FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER

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1 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER MONICA TORRES Abstract. We present in these notes some fine properties of functions of bounded variation and sets of finite perimeter, which will be used in the first part of the mini-course. We define 1. Radon and Hausdorff measures M(Ω) = {µ : µ is a Radon measure in Ω; i.e., µ (K) <, for any K Ω compact set }. We denote the total variation of µ as µ, which is denoted by µ (B) = sup{ µ(c i ) : B = C i, C i pairwise disjoint} We recall that if µ is a signed measure then, the positive and negative parts of µ, denoted as µ + and µ respectively, are given by Clearly, we have that We define now µ + = µ + µ, µ = µ µ. µ µ. Week convergence of measures Definition.1. Let µ k, µ M. We say that µ k µ weakly* if fdµ k fdµ, for all f C c (Ω). Ω The following result will be of great use to us Ω Theorem.. Let µ k M(Ω) such that sup k µ k (K) <, for each compact set K Ω. Then, there exists a sequence µ kh (which we denote again as µ k ) such that µ k µ weakly*. Lemma.3. Let c R and let f : Ω [c, ] not identically equal to. Define, for t > 0 f t (x) = inf{f(y) + td(x, y) : y Ω}, then f t is continuous, f t f and f t (x) f(x) as t whenever x is a lower semicontinuity point of f. Key words and phrases. unbounded divergence-measure fields, Gauss-Green formula, approximation of sets, distance function. 1

2 MONICA TORRES Proof. Let x be a lower semicontinuity point of f; i.e., (.1) f(x) lim inf f(x i ) whenever x i x. Let x t R N such that: (.) f(x t ) + td(x, x t ) < f t (x) + 1 t Clearly, f t (x) f(x). We will show that f t (x) f(x) as t. If f t (x) as t we are done. We now assume that f t (x) has a finite limit as t, say α. Thus, (.) implies that x t x 0. Since f is lower semicontinuos at x we obtain, using (.): (.3) f(x) lim inf f(x t ) lim f t (x) = α. Since f t (x) f(x), the reverse inequality also holds. f(x) lim f t (x) Theorem.4. Let µ, µ k M such that µ k µ weakly*. Then (a) If µ k are positive then for every lower semicontinuous function f : Ω [0, ] (.4) fdµ lim inf fdµ k. k In particular, if f = χ A, A open we have (.5) µ(a) lim inf µ k(a), k Ω and for every upper semicontinuous function g : Ω [0, ] with compact support we have (.6) lim sup gdµ k gdµ. k Ω Ω In particular, if g = χ K, K compact, then (.7) lim sup µ k (K) µ(k). k (b) If µ k λ weakly*, then µ λ. Moreover, if E is a µ-measurable set that satisfies λ( E) = 0, then (.8) µ k (E) µ(e). More generally: (.9) udµ = lim udµ k. Ω k Ω for any bounded Borel function u : Ω R with compact support such that the set of its discontinuities is µ- negligible. Proof. Let f : Ω [0, ] be a lower semicontinuous function and let f t be the approximation given by the previous lemma, such that f t (x) f(x). Choose a function ψ so that 0 ψ 1 and ψ C c (Ω). Since µ k µ weakly* we have that ψf t dµ = lim ψf t dµ k Ω k Ω lim inf f t dµ k. k Ω Ω

3 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 3 We now take the sup over all ψ and let t to obtain fdµ lim inf fdµ k. k Ω The proof of the case when g is upper semicontinuous is similar but using the corresponding approximation as in the previous lemma. Definition.5. If µ M(Ω) we define supp µ = {y Ω : µ(b(y, r)) > 0, for all r > 0} Theorem.6. Let ν be a non-negative locally finite Radon measure on Ω. Let J be a set with the following properties (a) For all y J there exist orthonormal coordinates x 1, x,..., x N such that C y := {8 x 1 (x, x 3,..., x N ) } satisfies (.10) lim ν(y + C y ) B(y, r) r N 1 = 0, and (b) For all y J ν(b(y, r)) (.11) lim inf r N 1 > 0. Then J is contained in a countably union of Lipschitz graphs. Definition.7. Let µ M(R N ). The rescalings or blow-ups of the measure µ around y is the sequence of measures (.1) µ y,r (A) := We have Ω µ(y + ra) r N 1, A R N Lemma.8. For H N 1 -almost every y R N, the sequence µ y,r has a weakly* convergence subsequence. (A limit will be denoted as µ y, ). Proof. We claim that (.13) lim sup µ(b(y, r)) r N 1 <, for H N 1 a.e. y R N. The claim implies that, for each compact set K, sup µ y,r (K) <, r and thus the Lemma follows. In order to prove the claim we proceed by contradiction and assume that there exists a bounded set K with H N 1 (K) > 0 and lim sup µ(b(y, r)) r N 1 =, for all y K. We choose an open set K ε, K K ε and µ(k ε ) < µ(k) + ε. Since K ε is open, for each y K ε, there exists an open ball B ry (y) K ε, r y r 0 and µ(b ry (y)) Mry N 1. The family of balls {B ry (y)} covers K ε. Thus, Vitali s covering Lemma implies that there exists a countable collection {B ri (y i )} such that B 5ri (y i ) covers K ε. Therefore, we have that µ(k ε ) µ(b ri (y i )) =

