On indecomposable sets with applications

DOI:10.1051/cocv/2013077 www.esaim-cocv.org

ON INDECOMPOSABLE SETS WITH APPLICATIONS

Andrew Lorent

Abstract.In this note we show the characteristic function of every indecomposable setFin the plane isBV equivalent to the characteristic function a closed setF,i.e.11_F −11_FBV(IR²₎= 0. We show by example this is false in dimension three and above. As a corollary to this result we show that for every >0 a set of finite perimeter S can be approximated by a closed subset S with finitely many inde- composable components and with the property thatH¹(∂^MS\∂^MS) = 0 and11_S−11_SBV(IR2)< . We apply this corollary to give a short proof that locally quasiminimizing sets in the plane are BVl

extension domains.

Mathematics Subject Classification. 28A75.

Received May 14, 2013. Revised October 29, 2013.

Published online March 28, 2014.

1. Introduction

Sets of finite perimeter are the largest class of sets that permit a broad theory of analysis. They have wide application in the Calculus of Variations, PDE, image processing and fracture mechanics. In some sense the theory of sets of finite perimeter is an analogue for sets of what the theory of Sobolev functions is for functions.

A very useful notion for Sobolev functions is the notion of precise representative for a function f ∈ W^1,p, this is a function ˜f ∈W^1,p with f −f˜_W^1,p = 0 and ˜f has additional smoothness and regularity properties.

Another very useful result is Whitney’s theorem that for any >0 there exists a smooth function ˆf such that fˆ−f_W^1,p < . We prove analogues results for sets of finite perimeter in the plane and we apply them to give a short proof that quasiminimizing sets areBVl extension domains.

Recall a setE is said to be offinite perimeter in domainΩif it is measurable and Per (E, Ω) := sup

Edivφ:φ∈

C_c¹(Ω)_N

,φ_∞≤1

<∞.

A set is simply called a set of finite perimeter if it is of finite perimeter in IRⁿ and we define Per(E) :=

Per(E,IRⁿ). When dealing with sets of finite perimeter it is common to work with the representative that has negligible points removed. Given a set E we define Neg(E) := {x∈E:|B_r(x)∩E|= 0 for somer >0}. Let E=E\Neg(E). By Proposition 3.3 [1] we know11E−11_EBV(IR²₎= 0. If a setF is such that Neg(F) =∅ we sayF isshaved.

Keywords and phrases.Sets of finite perimeter, indecomposable sets.

1 Mathematics Department, University of Cincinnati, 2600 Clifton Ave. Cincinnati OH 45221.lorentaw@uc.edu

Article published by EDP Sciences c EDP Sciences, SMAI 2014

A set of finite perimeter is calledindecomposable iff for any disjoint subsetsA, B⊂E such that E=A∪B we have Per(E) = Per(A) + Per(B) then either|A|= 0 or|B|= 0.

The main theorem we will establish in this note is the following.

Theorem 1.1. If F ⊂IR² is a shaved indecomposable set then the closureF¯ of F has the property 11_F−11_F¯BV(IR²₎= 0.

Consequently the characteristic function of any indecomposable set is BV equivalent to the characteristic func- tion of a closed set.

A straightforward corollary to this is:

Corollary 1.2. SupposeE ⊂IR² is a set of finite perimeter, then for any > 0 we can find a closed subset E⊂E with finitely many indecomposable components such thatH¹(∂^ME\∂^ME) = 0and

11_E−11_EBV(IR²₎< . (1.1) We will show by example that Theorem 1.1 is false for indecomposable sets of finite perimeter in IRⁿ for n≥3. Our example also shows that Theorem 7 of [2] for saturated indecomposable sets in the plane does not hold true in dimension three and above. It is unclear to us if Corollary1.2is true in dimension three and above.

Theorem 1.3. There exists a shaved set of finite perimeter S ⊂ Q = (−1,1)×(−1,1)×(−1,1) with the following properties

(i) S is connected and S^c is connected. HenceS is an indecomposable saturated set.

(ii) S¯\S> ³⁹⁹⁹₄₀₀₀. (iii) H² ∂^MS\φ(S²)

>0 for any Lipschitz mapφ:S²→IR³.

Our main application of Theorem1.1will be to show that local quasiminimizers in the plane areBVlextension domains. This is already known as a consequence of the work of David−Semmes [5], however our proof is much shorter. Specifically we say a setE of finite perimeterEis aK-quasiminimal set inΩ iff for all openU ⊂⊂Ω and all Borel setsF, G⊂U we have Per(E, U)≤KPer((E∪F)\G, U). And we say a setE of finite perimeter is locally K-quasiminimizing if there exists δ > 0 such that for any x ∈ ∂E the set E is a K-quasiminimal in Bδ(x). In the case whereE is boundedδcan be chosen depending onx.

