {x : µ(B(x, r)) > 0 for all r > 0}.

(1) The upper density is

Dµν(x) =

{lim supr→0

ν(B(x,r))µ(B(x,r))

if x ∈ spt µ

+∞ otherwise

(2) The lower density is

Dµ(ν(x)) =

{lim infr→0

ν(B(x,r))µ(B(x,r))

if x ∈ spt µ

+∞ otherwise

If Dµν(x) = Dµν(x), then define Dµν(x) := Dµν(x) to be the density. Notice that Dµν andDµν are Borel measurable.

Lemma 39. For 0 < α <∞,

(1) If A ⊂ {x : Dµν(x) ≤ α}, then ν(A) ≤ αµ(A)

(2) If A ⊆ {x : Dµν(x) ≥ α}, then ν(A) ≥ αµ(A).

Proof. For a complete proof, see Lectures on Geometric Measure Theory by Leon Simon page26. The proof is similar to the proof of Theorem 32 but use Besicovitch’s covering lemmainstead of Vitali’s covering lemma. �

Theorem 40. Let µ and ν be Radon measures in Rn. Then Dµν exists and is finite µ-a.eand is µ-measurable.

Proof. We can assume without loss of generality that µ(Rn) < ∞ and ν(Rn) < ∞ since µand ν are Radon measures and Rn is locally compact and separable.

Define I = {x : Dµν(x) = +∞} ⊂ {x : Dµν(x) ≥ α}. By Lemma (39) for all α > 0,αµ(I) ≤ ν(I). Hence, µ(I) ≤ 1

αν(Rn)→ 0 as α→∞. As a result, µ(I) = 0.

Let 0 < a < b < ∞ and R(a, b) = {x : Dµν(x) < a < b < Dµν(x)}. Note this is a similaridea as proving the Lebesgue differentiation theorem. Next, note that

{Dµν(x) < Dµν(x)} =⋃

0<a,b∈Qa<b

R(a, b).

By Lemma (39), bµ(R(a, b)) ≤ ν(R(a, b)) and ν(R(a, b)) ≤ aµ(R(a, b)). Since a < b, thisimplies that µ(R(a, b)) = 0. Consequently, we showed that for µ a.e. x, Dµν(x) = Dµν(x)

so Dµν(x) exists µ-a.e. and since Dµν(x) <∞ µ-a.e. then Dµν(x) is finite µ-a.e.

It remains to show that Dµν(x) is measurable. To do this, we use the following claim:

Claim 41. For any r > 0 and xi → x, the lim supi→∞ µ(B(xi, r)) ≤ µ(B(x, r)).

13

Proof. (Proof of Claim) Let fi = χB(xi,r) and f = χB(x,r). We claim that lim supi→∞ fi(y) ≤f(y). First, lim sup can only take the values 0 or 1. Next, there is only one case to considerfor if lim supi→∞ fi(y) = 0, then the result holds trivially. Suppose lim supi→∞ fi(x) = 1.Then for all k there exists a jk such that fjk(y) = 1 and |x − xjk | < ε. In particular,y ∈ B(xjk , r). Suppose on way to a contradiction, f(y) = 0. Hence, y 6∈ B(x, r). SinceB(x, r) is closed, there exists an s > 0, such that B(y, s) ⊂ B(x, r)c. Thus, |x− y| ≥ r + s,but

|x− y| ≤ |y − xjk |+ |xjk − x| < r + ε.

By choosing ε small, we get a contradiction. Therefore, lim supi→∞ fi(y) ≤ f(y). Thisimplies that lim infi→∞(1− fi(y)) ≥ 1− f ≥ 0. By Fatou’s Lemma,∫

B(x,2r)

(1− f) dµ ≤ lim infi→∞

∫B(x,2r)

(1− fj) dµ.

As a result,

µ(B(x, 2r))− µ(B(x, r)) ≤ µ(B(x, 2r))− lim supi→∞

µ(B(xi, r)).

