an introduction to geometric measure theory part...

An introduction to Geometric Measure TheoryPart 3: Covering Theorems

Toby O’Neil, 10 October 2016

TCON (Open University) An introduction to GMT, part 3 10 October 2016 1 / 28

Last week. . .

• Talked a lot about Hausdorff measure and dimensions• Determined the Hausdorff dimension of a simple set.

(Eventually!)• Developed some general theory to do with outer measures

and limits of sets.• Introduced t-energies.


Today

1 Finish off the energy discussion.2 Talk about covering theorems.3 Give a classical application of covering theorems.


Nice measures

DefinitionThe family of Borel sets in a metric space X is the smallest�-algebra that contains the open subsets of X .

DefinitionA measure µ is:

1 a Borel measure if the Borel sets are µ-measurable2 Borel regular if it is a Borel measure and for each A ✓ X ,

there is a Borel set B with A ✓ B and µ(A) = µ(B).


TheoremLet µ be a measure on X. Then µ is a Borel measure if, andonly if,

µ(A [ B) = µ(A) + µ(B),

whenever inf{d(x , y) : x 2 A, y 2 B} > 0.

TheoremHs is a Borel regular measure for each s � 0.


Approximating sets

TheoremLet µ be a Borel regular measure on X, let A be a µ-measurableset and let ✏ > 0.

1 If µ(A) < 1, then there is a closed set C ✓ A for whichµ(A \ C) < ✏.

2 If there are open sets V1,V2, . . . with A ✓ S1i=1 and

µ(Vi) < 1 for each i, then there is an open set V withA ✓ V and µ(V \ A) < ✏.


Radon measuresDefinitionA Borel measure µ is a Radon measure on X if

1 all compact subsets of X have finite µ-measure2 for open sets V ,

µ(V ) = sup{µ(K ) : K ✓ V is compact}3 for each set A ✓ X ,

µ(A) = inf{µ(V ) : V is open and A ✓ V}.

CorollaryA measure µ on Rn is a Radon measure if and only if it is locallyfinite and Borel regular.


Energies

DefinitionFor A ✓ Rn, let M(A) denote the set of all compactly supportedRadon measures µ with 0 < µ(A) < 1 and with supportcontained in A.

Definition (t-energy)For a Radon measure µ on Rn and t � 0, we define the t-energyof µ by

It(µ) =ZZ

1|x � y |t dµ(x)dµ(y).

(This may be infinite.)


Aside: integration

Theorem (Fubini)Let X and Y be separable metric spaces with µ and ⌫ locallyfinite Borel measures on X and Y , respectively.If f is a non-negative Borel function on X ⇥ Y, then

ZZf (x , y) dµ(x)d⌫(y) =

ZZf (x , y) d⌫(y)dµ(x).

In particular, if f is the characteristic function of a Borel set, thenZ

µ({x : (x , y) 2 A}) d⌫(y) =Z

⌫({y : (x , y) 2 A}) dµ(x).


More integration: a useful equation

TheoremLet µ be a Borel measure and f a non-negative Borel functionon a separable metric space X. Then

Zf dµ =

Z 1

0µ({x 2 X : f (x) � t}) dt .


Another view of It(µ)

It(µ) = tZZ 1

0r�t�1µ(B(x , r)) dr dµ(x)

= tZ 1

0r�t�1

Zµ(B(x , r)) dµ(x) dr

TheoremIf A is a Borel set in Rn, then

dimH(A) = sup{t : There is µ 2 M(A) with It(µ) < 1}.(M(A) denotes the collection of non-zero Radon measures withcompact support that is contained in A.)


An application of energies

TheoremFix integers 0 m n. and let A be a Borel set in Rn. Then foralmost every m-dimensional linear subspace V ,

dimH(PV (A)) = min{dimH(A),m}.

(Here PV denotes orthogonal projection onto V .)


A basic covering theorem

TheoremSuppose that B is a family of closed balls in a metric space X forwhich

sup{diam(B) : B 2 B} < 1.

Then there is a disjoint subfamily C ✓ B such that for eachB 2 B, there is C 2 C with C \ B 6= ; and diam(B) 2diam(C).(So

SB B ✓ SC2C 5C.)

Note that if X is separable, then we can assume that C is atmost countable.


Proof of basic covering theoremLet ⌦ be the collection of all disjoint subfamilies C of B with thefollowing property:

Whenever B 2 B• either B \ C = ; for each C 2 C;• or for some C 2 C, B \ C 6= ; and diam(B) 2diam(C).

Ordering ⌦ by inclusion, we now apply Hausdorff’s maximalprinciple to find D 2 ⌦ that is not a proper subset of any memberof ⌦.Let E = {B 2 B : B \SD = ;}. If E 6= ;, then we could findC 2 E so that

2diam(C) � sup{diam(B) : B 2 E},meaning that D [ {C} 2 ⌦, contradicting the definition of D.


Besicovitch covering theorem

TheoremFix n 2 N. There are integers P(n) and Q(n) such that thefollowing holds.Let A be a bounded subset of Rn and let B be a family of closedballs such that each point of A is the centre of some ball in B.