Finally a setE of finite perimeter is a BVl extension domain if and only if there are constants c ≥1 and δ >0 such that wheneveru∈BV(E) is such that the diameter of the support ofuis smaller thanδ, then there is a functionT u∈BV such thatDT u ≤cDu(E) andT u=uonE.

Corollary 1.4. IfE ⊂IR² is a locally K-quasiminimizing set then it is aBV_l extension domain.

Note also that if Corollary1.2were true in dimension three and above then the proof of Corollary1.4would work in these dimensions too. A generalization of Corollary 1.2 to higher dimensions could potentially be a useful technical tool in the study of sets of finite perimeter.

In addition we will obtain the following corollary which is also an easy corollary to Theorem 7 [2]. Firstly some definitions, Ambrosioet al.[2] (Def. 2) define aholeof a set of finite perimeterSto be an indecomposable component ofS^c with finite measure. A setSis calledsaturated(again see Def. 2, [2]) if it is the union of itself and all its holes.

Corollary 1.5 (To Thm.1.1). Suppose S ⊂ IR² is a indecomposable saturated set then there exists an open setS such that 11_S−11_SBV(IR²₎= 0.

ANDREW LORENT

2. Sketch of proof of main theorem

The proof of the Theorem1.1 follows from three basic steps. Each follows from the last in a fairly natural way. We will firstly state the steps then sketch the reasons they hold afterwards.

SinceE is a shaved indecomposable set it has the property

|Br(x)∩E|>0 for anyx∈E, r >0. (2.1) Step 1.LetZ :=

z∈R²: lim sup_r→0^|E∩B_πr^r2^(x)|≥ ¹₂

. We will show we can find a countable collection of balls {B_rn(x_n) :x_n∈Z}with the following properties.

(i)

B^rn₅ (xn) :n∈N

are disjoint.

(ii) Z⊂

nB_rn(x_n).

(iii) |Brn(xn)∩E| ≥ ^πr₄ⁿ² for eachn.

(iv) Π :=

nBrn(xn) is connected.

Step 2.We will show that forH¹a.e.x∈∂E\Z there exists r_x>0 such that H¹(∂E∩Br(x))> r

1600 for allr∈(0, rx]. (2.2)

Step 3.We will show thatH¹(∂E\Z) = 0.

Sketch of Step 1. By definition of Z for any z ∈ Z we can find r_z > 0 such that |B_r(z)∩Z| ≥ ^πr₄² for all r ∈ (0, rz]. So by the 5r Covering Theorem we can find a subcollection

Brzn(zn) :n∈N

such that Brzn

5 (z_n) :n∈N

are pairwise disjoint and Z ⊂

nB_rzn(z_n). The only remaining issue is to show that Π :=

nBr_zn(zn) is connected. Suppose it is not, so there are two non-empty disjoint connect components of Π₀ and Π₁ such that Π =Π₀∪Π₁. LettingQ₀ =E∩Π₀ and Q₁ = E∩Π₁ it is possible to show that H¹(∂^MQ₀∩∂^MQ₁) = 0 and this impliesE is not an indecomposable set, contradiction. ThusΠ is connected.

Sketch of Step 2.Now we can assume without loss of generality thatx∈E⁰. So we can findpx>0 such that

|E∩Br(x)|< ₁₀₀₀₀^r2 for allr∈(0, px). Note also by (2.1) for anyx∈∂E we have thatE∩B₁₀₀₀^r (x)>0 for any r >0. So letr ∈(0, px), by Step 1 we have a countable collection of balls{Brn(xn) :n∈N} that satisfy (i),(ii),(iii),(iv). A subcollection of these balls

B_rpk(x_pk) :k∈N

is such that B_rpk(x_pk)∩B_r(x) =∅ for any k∈N, E∩B_r(x)⊂

kB_rpk(x_pk) and Π :=

kB_rpk(x_pk) is a connected set. So Π is a ‘tentacle’ of balls that reaches from the outside ofB_r(x) to B₁₀₀₀^r (x). And note that any ballB_rpk(x_pk) has at least a quarter of its area is filled byE. On the other hand most ofBr(x) is empty ofE. So assume for the moment for simplicity that the tentacle reaches intoB₁₀₀₀^r (x) in something like a line, pick a directionvthat is roughly orthogonal to the line. Now there must be a large set of lines in directionv running throughBr(x) that start at some point in E∩B_r(x) and end in some point in E^c∩B_r(x). The variation of 11_E restricted to any of these lines is at least 1, so integrating across these lines gives (2.2). Note the reason our proof works in IR² and does not work in higher dimension²is that a ‘tentacle’ in higher dimension has arbitrarily small surface area, where as in two dimension the surface area of a tentacle isO(1).