Because µ(B(x, 2r)) < ∞ for we assumed the measure of the whole space was finite,lim supi→∞ µ(B(xi, r)) ≤ µ(B(x, r)). �

A function f , defined almost everywhere on Rn is µ-measurable if and only if there existsa sequence of simple A-functions which converge to f almost everywhere where A is theσ-algebra for µ. Thus, we allow f to be undefined on a null set of µ. By the definitionof upper semi-continuous and Claim 41, x 7→ µ(B(x, r)) and x 7→ ν(B(x, r)) are an uppersemi-continuous functions. Recall that upper (lower) semi-continuous functions are Borelmeasurable functions. Since µ is a Radon measure, it contains the Borel sets and so these

functions are µ-measurable. Hence, x 7→ ν(B(x,r))µ(B(x,r))

is a µ-measurable function for each fixed r.

Note that we assume the balls are non-degenerate and have finite measure. Since Dµν(x) =

limr→0ν(B(x,r))µ(B(x,r)

exists for µ-a.e. x ∈ Rn, the definition of µ-measurable gives that Dµν(x) is

µ-measurable. �

Differentiation Theorem for Radon Measures

Theorem 42. Let µ and ν Radon measures with ν � µ. Then

(43) ν(A) =

∫A

Dµν dµ for all A µ−measurable.

Radon-Nikodym: For any µ and ν Radon, there exists νac and νs such that ν = νac + νs,νac � µ, and νs ⊥ µ. In addition, Dµνac = Dµν and Dµνs = 0 for µ-a.e. x and

ν(A) =

∫A

Dµν dµ+ νS(A).

Proof. (Proof of (43)) Let Z = {x : Dµν(x) = 0} and I = {x : Dµν(x) = +∞}. In theprevious proof, we showed that µ(I) = 0. Since ν � µ, ν(I) = 0.

14

Note that Z ⊂ {x : Dµν(x) < α} for all α > 0. By Lemma (39), ν(Z) ≤ αµ(Z) for allα > 0. Letting α→ 0 (and assuming µ(Z) <∞), then ν(Z)→ 0.

Let A be µ-measurable and 1 < t < ∞. The idea is to decompose A into sets where theDµν(x) is almost constant. Let m ∈ Z+ > 0. Define

Am = A ∩ {x : tm ≤ Dµν(x) < tm+1}

and so

A \∞⋃

m=−∞

Am ⊂ (Z ∪ I ∪ {Dµν 6= Dµν})

From above, we showed that µ(Z) = µ(I) = 0. Moreover from Theorem (42), we showed inthe proof that µ({Dµν 6= Dµν}) = 0. Hence, µ

(A \

⋃∞m=−∞Am

)= 0, which implies that

ν(A \

⋃∞m=−∞Am

)= 0 (ν � µ). Because the Am’s are disjoint we have

ν(A) =∞∑−∞

ν(Am) ≤∞∑−∞

tm+1µ(Am) = t∞∑−∞

tmµ(Am) by Lemma (39)

= t∞∑−∞

∫Am

tm dµ

= t∞∑−∞

∫Am

Dµν dµ

= t

∫A

Dµν dµ.

On the other hand, using the definition of Am

ν(A) ≥∞∑−∞

tmµ(Am) =1

t

∞∑−∞

∫Am

tm+1 dµ ≥ 1

t

∫A

Dµν dµ.

Hence,1

t

∫A

Dµν dµ ≤ ν(A) ≤ t

∫A

Dµν dµ.

As this is true for all 1 < t <∞, letting t→ 1, gives the desired equation. �

Proof. (Proof of Radon-Nikodym) Assume µ(Rn) <∞ and ν(Rn) <∞. Define

E = {A ∈ Rn : A Borel, µ(Rn \ A) = 0}.

Choose Bk Borel such that Bk ∈ E and

ν(Bk) ≤ infA∈E

ν(A) + 1k.