4 4 MONICA TORRES M r N 1 i = C(N)H N 1 (K), which leads to the contradiction µ(k ε ) =, since M is arbitrary. Lemma.9. Let R N + = {x R N : x n > 0} and let µ be a nonnegative Radon measure on R N +. Let Ω R N 1 be a bounded open set. For r > 0 and y Ω, let then, for H N 1 -almost every y Ω, C R r (y) = (y + ( rr, rr) d ) (0, rr). µ(cr R (y)) lim r N 1 = 0 Proof. Let µ(cr R (y)) A k = {y Ω : lim sup r N 1 > 1 k }. It is sufficient to show that H N 1 (A k ) = 0 for each k. Given y A k and ε > 0, there exists a number r y < ε such that ( ) µ Cr R y (y) > 1 y. k rn 1 Using a covering argument we can choose a sequence y i Ω such that B(y i, r i ) := {y Ω : y y i < r i } where r i = r yi are disjoints and A k B(y i, 3r i ). Then H N 1 (A k ) ω N 1 On the other hand, since r i < ε, we have (3r i ) N 1 kω N 1 3 N 1 µ ( Cr R i (y i ) ). C R r i (y i ) Ω ε = Ω (0, εr) and hence H N 1 (A k ) < kω N 1 3 N 1 µ(ω ε ) for all ε > 0. Since µ is a Radon measure we have µ(ω ε ) 0 as ε 0 and H N 1 (A k ) = 0 3. Convolutions and representation of BV functions Definition 3.1. Let E, F sets in Ω and let x Ω. We say that x is a point of density α for E relative to F if lim B r (x) E F B r (x) F = α. Definition 3.. Let f be a measurable function. We say that β is the approximate limit of f(x) as x x 0 if ε > 0, x 0 is a point of density 1 for the set {x : f(x) β < ε}. It is denoted by alimf(x) x x 0 = β. The above definition says that if alim x x 0 f(x) = β then ε > 0: lim B r (x) { f(x) β < ε} B r (x) = 1. Definition 3.3. We say that alim f(x) = β if ε > 0, x 0 is a point of density x x 0,x E 1 for the set { f(x) β < ε}, relative to E; that is, if ε > 0 lim B r (x) { f(x) β < ε} E B r (x) E = 1.

5 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 5 and Let w Ω. We introduce the following notation: π w (x 0 ) = {(x x 0, w) > 0} f w (x 0 ) = alim f(x). x x 0,x π w(x 0) Definition 3.4. x 0 is a regular point for f(x) if there exists w R N such that f w (x 0 ) and f w (x 0 ) exist. The vector w is called a defining vector for f at x 0. It is easy to check that if w is a defining vector for f at x 0 and f w (x 0 ) = f w (x 0 ), then any w R N is also a defining vector and f w (x 0 ) = f w (x 0 ) = f w (x 0 ). On the other hand, if w is a defining vector for f at x 0 and f w (x 0 ) f w (x 0 ) then w is uniquely determined up to sign. We have the following. Theorem 3.5. Let f BV (R N ). Then, H N 1 a.e. x R N is a regular point for f. Let ρ be a standard symmetric mollifier and ρ r = 1 ( x ) r N ρ. r Definition 3.6. f r (x) = R N ρ r (x y)f(y) dy and f(x) = lim fr (x), x R N, whenever the above limit exists. We have: Lemma 3.7. If x is a regular point for f and w is a defining vector then: f(x) = 1 (f w(x) + f w (x)). Proof. Let us compute first f r (x) = ρ r (x y)f(y) dy R N = ρ r (x y)f(y) dy R N π w + ρ r (x y)f(y) dy. R N π w ρ r (x y)f(y) dy. R N π w We have: ρ r (x y)f(y) dy = ρ r (x y) (f(y) f w (x)) dy R N π w R N π w + ρ r (x y)f w (x) dy R N π w

6 6 MONICA TORRES because R N π w B r(x) ρ r (x y)f w (x) dy = f w (x) ρ r (y) dy B r(x) π w ρ r (y) dy = 1. = f w(x) ; Also, if we define A ε = {y : f(y) f w (x) < ε} we obtain ρ r (x y) (f(y) f w (x)) dy = ρ r (y) (f(y) f w (x)) dy R N π w B r(x) π w = ρ r (y) (f(y) f w (x)) dy + ρ r (y) (f(y) f w (x)) dy B r(x) π w A ε B r(x) π w (A ε ) c C r N ε B r(x) π w A ε + M r N B r(x) π w (A ε ) c. Since: B r (x) A ε π w lim = 1 B r (x) π w and B r (x) π w (A ε ) c lim = 0 B r (x) π w we conclude, since ε is arbitrary, that: lim ρ r (x y)(f(y) f w (x)) dy = 0 R N π w and thus: lim ρ r (x y)f(y) dy = f w(x). R N π w In the same way we can show that: lim ρ r (x y)f(y) dy = f w(x). R N π w We conclude: f(x) = 1 (f w(x) + f w (x)). 4. Sets of Finite Perimeter 5. Basic properties of sets of finite perimeter In what follows we will work in R N. We introduce now a few basic definitions and results on the theory of functions of bounded variations and sets of finite perimeter, for which we refer mainly to [1], [1] and [15] (see also [14] and [1]). Definition 5.1. A function u L 1 (R N ) is called a function of bounded variation if Du is a finite R N -vector valued Radon measure on R N. A measurable set E R N is called a set of finite perimeter in R N (or a Caccioppoli set) if χ E BV (R N ). Consequently, Dχ E is an R N -vector valued Radon measure on R N whose total variation is denoted as Dχ E.