Sketch of Step 3. SupposeH¹(∂E\Z)>0. Assume for simplicityH¹(∂E\Z)<∞. Let μ(A) :=H¹(∂^ME∪ (∂E\Z)). We can find an open setUwith∂^ME⊂U and a compact setC⊂∂^MEsuch thatμ(U)≤μ(∂^ME)+

andμ(C)≥μ(∂^ME)−. We can assumeis sufficiently small so thatμ(∂E\U)>^H1(∂E\Z)₂ .

2And recall indeed by the example constructed in Theorem1.3shows the result in false in higher dimension.

Now since δ = dist(∂U, C) > 0 we can find a countable collection of pairwise disjoint balls {B_rn(x_n) :x_n∈∂E\U, r_n< δ} such that

nr_n ≥ c₀μ(∂E\U) some constant c₀ > 0. By Step 2 we have that

nH¹(∂^ME∩Brn(xn))≥₁₆₀₀^c0 μ(∂E\U). SinceBrn(xn)∩C=∅ for allnwe have

n

H¹(∂^ME∩Brn(xn)) +H¹(C)≥ c0

1600μ(∂E\U) +H¹(∂^ME)− which is contradiction for small enough.

3. Sketch of proof of the application to quasiminimizing sets

As stated the main application of Theorem1.1is Corollary1.4. So to establish this we will use the criteria for BVl extension domains found in [3,4]. Namely E is aBVl extension domain if for every set of finite perimeter F ⊂E with diam(F)< γ we can find ˆF withF ⊂Fˆ and Per( ˆF ,IR²)≤CPer(F, E). Following the method of (the preprint form of) [6] we take ˆF to be equal toF and it will suffice to show that

Per( ˆF ,IR²)≤(1 +K)Per(F, E) (3.1) for any set of finite perimeterF ⊂E with diam(F)< γ. We will achieve this in the following way: for anyδ >0 we will find an open setΩ withF ⊂⊂Ωand

H¹(∂^ME∩Ω)≤H¹(∂^ME∩∂^MF) +δ. (3.2) Then by the fact thatE is a localK-quasiminimizer Per(E, Ω)≤KPer(E\F, Ω). Now

H¹(∂^ME∩∂^MF) ≤ H¹(∂^ME∩Ω)

= Per(E, Ω)

≤ KPer(E\F, Ω)

= K H¹(∂^MF∩E) +H¹

∂^ME\∂^MF

∩Ω

(3.2)

≤ KPer(F, E) +Kδ. (3.3)

As this holds for arbitraryδ > 0 we have inequality (3.1). Note this inequality can not work unless we can show (3.2) and hence establish H¹((∂^ME\∂^MF)∩Ω)< δ. Now for arbitrary sets of finite perimeterF ⊂E it is not true we can findΩ such thatF ⊂⊂ Ωand (3.2) holds true. For a counter example letE =B₁(0) and pickz0∈∂E, letα∈(0,1) and letζk be the set of points with rational coordinates inE∩Bα(z0). Then the set F =

kB_α2^−(1000+n)(ζn)∩Eis a set of finite perimeter for which (3.2) is false for any open setΩwithF ⊂⊂Ω.

However we will be able to carry out this argument by replacingF by the setFafforded to us by Corollary1.2.

The set F has almost the same characteristics of F and we can find an open set Ω with F ⊂⊂ Ω such that H¹(∂^ME∩Ω)≤H¹(∂^ME∩∂^MF) +δ. So we can carry out the chain of inequalities to establish (3.3). Having established (3.3) for F the same inequality follows forF with arbitrarily small error by (1.1). See Lemma 6.4 and Section6.2 for full details.

4. Preliminaries

As in Definition 3.60 [1] we everyt∈[0,1] we letE^tdenote the set of points oft density,i.e.

E^t:=

x∈IRⁿ : lim

r→0

|E∩Br(x)|

Γ(n)rⁿ =t

(4.1) whereΓ(n) =|B₁(0)|.

ANDREW LORENT

It is a fundamental result of Federer (see Thm. 3.61 [1]) that if E is a set of finite perimeter in Ω, Hⁿ⁻¹

Ω\

E⁰∪E¹² ∪E¹

= 0.

The measure theoretic boundary is defined by

∂^ME:=

x∈IRⁿ: lim sup

r→0

|E∩Br(x)|

|B_r(x)| >0 and lim sup

r→0

|E^c∩Br(x)|

|B_r(x)| >0

.