Set B =⋂k Bk. We want to show that B is in E . By countable subadditivity, µ(Rn \ B) ≤∑

k µ(Rn \Bk) = 0 since Bk ∈ E . Hence, B ∈ E .

By monotonicity, ν(B) ≤ ν(Bk) ≤ infA∈E ν(A) + 1k. As this is true for all k, let k → ∞ to

conclude that ν(B) = infA∈E ν(A).

15

Define

νac = ν B and νs = ν (Rn \B).

First, it is clear that ν = νac + νs by definition. To see that νs ⊥ µ. Let B and Rn \B be thedisjoint subsets. By definition of νs, it is clear that νs(A) = νs(A∩ (Rn \B)) for all A ∈ Rn.On the other hand, since B ∈ E , µ(Rn \B) = 0. Hence,

µ(A) = µ(A ∩B) + µ(A ∩ (Rn \B))︸ ︷︷ ︸set of measure 0

= µ(A ∩B).

It remains to show that νac � µ. Suppose on way to a contradiction that µ(A) = 0 andνac(A) > 0. One sees that µ(Rn \ (B \A)) ≤ µ(Rn \B) + µ(A) = 0, so B \A ∈ E . However,νac(B \ A) < νac(B) = ν(B). This is a contradiction for ν(B) was the infimum. Therefore,νac � µ.

From the previous proof and νac � µ,

νac(A) =

∫A

Dµνac dµ ⇒ ν(A) =

∫A

Dµνac dµ+ νs(A).

We want Dµνac = Dµν. Since the supports of νac and νs are different, Dµν = Dµ(νac + νs) =Dµνac +Dµνs. Hence, it suffices to show that Dµνs is 0 µ a.e. Set

Cα = {x ∈ B : Dµνs > α},

which implies by Lemma (39), νs(Cα) ≥ αµ(Cα). However since νs is supported on Rn \ Band Cα is a subset of B, νs(Cα) = 0 and so µ(Cα) = 0. This is true for every α > 0 and soit follows that Dµνs is 0 for µ a.e. �

In this section, X is always assumed to be locally compact and separable metric space.

Background: From real analysis, the Riesz Representation Theorem says:Let (X, d) be a locally compact and separable metric space.

(1) Let L : C+c (X,R) → R be a bounded positive linear functional (i.e. if f ∈ Cc(X,R)

and f ≥ 0, then L(f) ≥ 0). Then there exists a unique (finite) Radon measure µsuch that Lf =

∫f dµ.

(2) If T : Cc(X,R)→ R is a bounded linear functional (i.e. |Tf | ≤ C||f ||∞), then thereexists a signed Radon measure µ such that Tf =

∫f dµ.

From Lecture 1, recall that we proved the following theorem about Borel regular measures:Let µ be Borel regular on (X, d). Suppose X =

⋃∞j=1 Vj, Vj open, and µ(Vj) <∞. Then

(1) µ(A) = inf A⊂UU open

{µ(U)} for all A ⊂ A.

(2) µ(A) = sup C⊂AC closed

{µ(C)} for all A µ-measurable.

Important Remark: The finiteness of Radon measures µ on compact sets allows us tointegrate functions which are compactly supported. In fact, let Cc(X,Rm) be the set ofcompactly supported continuous functions from X → Rm, µ a Radon measure, and a µ-measurable function v : X → Rm such that |v| = 1 for µ a.e. x. Then we can associate with

16

each v and µ, a linear functional L : Cc(X,Rm)→ R defined by

Lf =

∫〈f, v〉 dµ,

where 〈·, ·〉 is the inner product. If supp f ⊂ K where K is compact, then

|Lf | ≤ ||f ||∞µ(K).

Hence, if you fix a compact set, this linear functional is bounded.

