Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 1, 3–12

A Note on Convergence of Level Sets

F. Camilli

Abstract. Given a sequence of functions f_{n} converging in some topology to a function f, in general the 0-level set of f_{n} does not give a good approximation of the one of f. In this paper we show that, if we consider an appropriate perturbation of the 0-level set of f_{n}, we get a sequence of sets converging to the 0-level set of f, where the type of set convergence depends on the type of convergence of f_{n} to f.

Keywords: Perturbed level sets, set convergence, capacity AMS subject classification: 28 A 12, 46 E 35

1. Introduction

In several fields (phase transition, free boundary problems, front propagation, etc.), a set of interest for the solution of the problem is represented by a level or a sublevel set of a function f. Let us suppose that by means of some approximation technique (f.e. discretization, regularization, rescaling of an order parameter) we get a sequence of functions converging in some topology to f. In general, no matter how strong is the convergence of f_{n} to f, the level sets of f_{n} do not give a good approximation of the ones of f.

Pursuing an idea used in Baiocchi and Pozzi [1], we show that appropriately perturbing the level sets of f_{n} (the same can be done for the sublevels or the superlevels), we get a sequence of sets defined by means of f_{n} converging to the level set of f. The type of set convergence is the convergence to zero of the measure of the symmetric difference between the level set of f_{n} and the correpsonding one of f, and the measure depends on the type of convergence of the sequence f_{n}.

We analyze the case of convergence in L^{p} and in W^{1,p}, but this technique could be useful in other situations.

The paper is organized as follows. In Section 2, we analyze the case of convergence in L^{∞} and W^{1,∞} and the associated convergence of perturbed level sets in set-theoretical sense. In Section 3 we first consider the case of convergence in L^{p}, which gives the convergence in the sense of Lebesgue measure. Then we analyze the case of convergence in W^{1,p} and the corresponding set convergence in the sense of capacity and Hausdorf measure.

F. Camilli: Univ. di Torino, Dip. di Matematica, Via Carlo Alberto 10, 10123 Torino (Italy)

New address: Univ. dell‘ Aquila, Dip. di Energetica, 67040 Roio Poggio (AQ), Italy e-mail: camilli@ing.univaq.it ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

2. The case p = ∞

In this section we will study (extending the result given in [1]) the case of the convergence in L^{∞}. We will see that the natural set convergence associated to the L^{∞} convergence is the convergence in set-theoretical sense.

Definition 2.1. Given a sequence of sets {A_{n}}_{n∈ℕ}, we set

lim sup_{n→∞} A_{n} = ⋂_{n=1}^{∞} ⋃_{m=n}^{∞} A_{m}  and  lim inf_{n→∞} A_{n} = ⋃_{n=1}^{∞} ⋂_{m=n}^{∞} A_{m}.

We say that {A_{n}}_{n∈ℕ} converges to A in set-theoretical sense and write A = lim_{n→∞} A_{n} if

A = lim sup_{n→∞} A_{n} = lim inf_{n→∞} A_{n}.

We have the following result.

Proposition 2.1. Let f_{n} and f be continuous functions on ℝ^{N} such that

∥f − f_{n}∥_{L^{∞}(ℝ^{N})} = ε_{n}  (2.1)

where ε_{n} → 0 for n → ∞. Let {δ_{n}}_{n∈ℕ} be a sequence such that

\begin{matrix}δ_{n} > & 0  (n ∈ ℕ) \\ δ_{n} → & 0  (n → ∞) \\ \frac{ε_{n}}{δ_{n}} → & 0  (n → ∞). \end{matrix} \}  (2.2)

Set, for any n ∈ ℕ,

\begin{aligned}Γ & = \{x ∈ ℝ^{N} : f(x) = 0\} \\ Γ_{n} & = \{x ∈ ℝ^{N} : |f_{n}(x)| ≤ δ_{n}\}. \end{aligned}  (2.3)

Then Γ ⊂ Γ_{n}, for n sufficiently large, and

Γ = lim_{n→∞} Γ_{n}.  (2.4)