It is well-known thatHⁿ⁻¹(∂^ME\E¹²) = 0 and|D11E|=H_∂ⁿ⁻¹ME, see Theorems 3.59 and 3.61 [1].

5. Preliminary lemmas

Lemma 5.1. SupposeE⊂IR² is indecomposable. Then we can find a countable collection {Brn(xn) :xn ∈E}

such that

n

B_rn(x_n)

is connected, (5.1)

|B_r(x_n)∩E|>πr²

4 for allr∈(0, r_n], n∈N. (5.2)

As a consequence forH¹ a.e.x∈∂E\(E¹∪∂^ME)there existsrx>0such that H¹(∂E∩B_r(x))> r

1600 for allr∈(0, r_x]. (5.3)

Proof of Lemma 5.1. LetZ =E¹∪E¹². By Theorem 3.61 [1]

H¹(E\Z) = 0. (5.4)

So for anyx∈Z there existsrx>0 such that

|B_r(x)∩E|> πr²

4 for allr∈(0, r_x]. (5.5)

Step 1. By the 5r Covering Theorem (see Thm. 2.11 [8]) we can find a disjoint sub-collection Brxn

5 (xn) :xn∈Z

such thatZ⊂

nBrxn(xn) =:Π. We will showΠ is connected.

Proof of Step 1. We argue by contradiction. Suppose Π is disconnected. Let Π₀ be a connected non-empty component of Π and let Π₁ = Π\Π₀. Now define Q₀ = E ∩Π₀ and Q₁ = E ∩Π₁. These are both the intersection of two sets of finite perimeter and hence are sets of finite perimeter.

We claim

H¹((E\Q₀)\Q₁) = 0. (5.6)

So letω=E\Z, noteH¹(ω)^(5.4)= 0 andE\(Q0∪Q1) =E\Π ⊂(Z∪ω)\Π ⊂ω. So equality (5.6) follows. We also claim

Q1\(E\Q0) =∅. (5.7)

Now (Q₁∪Q₀)\E=Π∩E\E=∅ so (5.7) is immediate. So we actually have

H¹(Q₁(E\Q₀)) = 0. (5.8)

By Proposition 3.38 [1] this is more than enough to conclude

Per(Q₁) = Per(E\Q₀). (5.9)

We will show

H¹

Q₀¹² ∩Q₁¹²

= 0. (5.10)

Now ifx∈Q₁¹² we must have

r→0lim

|E∩Π₁∩B_r(x)|

πr² = 1 2 so

r→0lim

|E∩B_r(x)|

πr² ≥1

2· (5.11)

Hence (5.11) together with Theorem 3.61 [1] implies H¹

Q₁¹²\Z

= 0 so

H¹(Q₁¹²\Π) = 0. (5.12)

Now note that ifx∈Q₁¹² ∩Π we can not havex∈Π0 because Π0 is open and soBδ(x)⊂Π0 for someδ >0.

By the factx∈Q₁¹² also we must be able to findy∈Bδ(x)∩Q1which contradicts the factΠ1,Π0are disjoint.

Thus

Q₁¹² ∩Π ⊂Q₁¹² ∩Π₁. (5.13)

And in the same way sinceΠ1 is open for all small enoughrwe have thatBr(x)⊂Π1. Thus ifx∈Q₁¹²∩Π1

by definition ofQ₁¹², lim_r→0^|(E∩Π1_πr^)c2^∩Br(x)|= ¹₂ we actually have

r→0lim

|E^c∩Br(x)| πr² =1

2 hencex∈E¹². Thus

Q₁¹² ∩Π₁⊂E¹². (5.14)

Hence

H¹ Q₁¹²\

E¹² ∩Π1

≤H¹

Q₁¹²\Π +H¹

Q₁¹² ∩Π\Q₁¹²∩Π₁ +H¹

Q₁¹²∩Π₁\E¹² ∩Π₁

(5.12),(5.13),(5.14)

= 0. (5.15)

Now going in the opposite direction ifx∈E¹² ∩Π1again since Π1 is open for all small enoughrwe have that

r→0lim

|B_r(x)∩E|

πr² = lim

r→0

|B_r(x)∩E∩Π₁| πr² =1

2 and

r→0lim

|B_r(x)∩E^c| πr² = lim

r→0

|B_r(x)∩(E∩Π₁)^c|

πr² =1

2·

SoE¹² ∩Π1⊂(E∩Π1)¹² =Q₁¹² hence putting this together with (5.15) we have established H¹

E¹² ∩Π1Q₁²¹

= 0·

ANDREW LORENT

In exactly the same way we can show thatH¹(E¹²∩Π2Q₂¹²) = 0. SinceΠ1,Π0are disjoint this completes the proof of (5.10).

