On the structure of divergence-free measures on ℝ²

Definition A.1.

A function f ∈ FV ⁡ ( ℝ d ) is said to be monotone if the sets { f > t } and { f ≤ t } are indecomposable for a.e. t ∈ ℝ .

Theorem A.2.

For any f ∈ FV ⁡ ( R d ) there exists an at most countable family { f i } i ∈ I ⊂ FV ⁡ ( R d ) of monotone functions such that

In particular,

Remark A.3.

Observe that from the embeddings of FV (see Theorem 2.3) the first series in (A.1) converges also in L 1 ⁣ * ⁢ ( ℝ d ) but, in general, we cannot improve this to convergence in L 1 ⁢ ( ℝ d ) . Secondly, we remark that the decomposition provided in Theorem A.2 is not unique: we refer the reader to the counterexample presented in the paper [7].

Lemma A.4.

Let φ , ψ ∈ FV ⁡ ( R d ) and assume 0 ≤ ψ ≤ φ .

1. If for a.e. t ∈ ℝ it holds

   (A.2) P ( { φ > t } ) = P ( { φ > t } ∖ { ψ > t } ) + P ( { ψ > t } ) ,

   then

   ∥ φ ∥ FV = ∥ φ - ψ ∥ FV + ∥ ψ ∥ FV .

2. If for a.e. t ∈ ℝ it holds

   (A.3) P ( { ψ > t } ) = P ( { φ > t } ∖ { ψ > t } ) + P ( { φ > t } ) ,

   then

   ∥ ψ ∥ FV = ∥ φ ∥ FV + ∥ φ - ψ ∥ FV .

Proof.

We present the proof of the two claims.

(1) Concerning the first point, it suffices to show

because the other inequality is trivial by the triangle inequality. Using the layer cake representation and Fubini’s Theorem we get

where the last equality follows from (A.2). Applying again the Coarea Formula, we obtain the conclusion.

(2) The proof of the second claim is similar to the proof of the first one. Notice that | D ⁢ w | = | D ⁢ ( - w ) | as measures for any w ∈ FV ⁡ ( ℝ d ) hence

which is equivalent to

It thus remains to show

By layer cake representation and Fubini, as in Point (1), we have

where the last equality follows from (A.3). Again the application of the Coarea Formula yields the desired conclusion. ∎

Lemma A.5 (From superlevel sets to function).

Let I ⊂ [ 0 , + ∞ ) be an interval and let ( A t ) t ∈ I be a family of sets such that t , s ∈ I with s < t implies A t ⊂ A s . Then there exists a measurable function w : R d → [ 0 , + ∞ ) such that { w > t } = A t (up to Lebesgue negligible subsets) for a.e. t ∈ I .

Proof.

Due to monotonicity of the family ( A t ) t ∈ I , the function h ( t ) := | A t | is non-increasing on I. Therefore there exists a Lebesgue negligible set N ⊂ I such that h is continuous at every t ∈ I ∖ N . Let Q ⊆ I ∖ N be a countable set, which is dense in I. For any x ∈ ℝ d we define

Clearly w is Lebesgue measurable. By definition of w for any s ∈ I ∖ N ,

Since for any s < t it holds | A t ∖ A s | = 0 and Q is countable, it follows that

On the other hand, let ε := | A s ∖ ⋃ t ∈ Q ∩ ( s , + ∞ ) ∩ I A t | . For any t ∈ Q ∩ ( s , + ∞ ) ∩ I we have A t ⊂ A s , hence

In particular, we can estimate

Since Q is dense in ( s , + ∞ ) ∩ I and h is continuous at s, the only possible case is ε = 0 . We have thus proved that | { g > s } △ A s | = 0 for a.e. s ∈ I and this concludes the proof. ∎

Lemma A.6 (Extraction Lemma I).

Let f ∈ FV ⁡ ( R d ) and assume f is not identically zero and non-negative. Then there exists g ∈ FV ⁡ ( R d ) with 0 ≤ g ≤ f and g ≢ 0 such that:

1. for a.e. t ≥ 0 the set { g > t } is indecomposable,

2. it holds ∥ f ∥ FV = ∥ f - g ∥ FV + ∥ g ∥ FV .

Proof.