7 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 7 By the polar decomposition of measures, we can write Dχ E = ν E Dχ E, where ν E is a Dχ E -measurable function such that ν E (x) = 1 for Dχ E -a.e. x R N. We define the perimeter of E as { } P (E) := sup div(ϕ) dx : ϕ Cc 1 (R N ; R N ), ϕ 1 E and it can be proved that P (E) = Dχ E (R N ). The notion of perimeter generalizes the idea of H N 1 -measure of the boundary of the set E. It is a well-known fact that the topological boundary of a set of finite perimeter can be very irregular, it can even have full Lebesgue measure. This suggests that for a set of finite perimeter is interesting to consider subsets of E instead. De Giorgi considered a set of finite H N 1 -measure on which Dχ E is concentrated, which he called reduced boundary. Definition 5.. We say that x E, the reduced boundary of E, if (1) Dχ E (B(x, r)) > 0, r > 0; 1 () lim ν E (y) d Dχ E (y) = ν E (x); Dχ E (B(x, r)) B(x,r) (3) ν E (x) = 1. It can be shown that this definition implies a geometrical characterization of the reduced boundary, by using the blow-up of the set E around a point of E. Theorem 5.3. If x E, then E x H ν + ε E (x) := {y R N : y ν E (x) 0} in L 1 loc(r N ) as ε 0 and (R N \ E) x Hν ε E (x) := {y R N : y ν E (x) 0} in L 1 loc(r N ) as ε 0. This theorem will be proved later. Formulated in another way, for ε > 0 small enough, E B(x, ε) is asymptotically close to the half ball Hν E (x) B(x, ε). Because of this result, we call ν E (x) measure theoretic unit interior normal to E at x E, since it is a generalization of the concept of unit interior normal. In addition, De Giorgi proved that Dχ E = H N 1 E, so that Dχ E = ν E H N 1 E and P (E) = H N 1 ( E) (see [1, Section 5.7.3, Theorem ]). For every α [0, 1] we set where E α := {x R N : D(E, x) = α}, B(x, r) E D(E, x) := lim, B(x, r) and we give the following definitions: (1) E 1 is called the measure theoretic interior of E. () E 0 is called the measure theoretic exterior of E. We recall (see Maggi [15, Example 5.17]) that every Lebesgue measurable set is equivalent to the set of its points of density one; that is, (5.1) L N (E E 1 ) = L N ((R N \ E) E 0 ) = 0. It is also a well-know result due to Federer that there exists a set N with H N 1 (N ) = 0 such that R N = E 1 E E 0 N (see [1, Theorem 3.61]).

8 8 MONICA TORRES The perimeter P (E) of E is invariant under modifications by a set of L N -measure zero, even though these modifications might largely increase the size of the topological boundary. In this paper we consider the following representative (5.) E := E 1 E. Given a smooth nonnegative radially symmetric mollifier ρ Cc (B 1 (0)), we denote the mollification of χ E by u k (x) := (χ E ρ εk )(x) for some positive sequence ε k 0. We define, for t (0, 1), (5.3) A k;t := {u k > t}. By Sard s theorem (for which we refer to [15, Lemma 13.15]), we know that, since u k : R N R is C, L 1 -a.e. t (0, 1) is not the image of a critical point for u k and so A k;t has a smooth boundary for these values of t. Thus, for each k there exists a set Z k (0, 1), with L 1 (Z k ) = 0, which is the set of values of t for which A k;t has not a smooth boundary. If we set Z := + k=1 Z k, then L 1 (Z) = 0 and, for each t (0, 1) \ Z and for each k, A k;t has a smooth boundary. It is a well-known result from BV theory (see for instance [1, Corollary 3.80]) that every function of bounded variations u admits a representative which is the pointwise limit H N 1 -a.e. of any mollification of u and which coincides H N 1 -a.e. with the precise representative u : 1 lim u(y) dy if this limit exists u (x) := B(x, r) B(x,r). 0 otherwise For any set of finite perimeter E, we denote the precise representative of the function χ E by u E, which is given by 1, x E 1 u E (x) = 0, x E 0. 1, x E Since H N 1 (R N \(E 1 E E 0 )) = 0, the function u E is well defined H N 1 -a.e.. In order to prove Theorem 9.1, we need to use the classical coarea formula, for which we refer to [1, Section 3.4, Theorem 1]. Theorem 5.4. Let u : R N R be Lipschitz. Then, for any L N -measurable set A, we have (5.4) u dx = H N 1 (A u 1 (t)) dt. A R 6. Structure of sets of finite perimeter Definition 6.1. Let E be a set of finite perimeter and x m E. We say that ν is the inner unit normal at x if: B r (x) E π ν (6.1) lim = 1 B r (x) π ν and B r (x) E π ν (6.) lim = 0 B r (x) π ν