Theorem 44. (Riesz Representation Theorem)Suppose X is locally compact and separable metric space. Let L : Cc(X,Rm)→ R be a linearfunctional such that for all K ⊂ X, K compact, and

sup{Lf : f ∈ Cc(X,Rm), ||f || ≤ 1, and spt f ⊂ K} <∞.Then there exists a unique Radon measure µ and a µ-measurable function v : X → Rm suchthat |v| = 1 for µ-a.e. x and

Lf =

∫〈f, v〉 dµ for all f ∈ Cc(X,Rm).

17

5. Lecture 5: April 14, 2014

In this section, we give a proof of the Riesz Representation Theorem, the main strategyfor which is as follows: define a set function µ on X on open subsets V ⊂ X by writing

µ(V ) := sup{Lf : f ∈ Cc(V,Rm) with |f | ≤ 1}extended to arbitrary A ⊂ X by

µ(A) := inf{µ(U) : A ⊂ U and U is open }Additionally, let C+

c (X) be the nonnegative functions of Cc(X,R) and write for f ∈ C+c (X)

λ(f) := sup{|Lg| : g ∈ Cc(X,Rm) and |g| ≤ f}We will prove the theorem in steps by showing:

(1) µ is a Radon measure,(2) if f1, f2 ∈ C+

c (X) then λ(f1 + f2) = λ(f1) + λ(f2),(3) if f ∈ C+

c (X) then λ(f) =∫Xfdµ,

(4) there exists ν : X → Rm µ-measurable such that for g ∈ Cc(X,Rm), we have

Lg =

∫X

〈g, ν〉dµ

(5) |ν(x)| = 1 for µ-a.e. x ∈ X.

Proof. (Proof of Riesz Representation Theorem) Noting that µ(∅) = 0 follows from thedefinition (f ≡ 0 provides the function). We prove that µ is a measure using the definition.We first show countable subadditivity relative to open sets and then we extend it to arbitrarysets.

Let V =⋃i≥1 Vi with each Vi open, and let g ∈ Cc(X,Rm) be supported in V with |g| ≤ 1.

As the Vi cover V, they cover the support of g, and thus for some k sufficiently large, wehave

spt(g) ⊂⋃ki=1 Vi

Let {ξi}ki=1 be a partition of unity subordinate to V1, . . . , Vk, so that

ξi ∈ Cc(Vi,Rm) with 0 ≤ ξi ≤ 1 and∑k

i=1 ξi(x) = 1 for x ∈ spt(g)

noting that this gives g ≡ 0 on (∪ki=1Vi)c and g =

∑ξig. Using this and the definition of µ

to estimate Lg, we obtain

Lg =∑k

i=1 L(ξi, g) ≤∑k

i=1 µ(Vi) ≤∑∞

i=1 µ(Vi)

so that taking the supremum over g ∈ Cc(V,Rm) with |g| ≤ 1 gives

µ(V ) ≤∞∑i=1

µ(Vi)

Now, for an arbitrary A ⊂ X, let A ⊂⋃∞i=1Ai. For each ε > 0, we can choose a sequence of

open sets Ui with µ(Ui) ≤ µ(Ai) + 2−iε, so that as µ is monotonic (by definition)

µ(A) ≤ µ(⋃

Ui) ≤∞∑i=1

µ(Ui) ≤ ε+∞∑i=1

µ(Ai),

18

using subadditivity relative to open sets. As ε is arbitrary, we let it go to 0 and obtain thedesired inequality, showing that µ is sub additive and thus a measure.