Proof. Let \overline{n} ∈ ℕ be such that δ_{n} ≥ ε_{n} for any n ≥ \overline{n} (recall that \frac{ε_{n}}{δ_{n}} → 0). If x ∈ Γ, then, for n ≥ \overline{n}, we have from (2.1)

|f_{n}(x)| ≤ |f(x)| + ∥f_{n} − f∥_{L^{∞}(ℝ^{N})} = ε_{n} ≤ δ_{n},

hence x ∈ Γ_{n}. Hence Γ ⊂ Γ_{n} for n ≥ \overline{n} and therefore Γ ⊂ lim inf_{n→∞} Γ_{n}. Let us prove yet that lim sup_{n→∞} Γ_{n} ⊂ Γ. If x ∈ lim sup_{n→∞} Γ_{n}, then by definition there exists a subsequence {Γ_{n_{k}}}_{k≥1} such that x ∈ Γ_{n_{k}} for any k ∈ ℕ. It follows that |f_{n_{k}}(x)| ≤ δ_{n_{k}} for any k ∈ ℕ and therefore f(x) = lim_{k→∞} f_{n_{k}}(x) = 0 which yields x ∈ Γ

Remark 2.1 Observe that if Γ_{n} and Γ are contained in a compact set K, then the previous proposition gives the convergence to zero of the Hausdorff distance between Γ_{n} and Γ.

In the next proposition we show that improving the convergence of f_{n} to f, we get some additional information on the type of convergence of Γ_{n} to Γ.

Proposition 2.2. Let f, f_{n} ∈ C^{1}(ℝ^{N}) (n ∈ ℕ) be such that
∥f − f_{n}∥_{W^{1,∞}(ℝ^{N})} = ε_{n}

where ε_{n} → 0 for n → ∞. Let δ_{n} and Γ and Γ_{n} be defined as in (2.2) − (2.3). Set

Γ^{reg} = {x ∈ ℝ^{N} : f(x) = 0 and ∇f(x) ≠ 0}
Γ^{sing} = {x ∈ ℝ^{N} : f(x) = 0 and ∇f(x) = 0}

and

Γ_{n}^{reg} = {x ∈ ℝ^{N} : |f_{n}(x)| ≤ δ_{n} and |∇f_{n}(x)| > δ_{n}}
Γ_{n}^{sing} = {x ∈ ℝ^{N} : |f_{n}(x)| ≤ δ_{n} and |∇f_{n}(x)| ≤ δ_{n}}.

Then

Γ^{reg} = lim_{n→∞} Γ_{n}^{reg}  and  Γ^{sing} = lim_{n→∞} Γ_{n}^{sing}.

Proof. Let \overline{n} ∈ ℕ be such that δ_{n} ≥ ε_{n} for n ≥ \overline{n}. Then, for n ≥ \overline{n}, Γ ⊂ Γ_{n} and, if x ∈ Γ^{sing}, we have

|∇f_{n}(x)| ≤ |∇f(x)| + ∥∇f_{n} − ∇f∥_{L^{∞}(ℝ^{N})} = ε_{n} ≤ δ_{n}.

Therefore Γ^{sing} ⊂ Γ_{n}^{sing} for n ≥ \overline{n}. If x ∈ lim sup_{n→∞} Γ_{n}^{sing}, then x ∈ Γ_{n_{k}}^{sing} for a subsequence Γ_{n_{k}}. It follows that |f_{n_{k}}(x)| ≤ δ_{n_{k}} and |∇f_{n_{k}}(x)| ≤ δ_{n_{k}} for any k ∈ ℕ and therefore

f(x) = lim_{k→∞} f_{n_{k}}(x) = 0  and  ∇f(x) = lim_{k→∞} ∇f_{n_{k}}(x) = 0.

Therefore x ∈ Γ^{sing} and Γ^{sing} = lim_{n→∞} Γ_{n}^{sing}. Since (2.4) holds, we get also Γ^{reg} = lim_{n→∞} Γ_{n}^{reg}

We conclude this section giving an estimate of the Hausdorff distance between Γ and Γ_{n} in the case that Γ is regular.