9 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 9 We know that if x E, then Dχ E (B r (x)) (6.3) ν(x) = lim Dχ E (B r (x)) exists and ν(x) = 1. We now proceed to show that ν(x) is the inner normal to E at x. We need first the following result: Lemma 6.. Let E be a set of finite perimeter and x E. Then, there exists r 0, that depends on x, such that for all r r 0 ; (a) Dχ E (B r (x)) C 1 r N 1 (b) E B r (x) C r N (c) (R N \ E) B r (x) C 3 r N, where C 1, C, C 3 are universal constants. Proof. We recall that if f BV (Ω), where Ω is an open set with Lipschitz boundary, then: (6.4) fdiv ϕ = ϕ Df + f tr < ϕ, n > dh N 1, Ω Ω Ω for all ϕ C0(R 1 N ). Here, f tr is the trace of f on Ω. If we apply (6.4) to f = χ E, Ω = B(x, r) = B r and ϕ 1 we have, for a.e. r, (6.5) 0 = Dχ E + B r χ E < 1, n > dh N 1 B r and thus: (6.6) Dχ E HN 1 (E B r ) B r On the other hand, since x E we have B (6.7) lim r Dχ E B r Dχ E = ν(x) = 1 which implies that for r small enough: (6.8) Dχ E 1 Dχ E B r B r From (6.8) and (6.6) we have: (6.9) Dχ E (B r ) H N 1 (E B r ) C 1 r N 1 which gives (a). The isoperimetric inequality implies: (6.10) E B r C ( Dχ E Br (R N ) ) N N 1 = C ( Dχ E (B r ) + H N 1 (E B r ) ) N N 1 C(H N 1 (E B r )) N N 1

10 10 MONICA TORRES by (6.9) We define (6.11) V (r) = E B r We have (6.1) V (r) = χ E = B r and thus r 0 B t χ E dh N 1 dt (6.13) V (r) = χ E dh N 1 = H N 1 (E B r ) B r From (6.10) and (6.13) we obtain: Which gives (b) V (r) C(V (r)) N N 1 C V (r)(v (r)) N 1 N C (V (r) 1 N ) V (r) C r N, We can obtain (c) in a similar way but using X RN \E instead of X E. We have the following Lemma 6.3. Let E be a set of finite perimeter and x E. Then, ν(x)is the inner unit normal. Proof. We can assume without loss of generality that x = 0. For 0 r 1 we define the sets (6.14) E r = E r. We fix R > 0. Using (a) from lemma 6., we have (6.15) Dχ Er (B r ) = r 1 N Dχ E (B rr ) Cr 1 N (rr) 1 N = CR 1 N Since 6.15 is true for any R > 0, it follows that there exists a set Q such that E r Q in L 1 loc. We now proceed to show that Q is a hyperplane orthogonal to ν. We now claim that Q = π ν. We proceed by contradiction and assume first that there exists α > 0 such that (6.16) Q B α =. In this case we have, using (b) from lemma 6., (6.17) E r B α = r N E B rα Cr N (rα) N = Cα N

11 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 11 Since E r B α Q B α and E r B α C it follows that Q B α C, which contradicts (6.16). If we now assume that there exists α > 0 now such that (6.18) (R N \ Q) B α =, procceeding in the same way but using (c) from lemma 6. we obtain a contradiction. This shows that Q = π ν. We can now show that ν is the inner unit normal to E at x. In fact we have, since Q = π ν, (6.19) E r B 1 π ν B 1 π ν On the other hand (6.0) E r B 1 π ν = E B r π ν r N. Thus, from (6.19) and (6.0) we have that: (6.1) lim E B r π ν r N = B 1 π ν, or equivalently, E B r π ν (6.) lim = 1. B r π ν In the same way we show that E B r π ν (6.3) lim = 0. B r π ν If E is a set of finite perimeter, we know that χ E BV (R N ) and thus, from Theorem 3.5 it follows that H N 1 a.e.x R N is a regular point for χ E. If u k is a sequence of mollifiers for χ E, Lemma 3.7 says that if x R N is a regular point for χ E with defining vector w then the following limit exists (6.4) χ E (x) = lim k u k(x) = 1 ((χ E) w (x) + (χ E ) w (x)). From now on we define (6.5) u E (x) := χ E (x). If x E 0 we have, for any ε > 0, B r (x) { χ E 0 < ε} B r (x) E c (6.6) lim = lim = 1, B r (x) B r (x) which by definition implies that a lim x x χ E = 0 and thus u E (x) = 0. If x E 1 we have, for any ε > 0, B r (x) { χ E 1 < ε} B r (x) E (6.7) lim = lim = 1, B r (x) B r (x) which by definition implies that a lim x x χ E = 1 and hence u E (x) = 1. If x E we recall that ν E (x) = ν(x) = ν is the inner unit normal to E at

12 1 MONICA TORRES x. Since, by Lemma 6.3, B r (x) { χ E 1 < ε} π ν (6.8) lim = 1 B r (x) π ν and B r (x) { χ E 0 < ε} π ν (6.9) lim = 1 B r (x) π ν it follows that a lim x x,x πν χ E = (χ E ) ν (x) and a lim x x,x π ν χ E = (χ E ) ν (x) = 0. Thus, we obtain (6.30) u E (x) = 1 ((χ E) ν (x) + (χ E ) ν (x)) = 1 (1 + 0) = 1. Since H N 1 ( m E \ E) = 0 and (6.31) R N = E 0 E 1 E ( m E \ E), We can finally give the following result Lemma 6.4. The function u E is defined H N 1 almost everywhere in R N We have the following result: Lemma 6.5. Let E be a set of finite perimeter. Then: (6.3) H N 1 ( m E \ E) = 0 Proof. We first note that if x R N satisfies (6.33) lim Dχ E (B r (x)) r N 1 = 0 then x E 0 or x E 1. In fact, from the isoperimetric inequality (6.34) ( Dχ E (B r (x))) N N 1 C min{ E B r (x), (R N \ E) B r (x) } it follows that (6.35) or (6.36) and from (6.33) we have ( DχE (B r (x)) r N 1 ( DχE (B r (x)) r N 1 ) N N 1 C E B r (x) r N ) N N 1 C (R N \ E) B r (x) r N ; E B r (x) (6.37) lim r N = 0 or (6.38) lim (R N \ E) B r (x) r N = 0. From (6.37) and (6.38) we conclude that x E 0 or x E 1. The previous analysis implies that (6.39) m E \ E = {χ R N : lim sup Dχ E (B r (x)) r N 1 > 0}