Now we show it is Borel regular, using Caratheodory’s criterion. Let V1, V2 be open setswith d(V1, V2) > 0, and take gi ∈ Cc(Vi,Rm) with |gi| ≤ 1. As |g1 + g2| ≤ 1 by the triangleinequality, we have

L(g1) + L(g2) = L(g1 + g2) ≤ µ(V1 ∪ V2)by the definition of µ, and so taking the supremum over g1 and g2 gives

µ(V1) + µ(V2) ≤ µ(V1 ∪ V2) ≤ µ(V1) + µ(V2)

by subadditivity, and it follows that µ is additive on metrically separated open sets. Ifd(A,B) > 0 there exist VA and VB open sets such that A ⊂ VA, B ⊂ VB and d(VA, VB) > 0.Using the corresponding result for metrically separated open sets we prove that µ is additiveon metrically separated Borel sets thus µ is Borel regular. Finally, we show that µ is finiteon compact sets, and therefore Radon: as X is locally compact, for every x ∈ X, there is anrx > 0 such that B(x, rx) has compact closure. Fixing K ⊂ X compact, we have

K ⊂⋃x∈K B(x, rx) and so K ⊂

⋃ni=1B(xi, rxi)

for some finite collection x1, . . . , xn ∈ K by compactness. Using subadditivity and thedefinition of µ on open sets, we obtain

µ(K) ≤n∑i=1

µ(∪B(xi, rxi)) ≤n∑i=1

sup{Lg : |g| ≤ 1 and spt(g) ⊂ B(xi, rxi)} <∞

by assumption on L. Thus, µ defines a Radon measure on X, proving claim (1)Now, letting λ be as above, so that for f ∈ C+

c (X)

λ(f) = sup{|Lg| : g ∈ Cc(X,Rm) and |g| ≤ f}If f1 ≤ f2 with fi ∈ C+

c (X) then we have |g| ≤ f1 =⇒ |g| ≤ f2 and so for every c ≥ 0

λ(f1) ≤ λ(f2) and λ(cf1) = cλ(f2)

Choose g1, g2 ∈ Cc(X,Rm) such that |gi| ≤ fi and this gives

|g1 + g2| ≤ f1 + f2

and without loss of generality we can assume that Lg1 and Lg2 are nonnegative (by replacingthem by their negative) and so

Lg1 + Lg2 = L(g1 + g2) = |L(g1 + g2)| ≤ λ(f1 + f2) so that λ(f1) + λ(f2) ≤ λ(f1 + f2)

by taking the supremum over g1, g2. Conversely, if g ∈ Cc(X,Rm) satisfies |g| ≤ f1 + f2 thenfor i = 1, 2 let

gi =

{fi

f1+f2g if f1 + f2 > 0

0 otherwisethen |gi| ≤ fi and g = g1 + g2

so that by definition of λ, we have

|Lg| ≤ |Lg1 + Lg2| ≤ |Lg1|+ |Lg2| ≤ λ(f1) + λ(f2)

so that taking the supremum over |g| ≤ f1 + f2 we obtain

λ(f1 + f2) ≤ λ(f1) + λ(f2)

proving claim (2)

19

Now, fix f ∈ C+c (X) and let ε > 0. Let 0 = t0 < t1 < . . . < tN with

ti − ti−1 < ε3

2‖f‖∞ < tN < 2‖f‖∞ µ(f−1{ti}) = 0

which is possible as f is continuous with compact support and µ ◦ f−1 is a finite measureand thus has at most countably many atoms. Let Ui = f−1

((ti−1, ti)

), so that the Ui are

open and disjoint sets of finite µ measure, and

µ(spt(f) r ∪Ni=1Ui

)= 0

Now, as µ is inner regular, for all U ⊂ X open we can write

µ(U) = sup{λ(h) : h ∈ C+c (U) and h ≤ 1}

using the definition of µ and moreover, for each ε > 0, we can find a compact set K ⊂ U withµ(U rK) < ε. Combining, we have for ε > 0 and each 1 ≤ i ≤ N, a function hi ∈ C+

c (Ui)with

µ(Ui r h−1i {1}

)<

ε

Nand µ(Ui) ≤ λ(hi) +

ε

NFrom the definition of λ,

λ(hi) ≤ sup{|Lg| : g ∈ Cc(Ui,Rm) and |g| ≤ 1} = µ(Ui)

and so we obtain

λ(hi) ≤ µ(Ui) ≤ λ(hi) +ε

NLetting A ⊂ X be the open set

A = {x : f(x)