Now E = Q₀∪Q₁ and Per(Q₀) = H¹(Q₀¹²), Per(Q₁) = H¹(Q₁¹²) and Per(E) = Per(Q₀) + Per(Q₁) ^(5.9)= Per(Q0) + Per(E\Q0). So as Q0, E\Q0 are both sets of finite perimeter this contradicts the fact E is indecomposable. This concludes the proof of Step 1.

Step 2.We will establish (5.3).

Proof of Step 2. Firstly by Theorem 3.61 [1] we can assume x∈E⁰. Now recall from the sketch of the proof (see property (2.1)), for anyy ∈E we have|Br(y)∩E|>0 for anyr >0. Thus sincex∈∂Eit has the same property. Now sincex∈E⁰ we can findp_x>0 such that

|E∩Br(x)|< r²

100000 for allr∈(0, px). (5.16)

However we must also have thatE∩B₁₀₀₀^r (x)>0. So by property (i) we have established in Step 1 we can find a countable collection

Brk(xk) :xk ∈E, rk < r 1000

such thatE¹∪E¹² ⊂

kB_rk(x_k) and (

kB_rk(x_k)) is connected. Now pick a point x₀ ∈E¹∩B₁₀₀₀^r (x) and a pointy₀∈E¹\B_r(x). Since

kB_rk(x_k) is open and connected it is path connected and so we must be able to find a path φ: [0, t]→

kB_rk(x_k) withφ(0) =x₀,φ(t) =y₀. Let s∈(0, t) be the smallest number such that φ(s)∈∂B^r₄(x), by compactness clearly this number exists. So

{φ(w) : 0≤w≤s} ⊂B^r₄(x). (5.17) Letv=_|φ(s)−x^φ(s)−x0

0|. For any vectorwletw:={λw:λ∈IR}. DefinePV(x) to be the orthogonal projection ofx onto subspaceV. Note that for everyk

H¹ P_v(Br(xk)∩E)

≥ r

16 for allr∈(0, rk] (5.18)

since if this was not true we would have that|Br(xk)∩E| ≤ ^r₈² which contradicts (5.2).

We knowφ([0, t])⊂

kBrk(xk), let

Br_pk(xpk) :k∈N

be a subcollection defined by

{φ(w) : 0≤w≤s} ∩Br_pk(xpk)=∅ for anyk. (5.19) So

{φ(w) : 0≤w≤s} ⊂

k

Br_pk(xpk). (5.20)

And

k

B_rpk(x_pk)(5.19),(5.17)

⊂ B^r₃(x). (5.21)

Now since

P_v({φ(w) :w∈[0, s]})^(5.20)⊂

k

P_v

B_rpk(x_pk)

(5.22) by the 5rCovering Theorem (see Thm. 2.11 [8]) we can find a subcollection

Br_qk(xqk) :k∈N

such that P_v

B_rpk(x_pk)

:k∈N

⊂ P_v

B_rqk(x_qk)

:k∈N

(5.23)

and

P_v Brqk

5 (x_qk)

:k∈N

are disjoint. (5.24)

Now as H¹ P_v({φ(w) :w∈[0, s]})

≥ ^r₅ so putting this together with (5.22) and (5.24) we have

k

rqk

(5.24)

=

k

2⁻¹H¹

P_v(Br_pk(xpk))

(5.22)

≥ 2⁻¹H¹ P_v({φ(w) :w∈[0, s]})

≥ r

10. (5.25)

So let O:=

kBrqk

5 (x_qk)∩E. By (5.18), (5.24) and (5.25) we have H¹ P_v(O)(5.24)

=

k

H¹ P_v

Brqk

5 (xqk)∩E

(5.18)

≥

k

r_qk 80

(5.25)

≥ r

800· (5.26)

Now by (5.21)O ⊂B^r₃(x). We claim we can find a subset Π⊂P_v(O) with|Π| ≥ ₁₆₀₀^r such that H¹

P_v⁻¹(ω)∩E^c∩A x,r

3, r

>0 for anyω∈Π. (5.27)

Suppose this is not true. So there is a setΛ⊂P_v(O) where|Λ| ≥₁₆₀₀^r such thatH¹(P_v⁻¹(ω)∩E^c∩A(0,^r₃, r)) = 0 for allω ∈Λ. Let Ξ=

ω∈ΛP_v⁻¹(ω)∩E∩A(0,^r₃, r) and by Fubini|Ξ| ≥ |Λ|^2r₃ ≥ ₂₄₀₀^r2 and|Ξ∩E^c|= 0 and this contradicts (5.16). We have established (5.27).