Proposition 2.3. Assume the same hypothesis as in Proposition 2.1, with δ_{n} and Γ, Γ_{n} defined as in (2.2) − (2.3). Moreover, assume that Γ is compact and that f is differentiable with ∇f ≠ 0 on Γ. Then there exists a constant C > 0 such that

d_{H}(Γ, Γ_{n}) ≤ C(ε_{n} + δ_{n})  (2.5)

for n sufficiently large, where d_{H} denotes the Hausdorff distance.

Proof. By the assumptions on f and Γ, there exist η_{0} > 0 and C_{0} > 0 such that |∇f(x)| ≥ C_{0} on Γ_{η_{0}} = {x : d(x, Γ) ≤ η_{0}}. For η ≤ η_{0}, consider y ∈ ∂(Γ_{η}) = ∂{x : d(x, Γ) ≤ η} and let x ∈ Γ be such that d(y, Γ) = |y − x| = η. Then

|(y − x) · ∇f(x)| = η|∇f(x)| ≥ C_{0}η.

Since f(x) = 0, if ω is a modulus of continuity of ∇f on Γ_{η_{0}}, then

|f(y)| ≥ |(y − x) · ∇f(x)| − ω(|y − x|)|y − x| ≥ η(C_{0} − ω(η)).  (2.6)

For n sufficiently large in such a way that C_{0} − ω(δ_{n} + ε_{n}) ≥ \frac{C_{0}}{2} and 2\frac{δ_{n}+ε_{n}}{C_{0}} ≤ η_{0}, from (2.6) with η = 2\frac{δ_{n}+ε_{n}}{C_{0}} we get |f(y)| ≥ δ_{n} + ε_{n} and therefore |f_{n}(y)| ≥ δ_{n} on ∂Γ_{η}. It follows that Γ_{n} ⊂ Γ_{η}. Since Γ ⊂ Γ_{n} for n sufficiently large, we finally get d_{H}(Γ, Γ_{n}) ≤ d_{H}(Γ, Γ_{η}) ≤ η and therefore (2.5), with C = \frac{2}{C_{0}}

All the results of this section have an analogue in the case of sub- and superlevel sets of f_{n} and f.

3. The case 1 ≤ p < ∞

We first analyze the case of convergence in L^{p}(ℝ^{N}). We prove that in this case an appropriate notion of set convergence is the convergence to 0 of the Lebesgue measure of ΓΔΓ_{n}. In the following, L^{N} denotes the Lebesgue measure on ℝ^{N}.

Proposition 3.1. Let f_{n}, f ∈ L^{p}(ℝ^{N}) (1 ≤ p < ∞; n ∈ ℕ) such that
∥f − f_{n}∥_{L^{p}(ℝ^{N})} = ε_{n}  (3.1)

where ε_{n} → 0 for n → ∞. Let {δ_{n}}_{n∈ℕ} be a sequence such that

0 < δ_{n} (n ∈ ℕ)  and  δ_{n} → 0 (n → ∞).  (3.2)

Define, for any n ∈ ℕ,

\begin{aligned}Γ & = \{x ∈ ℝ^{N} : f(x) = 0\} \\ Γ_{n} & = \{x ∈ ℝ^{N} : |f_{n}(x)| ≤ δ_{n}\}. \end{aligned}  (3.3)

Then:

(i) If lim_{n→∞} \frac{ε_{n}}{δ_{n}} = 0, we have

lim_{n→∞} L^{N}(ΓΔΓ_{n}) = 0  (3.4)
L^{N}(Γ△ lim sup_{n→∞} Γ_{n}) = 0.  (3.5)

(ii) If

∑_{n} (\frac{ε_{n}}{δ_{n}} )^{p} < ∞,  (3.6)

we also have

L^{N}(Γ△ lim inf_{n→∞} Γ_{n}) = 0.  (3.7)

Therefore Γ = lim_{n→∞} Γ_{n} up to a set of 0-Lebesgue measure.