For any t ≥ 0 let E t := { f > t } . Since f ∈ FV ⁡ ( ℝ d ) , there exists a Lebesgue negligible set N ⊆ ( 0 , + ∞ ) such that for any t ∈ ( 0 , + ∞ ) ∖ N the set E t has finite perimeter. Let E t k denote the k-th M-connected component of E t , t ∈ ( 0 , + ∞ ) ∖ N .

Fix some a > 0 such that | E a | > 0 . Let R be some M-connected component of E a . For any t ∈ ( 0 , a ) ∖ N we have E t ⊇ E a ⊇ R , and R is indecomposable, hence by Theorem 2.10 there exists a unique j = j ⁢ ( t ) such that

Let R t := E t j ⁢ ( t ) , t ∈ ( 0 , a ) ∖ N . Note that for any s , t ∈ ( 0 , a ) ∖ N with s < t it holds that

Indeed, E s ⊇ E t ⊇ R t and R t is indecomposable, hence again by Theorem 2.10 there exists a unique k such that | R t ∖ E s k | = 0 . But | R ∖ R t | = 0 , hence

is Lebesgue negligible. Therefore k = j ⁢ ( s ) by the uniqueness of j ⁢ ( s ) . Applying now Lemma A.5, we can construct a function g : ℝ d → [ 0 , a ] such that { g > s } = R s (up to Lebesgue negligible subsets) for a.e. s ∈ ( 0 , a ) . (See Figure 2.)

Observe that ∥ f ∥ FV = ∥ f ¯ ∥ FV + ∥ f ^ ∥ FV , where f ¯ ⁢ ( x ) = min ⁡ ( a , f ⁢ ( x ) ) and f ^ := f - f ¯ . For a.e. t ∈ ( 0 , a ) we have

hence by the construction of g and Proposition 2.12,

Hence by Lemma A.4 we have

Then by the triangle inequality

and ∥ f ∥ FV = ∥ g + f ¯ - g + f ^ ∥ FV ≤ ∥ g ∥ FV + ∥ f ¯ - g + f ^ ∥ FV , hence property (ii) follows. ∎

Lemma A.7 (Extraction Lemma II).

Let f ∈ FV ⁡ ( R d ) and assume f is not identically zero and non-negative. Then there exists h ∈ FV ⁡ ( R d ) with h ≢ 0 such that:

1. for a.e. t ≥ 0 the set { h > t } is simple,

2. it holds ∥ f ∥ FV = ∥ f - h ∥ FV + ∥ h ∥ FV .

Proof.

First of all, we apply Lemma A.6 and we obtain a function g ∈ FV ⁡ ( ℝ d ) such that G t := { g > t } is indecomposable for a.e. t ≥ 0 and

Let us now work on the function g. By the construction of g, for a.e. t ≥ 0 the set G t is indecomposable. Fix some a > 0 such that | G a | > 0 and G a is not simple (otherwise there is nothing to prove): let us denote by { F t i } i ∈ I t the non-empty family of holes of G t (i.e. 𝒞 ⁢ 𝒞 M ⁢ ( ℝ d ∖ G t ) = { F t i } i ∈ I t ).

Observe that, if H is an hole of G a , for any t ∈ ( a , + ∞ ) ∖ N , we have G t ⊆ G a , and hence G t c ⊇ G a c ⊇ H : this means that H is an hole of G t for any t ∈ ( a , + ∞ ) ∖ N : by the uniqueness claim in Theorem 2.10 there exists a unique j = j ⁢ ( t ) such that | H ∖ F t j ⁢ ( t ) | = 0 .

For any t ∈ ( 0 , a ) define S t := sat ⁡ ( G t ) . Observe that the sequence ( S t ) t ∈ ( 0 , a ) is monotone [1, Proposition 6 (iii)] and thus, applying Lemma A.5, we obtain a function h : ℝ d → ℝ such that { h > r } = S r (up to Lebesgue negligible subsets) for a.e. r ∈ ( 0 , a ) . By construction the function h is non-negative and { h > r } is simple for a.e. r ∈ ( 0 , a ) , because the saturation of an indecomposable set is simple. It thus remains to show property (ii) of the statement. For, notice preliminarily, that h - g ≥ 0 by the construction of h; by [1, Proposition 9], it holds for any t ∈ ( a , + ∞ ) ∖ N ,

which can be also written as

We are now in a position to apply Lemma A.4 (ii), choosing φ := h and ψ := g (which is possible since h ≥ g ): we obtain

It is now easy to check that property (ii) follows combining (A.4) with (A.5) – and the triangle inequality:

and this completes the proof. ∎

Lemma A.8 (Extraction Lemma III).