Now by Theorem 3.103 [1] we know that r 1600 ≤

w∈ΠV

11_E, P_v⁻¹(w)∩Br(x)

dw

≤V (11_E, B_r(x))

=H¹ ∂^ME, Br(x) .

So this establishes (5.3).

Proof of Theorem 1.1. Firstly as before, without loss of generality we can assume that for any x ∈ F,

|F∩Bδ(x)|>0 for allδ >0.

Step 1.We will showH¹( ¯F\(∂^MF∪F¹)) = 0.

Proof of Step 1. Suppose

H¹( ¯F\(∂^MF∪F¹))>0. (5.28)

LetZ = ¯F\(∂^MF ∪F¹). Note that Z∩Int(F) =∅because if x∈Z∩Int(F) then x∈F¹ which contradicts the definition ofZ. So asZ⊂F¯ we know thatZ⊂∂F and thusZ⊂∂F\∂^MF.

ANDREW LORENT

Let > 0. If H¹(Z) = ∞ pick B ⊂ Z with H¹(B) = 1 and define S = B ∪∂^MF otherwise define S= ( ¯F\F¹)∪∂^MF. Letμ(A) :=H¹(A∩S). Note

μ(∂F\∂^MF)≥min

H¹( ¯F\(F¹∪∂^MF)), H¹(B)

=β >0. (5.29)

Measureμis Radon so we can find an open setU such that∂^MF ⊂U such that

μ(U)< μ(∂^MF) + (5.30)

and we can find a compact setC⊂∂^MF such that

μ(C)> μ(∂^MF)−. (5.31)

And so

dist(C, ∂U) =δ >0. (5.32)

We can take >0 and a subsetΓ0⊂∂F\U with

H¹((∂F\U)\Γ₀)< (5.33)

and for anyr∈(0, ), x∈Γ₀ we have thatμ(B_r(x)) =H¹(B_r(x)∩S)≤2r. By Lemma5.1we can findσ >0, Γ₁⊂Γ₀ such that

H¹(∂^MF∩Br(x))> r

1600 for allx∈Γ1, r∈(0, σ). (5.34) And

H¹(Γ₀\Γ₁)< . (5.35)

Now by Vitali covering theorem (see Thm. 2.8 [8]) we can find a pairwise disjoint collection {B_rk(x_k) :x_k ∈Γ₁}

such that

μ(Γ₁\(^∞

k=1

B_rk(x_k))) = 0 (5.36)

and

sup{rk:k∈IN}<min δ

2, , σ

. (5.37)

Note

β ≤ μ(∂F\∂^MF)

= μ(∂F\U) +μ(∂F∩U\∂^MF)

≤ μ(∂F\U) +μ(U\∂^MF)

(5.30)

≤ μ(∂F\U) +. Soμ(∂F\U)> β−. Now

k

μ(Brk(xk))(5.36),(5.35),(5.33)

≥ μ(∂F\U)−2^(5.30)≥ μ(∂F\∂^MF)−3. (5.38) By (5.34) and the factx_k∈Γ₁,r_k< σ we have

1600H¹(∂^MF∩B_rk(x_k))≥r_k for allk.

Since we chooser_k < andx_k∈Γ₁⊂Γ₀be definition ofΓ₀,μ(B_rk(x_k))≤2r_k so

μ(B_rk(x_k))≤3200H¹(∂^MF∩B_rk(x_k)) (5.39) thus putting this together with (5.38) we have

k

3200H¹(∂^MF∩Brk(xk))≥μ(∂F\∂^MF)−3^(5.29)≥ β

2. (5.40)

Now sincexk ∈Γ1⊂∂F\U by (5.32) and (5.37) we know thatC∩Brk(xk) =∅ for anyk, so H¹(∂^MF)≥H¹(C) +

k

H¹(Brk(xk)∩∂^MF)

(5.40),(5.31)

≥ H¹(∂^MF) + β

6400− (5.41)

which is a contradiction assumingis small enough.

Step 2.We will show that ¯F is a set of finite perimeter andD11F¯ =D11_F.

Proof of Step 2. By Step 1F¯\∂^MF∪F¹=F¯\F= 0. So by Proposition 3.38 [1] we have Per( ¯F) = Per(F)

and hence ¯F is a set of finite perimeter andD11F¯=D11F.

6. The applications

6.1. Quasiminimizing sets

The following lemmas hold true in IRⁿ without additional complexity, so we state them in IRⁿ. Lemma 6.1. Given as set of finite perimeter S, supposeHⁿ⁻¹(A) = 0 thenPer(S, A) = 0.