13 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 13 Moreover, we can write (6.40) m E \ E = k=1a k, where (6.41) A k = {χ m E \ E : lim sup Dχ E (B r (x)) r N 1 > 1 k }. We now proceed to show that H N 1 (A k ) = 0. We fix k and let η > 0. Let V be an open set such that A k V and (6.4) Dχ E (V ) Dχ E (A k ) + η For any x A k, using (6.41), we can find r x < η such that B r(x) Dχ E rn 1 x k and B rx (x) V. We have (6.43) A k B rx (x) A covering argument implies that there exists a countable collection {x i } A k such that, setting r i := r xi, (6.44) A k B 3ri (x i ), B ri (x i ) B rj (x j ) =, i j. We can now compute (3r i ) N 1 3 N 1 k Dχ E (B ri(x i)) = 3 N 1 k Therefore, since η is arbitrary, we have. Dχ E (B ri(x i)) C Dχ E (V ) C( Dχ E (A k ) + η) (6.45) H N 1 (A k ) C Dχ E (A k ), which implies that H N 1 (A k ) = 0 because Dχ E (A k ) = H N 1 (A k E) = 0. We conclude that H N 1 ( m E \ E) = 0. We have: Theorem 6.6. Let E be a set of finite perimeter. Then, E is a rectifiable set; that is, E = ( C i ) N, where H N 1 (N ) = 0 and each C i is contained in a C 1 manifold M i. Proof. Using Egorov s theorem, there exists, for each i, a set F i such that χ E ( E\F i ) < 1 i and the following limit is uniform for a.e. x E, (6.46) lim B r (x) E π ν r N = 0.

14 14 MONICA TORRES From Luzin s theorem, there exists a set C i such that and We now fix one of the sets C i. Since ν of Lemma 6., that χ E (F i \C i ) < 1 i ν C i is continuous. (6.47) E B r (z) Cr N, C i is continuous it follows, from the proof for all z C i and r small enough. We now fix ε > 0. From (10.1) and (10.) it follows that there exists σ ε > 0 such that for all r < σ ε and z C i, (6.48) and E B r (z) π ν B r (z) π ν < εn C ω N N (6.49) E B r (z) Cr N, where C is a universal constant. We consider the cone C ε with opening θ such that cos θ = ε. We take x C i. We claim that ( E B x (σ ε )) x + C ε. Indeed, if this is not true, then there exists by E, x y < σ ε such that B ε x y (y) π ν = φ or B ε x y (y) π ν = φ Without loss of generality we can assume that (6.50) B ε x y (y) π ν = φ Using (10.3) we have (6.51) B ε x y (y) E = B ε x y (y) E π ν + B ε x y (y) E π ν = B ε x y (y) E π ν B x y (x) E π ν < εn C w N N B x y π ν = εn C w N N N x y N w N = C x y N ε N On the other hand, (10.4) implies that: B ε x y (y) E Cε N x y N, which contradicts (10.6). Since ε is arbitrary, we have proved that for any cone C, E B r (x) x + C, for all x C i and r r 0 where r 0 depends on C. This implies, see??, that C i is the zero level set of a C 1 function f i, that is, C i is contained in the C 1 manifold f 1 i (0).

15 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 15 We have the following Theorem 6.7. (6.5) Dχ E = H N 1 E Proof. Let E be a set of finite perimeter and let B E. We first show that (6.53) H N 1 (B) C Dχ E (B). We let E,η > 0. We can find an open set. A such that B A and (6.54) Dχ E (A) Dχ E (B) + η For any x B, there exists r x such that r x < E, B rx (x) A and (6.55) Dχ E (B rx (x)) 1 w N 1r N 1 x. Thus, B rx (x) covers B and a covering argument implies that there exist a countable collection x i such that, if we denote r xi := r i, we have (6.56) B B 3ri (x i ) and B 3ri (x i ) B 3rj (x j ) = We now compute, using (6.55), (3r i ) N 1 3N 1 (6.57) Dχ E B rx (x) W N 1 (6.58) (6.59) C Dχ E (A) C ( Dχ E (B) + η). Since Theorem 6.6, we know that E can be writen as (6.60) E = (U C i ) N, where H N 1 (N) = 0 and each C i is a compact set contained in a C 1 manifold, say M i. We define (6.61) M i := H N 1 M If x C i, clearly M i (B r (x)) (6.6) lim r N 1 = w N 1 On the other hand, since x E, we have (6.63) lim Dχ E (B r (x)) r N 1 = w N 1. From (6.6) and (6.63) we obtain M i (B r (x)) (6.64) lim Dχ E (B r (x)) = 1, x C i