Proof. We first observe that, since we are considering only the measure of Γ and Γ_{n}, we can assume that these sets are defined by means of any element in the class of equivalence of f and f_{n}. We have

ΓΔΓ_{n} = (Γ \ Γ_{n}) ∪ (Γ_{n} \ Γ)

and

Γ \ Γ_{n} = {x ∈ ℝ^{N} : f(x) = 0 and |f_{n}(x)| > δ_{n}}
Γ_{n} \ Γ = {x ∈ ℝ^{N} : f(x) ≠ 0 and |f_{n}(x)| ≤ δ_{n}}

(the previous and all the others inclusions in this proof are intended up to sets of null Lebesgue measure).

Since Γ ⊂ Γ_{n} ⊂ {x ∈ ℝ^{N} : |f(x) − f_{n}(x)| > δ_{n}}, from the Cebycev inequality we get

L^{N}(Γ \ Γ_{n}) ≤ \frac{1}{δ_{n}^{p}} ∫_{ℝ^{N}} |f(x) − f_{n}(x)|^{p}dx = (\frac{ε_{n}}{δ_{n}} )^{p}  (3.8)

and therefore

lim_{n→∞} L^{N}(Γ \ Γ_{n}) = 0.  (3.9)

Let us prove that

L^{N}( lim sup_{n→∞}(Γ_{n} \ Γ)) = 0.  (3.10)

Set Γ̃ = lim sup_{n→∞}(Γ_{n} \ Γ) and let x ∈ Γ̃. Then there exists a subsequence {n_{k}}_{k≥1} such that |f_{n_{k}}(x)| ≤ δ_{n_{k}} for any k ∈ ℕ. It follows that, L^{N}-a.e. on Γ̃,

lim inf_{n→∞} f_{n}(x) = 0  and  lim inf_{n→∞} |f_{n}(x) − f(x)|^{p} = |f(x)|^{p}.

Applying the Fatou Lemma we get

∫_{Γ} |f(x)|^{p}dx ≤ lim_{n→∞} ∫_{ℝ^{N}} |f(x) − f_{n}(x)|^{p}dx = 0.

Since |f(x)| > 0 on Γ̃, we get (3.10).

Since for any sequence {A_{n}}_{n≥1} of measurable sets we have

lim sup_{n→∞} L^{N}(A_{n}) ≤ L^{N}( lim sup_{n→∞} A_{n})

it follows from (3.10) that lim_{n→∞} L^{N}(Γ_{n} \ Γ) = 0 and therefore, together with (3.9), also (3.4) holds. From (3.10) and

L^{N}(Γ \ lim sup_{n→∞} Γ_{n}) = L^{N}( lim inf_{n→∞}(Γ \ Γ_{n})) ≤ lim_{n→∞} L^{N}(Γ \ Γ_{n}) = 0

we get (3.5).

Let us prove now statement (ii). Estimate (3.8) gives

L^{N}( ⋃_{m=n}^{∞}(Γ \ Γ_{m})) ≤ ∑_{m=n}^{∞} (\frac{ε_{m}}{δ_{m}} )^{p},

and therefore, for any n ∈ ℕ,

L^{N}(Γ \ lim inf_{n→∞} Γ_{n}) = L^{N}( lim sup_{n→∞}(Γ \ Γ_{n})) ≤ ∑_{m=n}^{∞} (\frac{ε_{m}}{δ_{m}} )^{p}.

For (3.6) we get L^{N}(Γ\lim inf_{n→∞} Γ_{n}) = 0. Since (3.10) yields L^{N}(lim inf_{n→∞} Γ_{n}\Γ) = 0 we get (3.7)

Remark 3.1. Since we have

|L^{N}(Γ) − L^{N}(Γ_{n})| ≤ L^{N}(Γ△Γ_{n})

then L^{N}(ΓΔΓ_{n}) → 0 implies that L^{N}(Γ_{n}) → L^{N}(Γ). The vice versa in general is not true. The result L^{N}(ΓΔΓ_{n}) → 0 gives a more complete information respect to the convergence of the measure of Γ_{n} to the measure of Γ. In fact, it shows that the measure of the part of Γ_{n} which does not approximate Γ tends to 0, while the measure of Γ \ Γ_{n} can be estimated by means of (3.8).