Proof Lemma6.1.By Theorem 1.9 (2), Corollary 1.11 [8] measureμdefined by μ(H) :=Hⁿ⁻¹(∂^MS∩H)

is a Radon measure.

Suppose setAhas the propertyHⁿ⁻¹(A) = 0. Thenμ(A) = 0, so 0 = inf{μ(V) :A⊂V, V is open}

= inf{Per(S, V) :A⊂V, V is open }

= Per(S, A).

Using the factA→Per(S, A) is also Radon measure, see Proposition 3.38(a) and Proposition 1.43 [1].

Lemma 6.2. Let E, F be sets of finite perimeter inIRⁿ,F ⊂E. Then

Per(F, ∂^ME) =Hⁿ⁻¹(∂^MF∩∂^ME). (6.1)

and

Per(E, ∂^MF)≥Hⁿ⁻¹(∂^MF∩∂^ME). (6.2)

Hence

Per(F, ∂^ME)≤Per(E, ∂^MF). (6.3)

ANDREW LORENT

Proof of Lemma 6.2. Note that the measureν(A) :=Hⁿ⁻¹(∂^MF ∩A) is a Radon measure. Letting σ >0 be some small number. Pick openΩwith

∂^ME∩∂^MF ⊂Ω (6.4)

such that

ν(Ω) =Hⁿ⁻¹(∂^MF∩Ω)< Hⁿ⁻¹(∂^ME∩∂^MF) +σ=ν(∂^ME) +σ. (6.5) Step 1.We will establish (6.1).

Proof of Step 1. Define

B:=

x∈(∂^MF)^c : lim sup

r→0

Hⁿ⁻¹(∂^MF∩B_r(x)) rⁿ⁻¹ >0

(6.6) and

D:=

x∈∂^ME: lim inf

r→0 Hⁿ⁻¹(∂^ME∩B_r(x))/rⁿ⁻¹<1 .

By Theorem 6.2 [8] we have thatHⁿ⁻¹(B) = 0 and since∂^ME is rectifiable by Theorem 16.2 [8]Hⁿ⁻¹(D) = 0.

So using Lemma6.1for the last equality

Per(F, ∂^ME) = Per(F, ∂^ME∩∂^MF) + Per(F, ∂^ME\∂^MF)

= Per(F, ∂^ME∩∂^MF) + Per(F, ∂^ME\(∂^MF∪B∪D)). (6.7) NoteHⁿ⁻¹(∂^ME)< C. Let >0. We can find a decreasing sequence of numberδ_m→0 such that the sets

U_m:=

x∈∂^ME\(∂^MF∪B∪D) : ^Hn−1(∂_r^Mn−1^F∩Br(x)) <

and ^Hn−1(∂_r^M_n−1^E∩Br(x)) ≥¹₂ for allr∈(0, δ_m)

(6.8) have the property that∂^ME\(∂^MF∪B∪D)⊂_∞

m=1Um. Let U₁=U₁\(_∞

i=2U_i), U₂ =U₂\(_∞

i=3U_i),. . . U_k =U_k\ ^∞_i=k+1U_i

. NowU_l, U_k are disjoint for anyl, k and

∂^ME\(∂^MF∪B∪D) =^∞

i=1

U_i = ∞ i=1

U_i. (6.9)

Since{U₁, U₂, . . .}are pairwise disjoint

m

Hⁿ⁻¹(Um)^(6.9)< Hⁿ⁻¹(∂^ME\∂^MF). (6.10)

Pickm. Since by Section 5.1 [8] we haveS¹(Um)≤2H¹(Um) whereS¹denotes Spherical Hausdorff measure.

So we can find a collection B^rk

2 (z_k) :r_k< δ_m

such that U_m ⊂

kB^rk

2 (z_k) and

kΓ(n−1) ^r₂^k_n−1

≤ 2S¹(Um). Now for eachkwe can pickxk∈B^rk

2 (zk)∩Um and then we have a collection of balls

{Brk(xk) :xk∈Um, rk < δm} (6.11) such that

Um⊂

k

Brk(xk) =:Vm (6.12)

and

k

rⁿ⁻¹_k ≤cS¹(U_m)≤cH¹(U_m). (6.13)

Hence using the fact thatx_k∈U_m⊂ U_m we have

Hⁿ⁻¹(V_m∩∂^MF) ^(6.12)≤

k

Hⁿ⁻¹(∂^MF∩B_rk(x_k))

(6.11),(6.8)

≤ c

k

r_kⁿ⁻¹

(6.13)

≤ cHⁿ⁻¹(U_m). (6.14)

Now

Per(F, ∂^ME\(∂^MF∪B∪D))(6.9),(6.12)