16 16 MONICA TORRES Theorem 6.8. Let n > 1 and 0 < τ < 1. Suppose E is a set of finite perimeter E B such that lim sup r(x) B r(x) > τ whenever x E. Then there exists a constant C = C(r, n) and a sequence of closed balls B ri (x i ) with x i E such that E B ri (x i ) and r n 1 i CH n 1 ( E). Remark 6.9. If E is an open set, the proof of Theorem 6.8 actually shows that we can take τ = 1. Moreover, the covering {B r i } can be chosen in such a way that Bri/5 E = 1. Bri/5 7. Almost one-sided smooth approximation of sets of finite perimeter We now proceed to establish a fundamental approximation theorem for a set of finite perimeter by a family of sets with smooth boundary essentially from the measure-theoretic interior of the set with respect to any Radon measure that is absolutely continuous with respect to H N 1. That is, we prove that, for any Radon measure µ on R N such that µ << H N 1, the superlevel sets of the mollifications of the characteristic functions of sets of finite perimeter provide an approximation by smooth sets which are µ -almost contained in the measure-theoretic interior of E. It is a classical result in geometric measure theory that a set of finite perimeter E can be approximated with smooth sets E k such that (7.1) L N (E k ) L N (E) and P (E k ) P (E), where P (E) is the perimeter of E and L N is the Lebesgue measure in R N. The approximating smooth sets (see for instance Ambrosio-Fusco-Pallara [1, Remark 3.4] and Maggi [15, Theorem 13.8]) are the superlevel sets of the convolutions of χ E, which can be chosen for a.e. t (0, 1). The one-sided approximation refines the classical result in the sense that it distinguishes between the superlevel sets for a.e. t ( 1, 1) from the ones corresponding to a.e. t (0, 1 ), thus providing an interior and an exterior approximation of the set respectively (see Theorem 8.1 and Theorem 9.1). Indeed, in the first case, the difference between the level sets and the measure theoretic interior is asymptotically vanishing with respect to the H N 1 -measure; in the latter, we obtain the same result for the measure theoretic exterior. 8. The approximation of E with respect to any µ << H N 1 The one-sided approximation theorem allows to extend (7.1) to any Radon measure µ such that µ << H N 1. More precisely, for any bounded set of finite perimeter E, there exist smooth sets E k;i, E k;e, such that (8.1) µ(e k;i ) µ(e 1 ), P (E k;i ) P (E)

17 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 17 and (8.) µ(e k;e ) µ(e), P (E k;e ) P (E). The convergence of the perimeters in (8.1) and (8.) follows as in the standard proof of (7.1). However, the convergence with respect to µ is a consequence of the following result. Theorem 8.1. Let µ be a Radon measure such that µ << H N 1 and E be a bounded set of finite perimeter in R N. Then: (a) µ (E 1 A k;t ) 0, for 1 < t < 1; (b) µ (E A k;t ) 0, for 0 < t < 1. Proof. We have (8.3) u k (x) u E (x) for H N 1 -a.e. x. Since {0 < u k u E 1} E δ := {x R N : dist(x, E) δ}, for any k if δ > max ε k, and E δ is bounded, then we can apply the dominated convergence theorem with respect to the measure µ, taking 1 as summable majorant since µ is a Radon measure. Hence, for any ε > 0, there exists k large enough such that, if 1 < t < 1, we have ε u k (x) u E (x) d µ R n u k (x) u E (x) d µ + u E (x) u k (x) d µ A k;t \E 1 E 1 \A k;t (t 1 ) µ (A k;t \ E 1 ) + (1 t) µ (E 1 \ A k;t ) min {t 1 }, 1 t µ (A k;t E 1 ). Thus, for k large enough and 1 < t < 1, we obtain µ (A k;t E 1 ε ) min { t 1, 1 t}, which is (a). Analogously, for 0 < t < 1, we have ε u k (x) u E (x) d µ R n u k (x) u E (x) d µ + u E (x) u k (x) d µ A k;t \E E\A k;t t µ (A k;t \ E) + ( 1 t) µ (E \ A k;t) min {t, 1 } t µ (A k;t E). Thus, for large k and 0 < t < 1, which gives (b). µ (A k;t E) ε min { t, 1 t},

18 18 MONICA TORRES Remark 8.. The convergence in (8.1) follows easily from Theorem 8.1: we have and it is clear that (a) implies µ(e 1 ) µ(a k;t ) = µ(e 1 \ A k;t ) µ(a k;t \ E 1 ) µ(e 1 \ A k;t ) µ (E 1 \ A k;t ) 0, µ(a k;t \ E 1 ) µ (A k;t \ E 1 ) 0. One can show (8.) in a similar way using (b). We also notice that Theorem 8.1 has been proved for any t (0, 1 ) ( 1, 1). However, since the sets A k;t have smooth boundary only for almost every t, we shall consider only t / Z, where Z is the set of singular values defined in the introduction. Remark 8.3. With µ = H N 1 E, we obtain from Theorem 8.1: (a) H N 1 ( E A k;t ) 0 for 1 < t < 1; (b) H N 1 ( E (R N \ A k;t )) 0 for 0 < t < 1. Indeed, this is clear from the following identities E (E 1 A k;t ) = E [(E 1 \ A k;t ) (A k;t \ E 1 )] = E A k;t, E (E A k;t ) = E [(E \ A k;t ) (A k;t \ E)] = E (R N \ A k;t ). Remark 8.4. Using Remark 8.3 we can also show that we have: (a) H N 1 ( E u 1 k (t)) 0 for 1 < t < 1; (b) H N 1 ( E u 1 k (t)) 0 for 0 < t < 1. Indeed, u 1 k (t) A k;s for 1 < s < t < 1 and u 1 k (t) RN \ A k;s for 0 < t s < 1. In addition, we observe that µ (u 1 k (t)) = 0 for L1 -a.e. t, since µ is a Radon measure. It is in fact clear that u 1 k (t) = A k;t, that A k;t A k;s A k;0 if 0 < s < t < 1, with A k;0 bounded, and that the sets A k;t are pairwise disjoint. Hence, since µ is finite on bounded sets and additive, the set {t (0, 1) : µ ( A k;t ) > ε} is finite for any ε > 0. This implies that the set {t (0, 1) : µ ( A k;t ) > 0} is at most countable (see also the observation at the end of Section 1.4 of [1]). Then we obtain also: (a) H N 1 ( E u 1 k (t)) = 0 for a.e. 1 < t < 1; (b) H N 1 ( E u 1 k (t)) = 0 for a.e. 0 < t < The main approximation result The following theorem, together with Theorem 8.1, shows that indeed we have an interior approximation of E for a.e. t ( 1, 1). Theorem 9.1. Let E be a set of finite perimeter in R N. There exists a sequence ε k converging to 0 such that, if u k := χ E ρ εk, we have (9.1) lim k + HN 1 (u 1 k (t) \ E1 ) = 0 for a.e. t ( 1, 1).