(1−

N∑i=1

hi(x)

)> 0} =⇒ µ(A) ≤ µ(∪N1 Ui r h−1{1}) ≤ N

ε

N= ε

then from the definition,

λ(f − f∑

hi) ≤ sup{|Lg| : |g| ≤ ‖f‖∞χA} ≤ ‖f‖∞µ(A) ≤ ‖f‖∞ε

Using the second claim proved above, we can write

λ(f) = λ(f − fN∑i=1

hi) + λ(fN∑i=1

hi) ≤ ‖f‖∞ε+N∑i=1

λ(fhi)

As the hi are supported in Ui = {x : ti−1 < f(x) < ti} it follows that

λ(f) ≤ ‖f‖∞ε+N∑i=1

tiµ(Ui)

and as f − f∑hi ≥ 0, we have

N∑i=1

ti−1µ(Ui) ≤ λ(fN∑i=1

hi) ≤ λ(f)

so thatN∑i=1

ti−1µ(Ui) ≤ λ(f) ≤ ‖f‖∞ε+N∑i=1

tiµ(Ui)

20

Now, asN∑i=1

ti−1µ(Ui) ≤∫fdµ ≤

N∑i=1

tiµ(Ui)

we have for all ε > 0

|λ(f)−∫fdµ| ≤ 3ε‖f‖∞ + εµ(spt(f))

and letting ε→ 0 shows that if f ∈ C+c (X), we have λ(f) =

∫fdµ, verifying claim (3).

Now, fix e ∈ Sn−1 and for f ∈ Cc(X), let λe(f) := L(fe), so that

|λe(f)| ≤ sup{|Lg| : g ∈ Cc(X,Rm) and |g| ≤ |f |} ≤ λ(|f |) ≤∫|f |dµ ≤ ‖f‖∞µ(spt f).

Thus λe is a bounded linear functional on Cc(U) where U open and U compact, so that, byRiesz Representation on Cc(X), there is a signed Radon measure ϕ on X with λe(f) =

∫fdϕ.

As |ϕ| � µ the Radon-Nikodym theorem gives us (for each e ∈ Sn−1) a ϕ-measurablefunction ve ∈ L∞(X) with ϕ = veµ. Letting e1, . . . , em be an orthonormal basis of Rm, wedefine the function ν : X → Rm by ν(x) =

∑m1 vej(x)ej so that for g ∈ Cc(X,Rm) we have

L(g) = L(m∑j=1

〈g, ej〉ej) =m∑j=1

∫〈g, ej〉vejdµ =

∫〈g, ν〉dµ,

giving claim (4). We complete the proof by showing that |ν(x)| = 1 for µ-a.e. x ∈ X.If U is an open subset of X with µ(U) <∞, then by definition

µ(U) = sup{Lg =

∫〈g, ν〉dµ : g ∈ Cc(U,Rm) and |g| ≤ 1

}Choosing a sequence of functions fk ∈ Cc(U,Rm) such that |fk| ≤ 1 and fkν → |ν| µ-a.e.,Lebesgue dominated convergence gives∫

U

|ν|dµ = limk→∞

∫〈fk, ν〉dµ ≤ µ(U)

and moreover, as |f | ≤ 1 we have 〈f, ν〉 ≤ |ν| and as µ is a nonnegative measure, this gives∫〈f, ν〉dµ ≤

∫U

|ν|dµ =⇒ µ(U) ≤∫U

|ν|dµ

so that combining the above gives |ν| = 1 µ-a.e., completing the proof. �

21

6. Lecture 6: April 16, 2014

Differentiation of Lipschitz FunctionsDefinition 45.