If we know that ∥f_{n} − f∥_{W^{1,p}(ℝ^{N})} = ε_{n}, then we can prove a result similar to Proposition 2.2 for the convergence of regular and singular parts of Γ_{n} to Γ. In this case, a more accurate way of studying properties of sets defined through Sobolev functions is given by the notion of capacity. We will show that, in the case of convergence in W^{1,p}(ℝ^{N}) (1 ≤ p < N) we get convergence of Γ_{n} to Γ up to sets of 0 capacity. Let us recall the definition and some basic properties of the capacity we will need in the following (see [2 - 4] for more details).

Definition 3.1. Let 1 ≤ p < N and set

K^{p} = {φ : ℝ^{N} → ℝ| 0 ≤ φ ∈ L^{p^{∗}}(ℝ^{N}) with ∇φ ∈ L^{p}(ℝ^{N}, ℝ^{N})}

where p^{∗} = \frac{Np}{N−p}. For A ⊂ ℝ^{N}, we define

Cap_{p}(A) = inf { ∫_{ℝ^{N}} |∇φ| dy| φ ∈ K^{p} with A ⊂ {φ ≥ 1}^{∘}}.

It is possible to prove that Cap_{p} is an exterior measure on subsets of ℝ^{N}. For a function φ ∈ L_{loc}^{1}(ℝ^{N}), the precise representative φ^{∗} of φ is defined by

φ^{∗}(x) = \{ \begin{matrix}lim_{r→0^{+}} −_{∫_{B(x,r)}} φ(y) dy & if the limit exists \\ 0 & otherwise \end{matrix}

where −_{∫_{B(x,r)}} φ(y) dy = ∫_{B(x,r)} φ(y) dy/L^{N}(B(x, r)). We have (see [2: Theorem 4.8.1]) the following

Theorem 3.1. Let φ ∈ W^{1,p}(ℝ^{N}) (1 ≤ p < N). Then:

(i) There is a Borel set E ⊂ ℝ^{N} such that Cap_{p}(E) = 0 and

lim_{r→0^{+}} −_{∫B(x,r)} φ(y) dy = φ^{∗}(x) (x ∈ ℝ^{N} \ E).

(ii) In addition,

lim_{r→0^{+}} −_{∫B(x,r)} |φ(y) − φ^{∗}(x)|^{p^{∗}}dy = 0 (x ∈ ℝ^{N} \ E).

(iii) The precise representative φ^{∗} is quasi-continuous.

Because of the previous theorem, any function in the space W^{1,p}(ℝ^{N}) admits a quasi-continuous representative. We have the following convergence result for the perturbed level sets.

Proposition 3.2. Let f, f_{n} ∈ W^{1,p}(ℝ^{N}) (1 ≤ p < N) be such that
∥f − f_{n}∥_{W^{1,p}(ℝ^{N})} = ε_{n}

where ε_{n} → 0 for n → ∞. Let δ_{n} and Γ, Γ_{n} be defined as in (3.2) − (3.3) by means of the precise representatives of f and f_{n}. Then:

(i) If lim_{n→∞} \frac{ε_{n}}{δ_{n}} = 0, then

Cap_{p}( lim sup_{n→∞} Γ_{n}△Γ) = 0.  (3.11)

(ii) If

∑_{n} (\frac{ε_{n}}{δ_{n}} )^{p} < ∞,  (3.12)

then we have also

Cap_{p}(Γ△ lim inf_{n→∞} Γ_{n}) = 0  (3.13)

and therefore Γ = lim_{n→∞} Γ_{n} up to a set of zero capacity.

Proof. Let us prove statement (i). Since the sets Γ and Γ_{n} are defined by means of the precise representatives of f and f_{n}, then they are well defined, i.e. up to sets of zero capacity. In the following all the relations involving Γ and Γ_{n} are intended to be satisfied Cap_{p}-a.e. We have

Γ \ Γ_{n} = {x ∈ ℝ^{N} : f(x) = 0 and |f_{n}(x)| > δ_{n}}.