Let f ∈ FV ⁡ ( R d ) and assume f is not identically zero. Then there exists m ∈ FV ⁡ ( R d ) with m ≢ 0 such that:

1. m is monotone and sign ⁡ m = constant a.e.,

2. it holds ∥ f ∥ FV = ∥ f - m ∥ FV + ∥ m ∥ FV .

Proof.

Let us decompose f = f + - f - . Suppose ∥ f + ∥ FV > 0 . Since f + ≥ 0 , we can apply Lemma A.6 to f + , thus obtaining a function u ≥ 0 such that { u > t } is indecomposable for a.e. t > 0 and it holds

Applying now Lemma A.7 to u ≥ 0 , we obtain a function m ∈ FV ⁡ ( ℝ d ) such that for a.e. t ≥ 0 the set { m > t } is simple and it holds

By the triangle inequality,

hence property (ii) holds true. Since the function m is monotone, this concludes the proof in the case ∥ f + ∥ FV > 0 . It remains to consider the case in which f + ≡ 0 . If f - ≡ 0 , there is nothing to prove; if ∥ f - ∥ FV > 0 , then we repeat the same argument above for the function f ~ := - f ∈ FV ⁡ ( ℝ d ) . We end up with a monotone function m ~ of constant sign such that

which is clearly equivalent to property (ii) (renaming - m ~ as m). ∎

Proof of Theorem A.2.

Let X := { g ∈ FV ⁡ ( ℝ d ) : g  is monotone and  ∥ g ∥ FV > 0 } . For any h ∈ FV ⁡ ( ℝ d ) let

Note that by Lemma A.8 Y ⁢ ( h ) = ∅ if and only if h ≡ 0 . Ultimately, let 𝔰 : 𝒫 ⁢ ( FV ⁡ ( ℝ d ) ) → FV ⁡ ( ℝ d ) denote a choice function (given by the Axiom of Choice).

Let us define, for any ordinal α < ω 1 (where ω 1 is the first uncountable ordinal) and any transfinite sequence { g ξ } ξ < α ⊂ X ∪ { ∞ } ,

By transfinite recursion (see e.g. [11, p. 21]) there exists a transfinite sequence { g α } α < ω 1 such that

for any α < ω 1 .

Note that for any α < ω 1 the following properties hold:

(A.8)

Observe that (A.8b) follows from (A.8c), but without (A.8b) the term ∑ ξ < α g ξ in (A.8a) is not well-defined. Indeed, these properties trivially hold for α = 0 . Let β < ω 1 and suppose that these properties hold for any α < β . In order to show that (A.8a)–(A.8c) hold with α = β , we consider two cases.

First, if β is not a limit ordinal, then β = γ + 1 for some ordinal γ, so by the definition of { g ξ } ξ < ω 1 we have

Hence

and it follows that (A.8a)–(A.8c) hold with α = γ + 1 .

Second, if β is a limit ordinal, then β = ⋃ α < β α . Consequently,

hence condition (A.8a) holds with α = β . Furthermore, since β is at most countable, we can enumerate it as β = { α n } n ∈ ℕ . Let A n := α 1 ∪ … ∪ α n (note that for any n ∈ ℕ there exists m ∈ { 1 , … , n } such that A n = α m ). Since β = ⋃ α < β α = ⋃ n ∈ ℕ A n , we have

hence (A.8b) holds with α = β . Consequently,

and

Writing (A.8c) with α = A n and passing to the limit as n → ∞ , we conclude that (A.8c) holds with α = β . We have thus shown that (A.8a)–(A.8c) hold with α = β . Hence by transfinite induction (A.8a)–(A.8c) hold for any α < ω 1 .

By (A.8b) for any ε > 0 the set { α < ω 1 : ∥ g α ∥ FV > ε } is finite and thus the set A := { α < ω 1 : ∥ g α ∥ FV > 0 } is at most countable. Setting γ := sup ⁡ A , we have g γ + 1 = 0 . As already noted above, by Lemma A.8 this means that

and

by (A.8c).

By the triangle inequality,

If this inequality were strict, we would have

which is a contradiction. ∎