Let A ⊂ Rn, and f : A→ Rm. We say f is Lipschitz if there is an L > 0 so that

|f(x)− f(y)| ≤ L|x− y| for x, y ∈ A and Lip(f) := supx 6=y∈A

|f(x)− f(y)||x− y|

is the Lipschitz constant of f. We say f is locally Lipschitz provided for every compact setK ⊂ A, the function f |K is Lipschitz; here, the (local) Lipschitz constant may depend on K.

The following theorem shows that we can assume A = Rn without any loss of generality.

Theorem 46 (Extension of Lipschitz Functions).If A ⊂ Rn and f : A→ Rm is Lipschitz, then there exists f : Rn → Rm Lipschitz with

Lip(f) ≤√mLip(f) and f |A = f

Proof. First, assume f : A→ R (m = 1) and write L := Lip(f). For x ∈ Rn define

f(x) := inf{f(a) + L|x− a| : a ∈ A}

so that f |A ≤ f by definition. For every a ∈ A, we have

f(b)− f(a) ≤ |f(b)− f(a)| ≤ L|b− a| =⇒ f(b) ≤ f(a) + L|b− a|

so that, taking the infimum over a ∈ A we obtain f ≤ f |A, and f extends f, as desired.Now,

f(x) = infa∈A{f(a) + L|x− a|} ≤ inf

a∈A{f(a) + L|y − a|}+ L|x− y| = f(y) + L|x− y|

so that f is Lipschitz with Lip(f) = L.For f : A → Rm Lipschitz, the coordinate functions fj := 〈f, ej〉 : A → R and Lipschitz,

and so we apply the above argument to the components fj to obtain the general result. �

The following theorem allows us to extend f without increasing the Lipschitz constant,and is stated without proof.

Theorem 47 (Kriszbraun).If f : A→ Rm is Lipschitz, it has a Lipschitz extension f : Rn → Rm with Lip(f) = Lip(f).

We now discuss differentiability for Lipschitz functions.

Definition 48. The function f : Rn → Rm is differentiable at x if there exists L : Rn →Rm linear such that

limy→x

|f(y)− f(x) + L(x− y)||x− y|

= 0 or equivalently f(y) = f(x) + L(y − x) + o(|x− y|)

If such an L exists it is unique, and we write L = Df(x) for the derivative of f at x.

Theorem 49 (Rademacher).If f : Rn → Rm is locally Lipschitz, then f is differentiable Hn-ae.

22

Proof. By Kriszbraun’s or the Lipschitz extension theorem, it suffices to assume that f isLipschitz on Rn, and as a function into Rn is differentiable if and only if its components aredifferentiable, we may assume without loss of generality that m = 1; let f : Rn → R beLipschitz with L = Lip(f).

For each ν ∈ Sn−1, define

Dνf(x) := limt→0

f(x+ tν)− f(x)

t

We claim that Dνf(x) exists for Hn-ae x ∈ Rn. Writing

Dνf(x) := lim supt→0

f(x+ tν)− f(x)

tand Dνf(x) := lim inf

t→0

f(x+ tν)− f(x)

t

then f locally Lipschitz implies both |Dνf | and |Dνf | are bounded by L. Let

Aν := {x ∈ Rn : Dνf(x) does not exist} = {x ∈ Rn : Dνf(x) < Dνf(x)}Now, as the map t 7→ f(x+ tν) is Lipschitz from R→ R, it is of bounded variation and thusis differentiable Hn-ae. Thus for any line L parallel to ν, we have H1(L ∩ Aν) = 0, so thatintegrating over the lines parallel to ν, we obtain Hn(Aν) = 0, and Dνf(x) exists Hn-ae,verifying the claim.