1 There is a finite or countable collection of balls Bi 2 Bcovering A and such that each point of Rn lies in at mostP(n) balls Bi . That is, 1A Pi 1Bi P(n).

2 There are families B1,B2, . . . ,BQ(n) ✓ B such that each Bi isa disjoint collection of balls and for which the union over allthe collections covers A.


Basic observationThere is a positive integer N(n) such that whenever pointsa1, . . . , ak 2 Rn and r1, . . . , rn > 0 have the property that

ai 62 B(aj , rj) for i 6= j andk\

i=1

B(ai , ri) 6= ;,

then k N(n).


Vitali covering theorem

TheoremLet µ be a Radon measure on Rn and A ✓ Rn and suppose thatB a family of closed balls such that each point in A is the centreof arbitrarily small balls of B.Then there is a countable (or finite) sequence of balls Bi 2 Bsuch that

µ

A \

[

i

Bi

!= 0.


Sketch proof 1We may assume that µ(A) > 0.We also suppose that A is bounded.Since µ is Radon, can find an open set U ◆ A for whichµ(U) (1 + ()4Q(n))�1)µ(A).Hence by BCT, can find disjoint collections B1, . . . ,BQ(n) ✓ Bsuch that

A ✓Q(n)[

i=1

[Bi ✓ U.

Hence there is i for which µ(SBi) � µ(A)/Q(n) and so we can

find a finite subfamily B0i ✓ Bi for which

µ([

B0i ) � µ(A)/(2)Q(n)).


Sketch proof 2So if A1 = A \SB0

i , then

µ(A1) µ⇣

U \[

B0i

⌘= µ(U)� µ

⇣[B0

i

⌘

(1 + (4Q(n))�1 � (2Q(n))�1)µ(A) = uµ(A),

where u < 1.Now A1 is contained in the open set U \SB0

i and so can find anopen set U1 such that A ✓ U1 ✓ U \SB0

i andµ(U1) (1 + (4Q(n))�1)µ(A1).Proceeding as above, we obtain a set A2 equal to A1 minusfinitely many balls from B that lie in U1 and for whichµ(A2) uµ(A1) u2µ(A).Iterating (countably many times) gives the result.


Application of the Vitali covering theorem

LemmaLet µ and ⌫ be Radon measures on Rn, 0 < t < 1 and A ✓ Rn.

1 If lim infr&0µ(B(x ,r))⌫(B(x ,r)) t for all x 2 A, then µ(A) t⌫(A).

2 If lim supr&0µ(B(x ,r))⌫(B(x ,r)) � t for all x 2 A, then µ(A) � t⌫(A).

(B(x , r) denotes the closed ball here.)


Differentiation of measures

Definition (Derivatives)Suppose that µ and ⌫ are locally finite Borel measures on Rn.We define the upper and lower derivatives of µ with respect to µat x 2 Rn by

D(µ, ⌫, x) = lim supr&0

µ(B(x , r))⌫(B(x , r))

andD(µ, ⌫, x) = lim inf

r&0

µ(B(x , r))⌫(B(x , r))

.

If the limits are the same, then we denote the common value byD(µ, ⌫, x) and call it the derivative of µ with respect to ⌫ at x .


TheoremLet µ and ⌫ be Radon measures on Rn.

1 The derivative exists and is finite for ⌫-almost very x 2 Rn.2 For all Borel sets B ✓ Rn,

Z

BD(µ, ⌫, x) d⌫(x) µ(B)

with equality if, and only if, µ is absolutely continuous withrespect to ⌫.

3 µ is absolutely continuous with respect to ⌫ if, and only if,D(µ, ⌫, x) < 1 for µ-almost every x.


Corollary (Lebesgue density theorem)Let µ be a Radon measure on Rn and A ✓ Rn.For µ-almost every x 2 A,

limr&0

µ(A \ B(x , r))µ(B(x , r))

= 1.

If, furthermore, A is µ-measurable, then for µ-almost everyx 2 Rn \ A

limr&0

µ(A \ B(x , r))µ(B(x , r))

= 0.


Some references

K. J. Falconer, The geometry of fractal sets, CambridgeUniversity Press, 1985.

K. J. Falconer, Fractal geometry, Wiley 2nd Edition, 2003.

H. Federer, Geometric Measure Theory, Springer, 1969.

P. Mattila, Geometry of sets and measures in Euclidean spaces,Cambridge University Press, 1995.


Problems

Homework, if interested.1 Provide the missing details in the proof of the Vitali

Covering Theorem.2 Let µ and ⌫ be Radon measures on Rn, 0 < t < 1 and

A ✓ Rn. Prove:1 If lim infr&0

µ(B(x ,r))⌫(B(x ,r)) t for all x 2 A, then µ(A) t⌫(A).

2 If lim supr&0µ(B(x ,r))⌫(B(x ,r)) � t for all x 2 A, then µ(A) � t⌫(A).


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