19 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 19 Proof. We take s > 1 and a sequence ε k, with ε k 0, and we consider the set A k;s := {u k > s}. By the coarea formula (5.4), we have 1 u k dx = H N 1 (u 1 k (t) (A k;s \ E 1 )) dt A k;s \E 1 0 (9.) = 1 s H N 1 (u 1 k (t) \ E1 ) dt, since, for t s, u 1 k (t) (A k;s \ E 1 ) =, while, for t > s, u 1 k (t) (A k;s \ E 1 ) = u 1 k (t) \ E1. We claim that (9.3) u k L 1 (A k;s \E 1 ) 0. In order to prove the claim, we observe that, for any x R N, u k (x) = χ E (y) x ρ εk (x y) dy = χ E (y) y ρ εk (x y) dy R N R N = ρ εk (x y)ν E (y) d Dχ E (y) = (ρ εk Dχ E )(x). R N Hence, u k = (Dχ E ρ εk ) = ( Dχ E ν E ρ εk ), which implies (9.4) u k Dχ E ρ εk. Recalling from (5.1) that L N (E E 1 ) = 0, (9.4) leads to u k L 1 (A k;s \E 1 ) R = u k χ Ak;s \E dx N ( Dχ E ρ εk ) χ Ak;s \E dx = (ρ εk χ Ak;s \E) d Dχ E = R N R N = (ρ εk χ Ak;s \E) dh N 1. E Thus, we need to investigate, for any x E, the behaviour of (ρ εk χ Ak;s \E)(x) as k +. We have ( ) x y (ρ εk χ Ak;s \E)(x) = ε N k ρ χ Ak;s (y) χ R ε (R N \E)(y) dy N k = [y = x + ε k z] = ρ(z) χ Ak;s (x + ε k z) χ (R N \E)(x + ε k z) dz. B 1(0) We observe that x + ε k z R N \ E if and only if z (RN \E) x ε k, hence χ (R N \E)(x + ε k ) = χ (R N \E) x ( ) χ H νe (x) ( ) in L1 (B 1 (0)) as k +. ε k In particular, this means that the L 1 limit of χ (R N \E)(x+ε k z) is not L N -a.e. zero only if z ν E (x) 0, so we can restrict the integration domain to B 1 (0) H ν E (x). On the other hand, x + ε k z A k;s = {u k > s} if and only if u k (x + ε k z) > s. We see that u k (x + ε k z) = ρ εk (x + ε k z y) χ E (y) dy R N = [y = x + ε k z + ε k u] = ρ(u) χ E (x + ε k (u + z)) du. B 1(0)

20 0 MONICA TORRES Arguing as before, we obtain χ E (x + ε k (z + )) χ H + νe (x) (z + ) in L1 (B 1 (0)) as k +, for any x E and z B 1 (0). Now, we recall that z ν E (x) 0, and, since we have χ H + νe (x) (z + u) = 1 if and only if 0 (z + u) ν E(x), we conclude that 0 z ν E (x) u ν E (x) 1; that is, u belongs to the half ball B 1 (0) H ν + E (x). This implies that, for any x E and z B 1 (0) Hν E (x), (9.5) lim u k(x + ε k z) := v(x, z) = ρ(u) χ k + H + νe (x) (z + u) du 1. B 1(0) Therefore, these calculations yield (ρ εk χ Ak;s \E)(x) = ρ(z) χ Ak;s (x + ε k z) χ (RN \E)(x + ε k z) dz (9.6) B 1(0) B 1(0) ρ(z) χ {v(x,z)>s} (z) χ H νe (x)(z) dz, for any x E. Equation (9.5) shows then that the limit in (9.6) is identically zero, since { z R N : v(x, z) > s > 1 } B 1 (0) Hν E (x) =, for any x E. We can now apply the Lebesgue dominated convergence theorem with respect to the measure H N 1 E and the sequence of functions ρ εk χ Ak;s \E (since the constant 1 is clearly a summable majorant), thus obtaining (9.3). Finally, up to passing to a subsequence (which we shall keep calling ε k with a little abuse of notation), (9.) and (9.3) yield (9.1), for a.e. t > s. Since s > 1 is fixed arbitrarily, we can conclude that (9.1) is valid for a.e. t > 1. For the case t = 1 we have: Lemma 9.. Let µ be a Radon measure on R N such that µ << H N 1. Let E be a set of finite perimeter, and let u k be the mollification of χ E. Then, for t = 1 and ε > 0, there exists k = k (ε) such that (9.7) µ (E 1 \ A k; 1 ) < ε and µ (A k; 1 \ E) < ε. Proof. Since u k (y) u E (y) for H N 1 -a.e. y, the dominated convergence theorem implies that u k u E in L 1 (R N, µ ). Thus, given any ε > 0, for k large enough, we have (9.8) ε R u E u k d µ (u E u k )d µ (1 1 N E 1 \A k; ) µ (E1 \ A k; 1 ), 1 which implies (9.9) µ (E 1 \ A k; 1 ) ε. In the same way, we compute ε (9.10) u E u k d µ ( 1 0) µ (A k; 1 \ E), A k; 1 \E