Let us prove that, defining

B_{n} = {x ∈ ℝ^{N} | −_{∫B(x,r)} |f_{n} − f| dy > δ_{n} for some r > 0},  (3.14)

then

Cap_{p}(Γ \ Γ_{n}) ≤ Cap_{p}(B_{n}).  (3.15)

In fact, if x ∈ Γ \ Γ_{n}, then, up to a set of zero capacity, we have

lim_{r→0} −_{∫B(x,r)} |f_{n} − f| dy = |f(x) − f_{n}(x)| > δ_{n}.

Therefore there exists r_{0} > 0 such that −_{∫_{B(x,r)}} _{B(x,r_{0})}|f_{n} −f| dy > δ_{n} and so (3.15) holds. Recall that (see [2: Lemma 4.8.1]), if φ ∈ K^{p}, then there exists a constant C, depending only on N and p, such that for any η > 0

Cap_{p}(\{x ∈ ℝ^{N} | −_{∫B(x,r)} φ(y) dy > η for some r > 0\}) ≤ \frac{C}{η^{p}} ∫_{ℝ^{N}} |Dφ|^{p}dy. (3.16)

From (3.14) and (3.16) we get

Cap_{p}(Γ \ Γ_{n}) ≤ \frac{C}{δ_{n}^{p}} ∫_{ℝ^{N}} |∇f − ∇f_{n}|^{p}dy ≤ C(\frac{ε_{n}}{δ_{n}} )^{p}  (3.17)

and therefore lim_{n→∞} Cap_{p}(Γ \ Γ_{n}) = 0. From the previous equality and the properties of the capacity, we get

Cap_{p}(Γ \ lim sup_{n→∞} Γ_{n}) = Cap_{p}( lim inf_{n→∞}(Γ \ Γ_{n})) ≤ lim_{n→∞} Cap_{p}(Γ \ Γ_{n}) = 0. (3.18)

Let A be the set

A = \{x ∈ ℝ^{N} | lim sup_{r→0^{+}} \frac{1}{r^{n−p}} ∫_{B(x,r)} |∇f(y)|^{p}dy > 0\} .

Then Cap_{p}(A) = 0 (see [2: Theorem 2.4.3]) and from the Poincaré inequality we have

lim_{r→0} −_{∫B(x,r)} |f(y) − (f)_{x,r}|^{p^{∗}}dy = 0  (x ∈ ℝ^{N} \ A)  (3.19)

where (f)_{x,r} = −_{∫_{B(x,r)}} f(y) dy. From Theorem 3.1, for any n ∈ ℕ there exists a Borel set E_{n} such that Cap_{p}(E_{n}) = 0 and

lim_{r→0^{+}} −_{∫B(x,r)} |f_{n}(y) − f_{n}(x)|^{p^{∗}}dy = 0  (x ∈ ℝ^{N} \ E_{n}).  (3.20)

Set Δ_{n} = B_{n} ∪ E_{n} ∪ A, where B_{n} has been defined in (3.14). If x ∈ Γ_{n} \ Δ_{n}, then from Theorem 3.1, (3.14) and (3.19) - (3.20) we get

lim sup_{r→0} |(f)_{x,r}| ≤ lim sup_{r→0} |(f)_{x,r} − f_{n}(x)| + δ_{n}

≤ lim sup_{r→0} { −_{∫B(x,r)} |f − (f)_{x,r}| dy

+ −_{∫B(x,r)} |f − f_{n}| dy + −_{∫B(x,r)} |f_{n} − f_{n}(x)| dy} + δ_{n}

≤ 2δ_{n}.