≤ Per

F,

m

Vm

≤

m

Hⁿ⁻¹(Vm∩∂^MF)

(6.14)

≤

m

cHⁿ⁻¹(U_m)

(6.10)

≤ cHⁿ⁻¹(∂^ME\∂^MF). (6.15)

Taking the limit as→0 we have Per(F, ∂^ME\(∂^MF∪B∪D)) = 0. So putting this together (6.7) we have

Per(F, ∂^ME) = Per(F, ∂^ME∩∂^MF). (6.16)

And

Per(F, ∂^ME∩∂^MF)^(6.4)≤ Per(F, Ω)

≤ Hⁿ⁻¹(∂^MF∩Ω)

(6.5)

≤ Hⁿ⁻¹(∂^MF∩∂^ME) +σ. (6.17)

Now as σ is arbitrary, from (6.17) and (6.16), Per(F, ∂^ME)≤Hⁿ⁻¹(∂^MF ∩∂^ME). Conversely for any open set Ω with ∂^ME ⊂ Ω we have Per(F, Ω) ≥ Hⁿ⁻¹(∂^MF∩∂^ME). Thus by taking the infimum over all such open sets we have Per(F, ∂^ME)≥Hⁿ⁻¹(∂^MF∩∂^ME) and this completes the proof of Step 1.

Step 2.We will establish (6.2).

Proof of Step 2.First note that∂^ME∩∂^MFis an (n−1) rectifiable set. Letδ >0, forHⁿ⁻¹a.e.x∈∂^ME∩∂^MF there existsr_x>0 such that

Hⁿ⁻¹(∂^MF∩∂^ME∩Bh(x))−nΓ(n)hⁿ⁻¹< δhⁿ⁻¹for any h∈(0, rx). (6.18) Letα >0. Defining Γ0:=

x∈∂^MF∩∂^ME :α < rx

for all small enoughα >0 we have that

Hⁿ⁻¹(∂^MF∩∂^ME\Γ0)< δ. (6.19) SinceA→Per(E, A) is a Radon measure we can extract a compact subsetΓ₁⊂Γ₀ such that

Per(E, Γ₁) +δ >Per(E, Γ₀).

ANDREW LORENT

And we can find a compact subsetΓ₂⊂Γ₀ such thatHⁿ⁻¹(Γ₂) +δ > Hⁿ⁻¹(Γ₀). LetΓ₃=Γ₁∪Γ₂. SoΓ₃ is compact and has the properties

Per(E, Γ₃) +δ >Per(E, Γ₀) andHⁿ⁻¹(Γ₃) +δ > Hⁿ⁻¹(Γ₀). (6.20) And

Γ₃⊂Γ₀⊂∂^MF∩∂^ME. (6.21)

Again since we are dealing with a Radon measure we can find an open setU withΓ3⊂U and

Per(E, U)<Per(E, Γ3) +δ. (6.22)

Now dist(Γ3, U) =: β > 0 by Vitali covering theorem we can find a collection of pairwise disjoint balls {Brk(xk) :xk ∈Γ3, k∈IN}where

sup{rk :k∈IN}<min{β, α}

and

Hⁿ⁻¹(Γ3\(

∞ k=1

Brk(xk))) = 0. (6.23)

Now since_∞

k=1B_rk(x_k)⊂U we know Per

E,^∞

k=1

Brk(xk)

≤Per(E, U)^(6.22)< Per(E, Γ3) +δ. (6.24)

But asΓ3⊂Γ0

Per

E, ^∞

k=1

B_rk(x_k)

= ∞ k=1

Per (E, B_rk(x_k))

= ∞ k=1

Hⁿ⁻¹(∂^ME∩B_rk(x_k))

(6.18)

≥ ∞ k=1

(nΓ(n)−δ)r_kⁿ⁻¹. (6.25) Now recallΓ3⊂Γ0⊂∂^ME∩∂^MF so

∞ k=1

nΓ(n)(1 +δ)r_k^{n−1 (6.18)}≥ ∞ k=1

Hⁿ⁻¹(∂^ME∩∂^MF∩Brk(xk))

(6.21)

≥ ∞ k=1

Hⁿ⁻¹(Γ3∩Brk(xk))

(6.23)

= Hⁿ⁻¹(Γ₃)

(6.20)

≥ Hⁿ⁻¹(Γ0)−δ

(6.19)

≥ Hⁿ⁻¹(∂^ME∩∂^MF)−2δ. (6.26)

Now putting (6.24)–(6.26) together we have

Per (E, Γ₃)≥Hⁿ⁻¹(∂^ME∩∂^MF)−3δ−c^∞

k=1

δr_kⁿ⁻¹. (6.27)