21 FUNCTIONS OF BOUNDED VARIATION AND SETS OF FINITE PERIMETER 1 which implies (9.11) µ (A k; 1 \ E) ε. The following remark shows that, with t = 1 and with µ = HN 1 E 0, parts of Theorem 8.1 do not hold. Remark 9.3. If we define E := {y R N : y 1}, then u 1 k ( 1 ) RN \ E for all k, and therefore it is clear that (9.1) H N 1 ((A k; 1 \ E1 ) E) = H N 1 ( E) 0 as k. If we now define E := {y R N : y 1}, then u 1 k ( 1 ) E for all k and thus (9.13) H N 1 ((E \ A k; 1 ) E) = H N 1 ( E) 0 as k. Approximation of sets of finite perimeter completely from the inside. It is well-known that, a set of finite perimeter, E, cannot be approximated by smooth sets that lie completely in the interior of E. For example, consider the open unit disk with a single radius removed, and let U be the resulting open set. Then the Hausdorff measure of the boundary of U is π plus the measure of the radius, while the Hausdorff measure of the reduced boundary is π. Thus, if U k is an approximating open subset of U, then its boundary will be close to that of boundary U and so its the Hausdorff measure will be close to π plus 1. Adding more radii, say m of them, will force the approximating set to have boundaries whose Hausdorff measure close to π plus m. In general, if we let K denote any compact subset without interior and of infinite Hausdorff measure, then the approximating sets will have boundaries whose measures will necessarily tend to infinity. We have the following: Proposition 1. Let U R N be an open set with H N 1 ( U) <. Then there exists a sequence of bounded open sets U k U k U such that (i) U k = U k ; (ii) U k U ; (iii) H N 1 ( U k ) H N 1 ( U). Proof. By definition, for each integer k, there exists a covering of U by balls U B i (r i ), each with radius r i, such that H N 1( B i (r i ) ) = ω N 1 r N 1 i < H N 1 ( U) + 1 k, where ω N 1 is the H N 1 measure of the boundary of the unit ball in R N. Since U is compact, the covering may be taken as a finite covering, say by m of them, B 1 (r 1 ), B 1 (r ),..., B m (r m ). Then the open set V k := B i (r i ) has the property that V k m B i (r i ) and therefore that H N 1 ( V k ) H N 1 ( m B i (r i )) ω N 1 r N 1 i < H N 1 ( U) + 1 k.

22 MONICA TORRES Thus, the open sets U k := U \ V k U will satisfy our desired result, except that they are not smooth. Given an arbitrary set of finite perimeter, E, we know from 4 that E can be approximated by sets with smooth boundaries essentially from the measure-theoretic interior of E, that is, a one-sided approximation can almost be achieved (see Theorem 7.1(e)). On the other hand, the next result shows that, if E is sufficiently regular, there does, in fact, exist a one-sided approximation. The condition of regularity we impose is similar to Lewis s uniformly flat condition in potential theory [?]. Theorem 9.4. Suppose that E is a bounded set of finite perimeter with the property that, for all y E, there are positive constants c 0 and r 0 such that E 0 B(y, r) (9.14) c 0 for all r r 0. B(y, r) Then there exists t (0, 1) such that (9.15) A k;t E for large k. Proof. Choose a mollifying kernel ρ such that ρ = 1 on B(0, 1 ). If y E, we have v k (y) := χ R N \E ρ εk (y) = 1 χ ε N R N \E (x)ρ( x y ) dx k B(y,ε k ) ε k 1 χ ε N k B(y, ε R N \E (x) dx k ) = (RN \ E) B(y, ε k ) ε N k = E0 B(y, ε k ) ε N c 0 / N := c 0, k where 0 < c 0 < 1 depends only on the dimension N and is independent of the point y. Note that u k (y) + v k (y) = 1 for all y R N. Therefore, for all y E, u k (y) = 1 v k (y) 1 c 0. Thus, taking 1 c 0 < t < 1, we see that A k;t E =. Consequently, each connected component of the open set A k;t lies either in the interior of E or in its exterior, and thus must lie in its interior. Corollary 9.5. Let E be a bounded set of finite perimeter with uniform Lipschitz boundary. Then there exists T (0, 1) such that A k;t E. Proof. Since E has a uniform Lipschitz boundary, for each x E, there is a finite cone, C x, with vertex x that completely lies in the complement of E. Each cone C x is assumed to be congruent to a fixed cone C. This implies that the hypothesis of Theorem 9.4 is satisfied. Therefore, there exists 0 < T < 1 such that u k (y) < T for all k and all y E. 10. Interior approximations of open sets with C 0 and Lipschitz boundary. These will be discussed in class and used to obtain Generalized Gauss-Green formulas.

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