(3.21)

Moreover, inequality (3.16) gives

Cap_{p}(Δ_{n}) ≤ Cap_{p}(B_{n}) + Cap_{p}(E_{n}) + Cap_{p}(A) ≤ C(\frac{ε_{n}}{δ_{n}} )^{p}.  (3.22)

Set Δ = lim inf_{n→∞} Δ_{n} and Γ̃ = lim sup_{n→∞}(Γ_{n} \ Γ). From (3.21) - (3.22) it follows that if x ∈ Γ̃\Δ, then lim_{r→0^{+}}(f)_{x,r} = 0. Therefore from Theorem 3.1 we get Γ̃\Δ ⊂ Γ and, since Cap_{p}(Δ) ≤ lim inf_{n→∞} Cap_{p}(Δ_{n}) = 0, it follows also that

Cap_{p}( lim sup_{n→∞}(Γ_{n} \ Γ)) = 0.

Convergence of Level Sets 11

The previous equality and (3.18) imply (3.11).

Let us prove statement (ii). If x ∈ Γ \ B_{n}, then

lim sup_{r→0^{+}} −_{∫B(x,r)} |f_{n}| dy ≤ lim sup_{r→0^{+}} ( −_{∫B(x,r)} |f| dy + −_{∫B(x,r)} |f − f_{n}| dy) ≤ δ_{n}. (3.23)

Thus (3.23) yields Γ \ B_{n} ⊂ Γ_{n} for any n and therefore

lim inf_{n→∞}(Γ \ B_{n}) = Γ \ lim sup_{n→∞} B_{n} ⊂ lim inf_{n→∞} Γ_{n}.

Set B = lim sup_{n→∞} B_{n}. Then, for any n ∈ ℕ,

Cap_{p}(B) ≤ ∑_{m=n}^{∞} Cap_{p}(B_{m}) ≤ ∑_{m=n}^{∞} (\frac{ε_{m}}{δ_{m}} )^{p}

and, for hypothesis (3.12) , we get Cap_{p}(B) = 0 and (Γ \ B) ⊂ lim inf_{n→∞} Γ_{n}. From statement (i) we get (3.13)

Remark 3.2. For the capacity we do not have an analogy of property (3.4). While, as we have proved, lim_{n→∞} Cap_{p}(Γ \ Γ_{n}) = 0 in general, it is not true that lim_{n→∞} Cap_{p}(Γ_{n} \ Γ) = 0 as it can been easy seen taking f_{n} ≡ f.

Taking into account the relation between capacity and Hausdorff measure (see [2, 3]), from the previous proposition we get the following result about convergence in the sense of the Hausdorff measure.

Corollary 3.1. Under the same hypothesis of Proposition 3.2, we have the following:

(i) If lim_{n→∞} \frac{ε_{n}}{δ_{n}} = 0, then for any σ > 0
H^{N−p+σ}( lim sup_{n→∞} Γ_{n}△Γ) = 0.  (3.24)

(ii) If ∑_{n} (\frac{ε_{n}}{δ_{n}} )^{p} < ∞, then we also have, for any σ > 0,
H^{N−p+σ}(Γ△ lim inf_{n→∞} Γ_{n}) = 0.  (3.25)

If p = 1, then (3.24) − (3.25) hold also for σ = 0.

If p > N, since W^{1,p}(ℝ^{N}) ⊂ L^{∞}(ℝ^{N}) with continuous immersion, we can apply the results of Section 2 to the continuous representatives of f and f_{n}. Therefore, from the convergence of f_{n} to f we get the convergence in the set theoretical sense of Γ_{n} to Γ.

Acknowledgement. The author wishes to thank Prof. C.Baiocchi of the University of Roma “La Sapienza” for suggesting the problem.

References

[1] Baiocchi, C. and G. A. Pozzi: Error estimates and free-boundary convergence for a finite difference discretization of a parabolic variational inequality. RAIRO Anal. Numér. 11 (1977), 315 – 353.

[2] Evans, L. C. and R. F. Gariepy: Measure Theory and Fine p Properties of Functions. Boca Raton: CRC Press 1992.

[3] Federer, H. and W. Ziemer: The Lebesgue set of a function whose distribution derivatives are p-th power summable. Indiana Univ. Math. J. 22 (1972), 139 – 158.

[4] Ziemer, W.: Weakly Differentiable Functions. New York: Springer-Verlag 1989.

Received 22.03.1998