Next, let e1, . . . , en be an orthonormal basis of Rn, and write

∂f

∂xi:= Deif and grad(f) :=

(∂f

∂x1, . . . ,

∂f

∂xn

)We claim that for ν ∈ Sn−1, we have

Dνf(x) = 〈grad(f), ν〉Take ξ ∈ C∞c (Rn), and observe that by change of variables, one has∫

Rn

f(x+ tν)− f(x)

tξ(x)dx = −

∫Rn

f(x)ξ(x)− ξ(x− tν)

tdx

As the difference quotients 1t(f(x + tν) − f(x)) → Dνf(x) ≤ L < ∞, we let t → 0 in the

above and apply dominated convergence, obtaining∫Rn

Dνf(x)ξ(x)dx = −∫Rn

f(x)Dνξ(x)dx = −∫Rn

f(x)〈grad(ξ), ν〉dx

as ξ is a smooth function. Continuing the above equalities, we obtain∫Rn

Dνf(x)ξ(x)dx = −n∑i=1

(∫Rn

f(x)∂ξ

∂xi(x)dx

)〈ei, ν〉 =

n∑i=1

∫Rn

Deif(x)〈ei, ν〉ξ(x)dx

Which gives us ∫Rn

Dνf(x)ξ(x)dx =

∫Rn

〈grad(f), ν〉ξ(x)dx

and so Dνf(x) = 〈grad(f)(x), ν〉 for a.e. x /∈ Aν . As the set of x where this holds dependson ν, however, we cannot complete the proof until we show that there is a set of full measurefor which the directional derivatives exist and for which the desired identity holds.

As Sn−1 is separable, let D = {νk} be a countable dense subset of Sn−1, and write

A =⋂k≥1

Ak where Ak := {x ∈ Rn : grad(f)(x) exists, and Dνkf(x) = 〈grad(f)(x), νk〉}

23

and note that

Hn(Rn r A) ≤∞∑k=1

Hn(Rn r Ak) = 0

We complete the proof by showing that for all ν ∈ Sn−1 and x ∈ A, f is differentiable at xand Dνf(x) = 〈grad(f)(x), ν〉. For x ∈ A, t 6= 0 ∈ R and ν ∈ Sn−1, write

Q(x, ν, t) :=f(x+ tν)− f(x)

t− 〈grad(f), ν〉

As D is dense in Sn−1, for every ε > 0 and ν ∈ Sn−1 there is a k with

|ν − νk| < ε′ :=ε

2(1 +√n)L

As x ∈ A, we have Q(x, νk, t)→ 0 as t→ 0, so that for each νk ∈ D there is a δk > 0 with

|t| < δk =⇒ |Q(x, νk, t)| <ε

2

As Sn−1 is compact and D is dense, there is N <∞ such that ν1, . . . , νN form a ε′ separatednet in Sn−1; let δ = min1≤k≤N δk > 0.

Observe that for every ν ′ ∈ Sn−1 we have

|Q(x, ν, t)−Q(x, ν ′, t)| ≤ (1 +√n)L|ν − ν ′|

Thus, if x ∈ A, ν ∈ Sn−1 and |t| < δ, take k < N with |νk − ν| ≤ ε′ and

|Q(x, ν, t)| ≤ |Q(x, νk, t)|+ |Q(x, νk, t)−Q(x, ν, t)| ≤ ε

2+ε

2= ε,

and so Q(x, ν, t)→ 0 as t→ 0 for all x ∈ A and ν ∈ Sn−1, as desired. �

Thus, for f : Rn → Rm Lipschitz, we can write

f(x+ tν) = f(x) + t〈grad(f)(x), ν〉+ o(|t|)For y 6= x ∈ Rn, let ν = y−x

|y−x| and take t = |y − x|. Applying the above gives

f(y) = f(x) + 〈grad(f)(x), y − x〉+ o(|y − x|)and so Df(x) = grad(f)(x).

Theorem 50. Let f : Rn → Rm be Lipschitz. For all ε > 0 there exists a g ∈ C1(Rn) suchthat

Hn({x ∈ Rn : f(x) 6= g(x) and Df(x) 6= Dg(x)}

)< ε

Proof. Whitney’s Extension Theorem �