# A Smirnov-Type-decomposition on open simple regular curves and properties

Anéva Doliciane Tsafack, Laure Blanc-Féraud, Gilles Aubert

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Anéva Doliciane Tsafack, Laure Blanc-Féraud, Gilles Aubert. A Smirnov-Type-decomposition on open simple regular curves and properties. 2026. hal-05468332v2

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# A SMIRNOV-TYPE DECOMPOSITION ON OPEN SIMPLE REGULAR CURVES AND PROPERTIES *

ANÉVA DOLICIANE TSAFACK†, LAURE BLANC-FÉRAUD†, AND GILLES AUBERT‡

**Abstract.** We present a new decomposition of two-dimensional divergence-measure vector fields into elementary measures supported on simple open curves, parametrized at constant speed. Our construction builds on Smirnov's decomposition theorem but is formulated in terms of parametrized curves of class $C^3$, with constant speed; in particular, in the two-dimensional case, the non-constant curves are simple and open. The resulting structure facilitates the analysis of total variation measures and the study of their convergence, with potential applications in geometric measure theory and imaging.

**Keywords:** divergence-measure fields, Smirnov decomposition, simple curves, regular curves.

**1. Preliminaries.** Throughout this work, the following notations are adopted.

## 1.1. Notations.

* $\| \cdot \|$ denotes the Euclidean norm on $\mathbb{R}^d$. The dimension $d$ is always clear from the context.

* $\#A$ denotes the cardinality (i.e., the number of elements) of the set $A$ and $\bar{A}$ the closure of $A$.

* supp denotes the support of a function or a measure.

* a.e. stands for “almost everywhere” or “almost every,” with respect to the relevant measure.

* For vectors $u, v \in \mathbb{R}^2$, $u \wedge v$ denotes the determinant of the $2 \times 2$ matrix whose columns are $u$ and $v$.

* $\mathcal{H}^1$ denotes the one-dimensional Hausdorff measure; $\mathcal{H}^1 \llcorner A$ stands for its restriction to a measurable set $A$.

* $\mathcal{L}^1$ denotes the one-dimensional Lebesgue measure.

The following concepts and properties of Radon measures are recalled, forming the mathematical foundation of the functional framework. More details can be found in [1, 6].

**1.2. Radon Measures.** Let $d \in \mathbb{N}^*$ and let $\Omega$ be a locally compact Hausdorff metric space and $\mathcal{B}(\Omega)$ its Borel $\sigma$-algebra.

We denote by $C_c(\Omega; \mathbb{R}^d)$ the space of continuous functions $f: \Omega \to \mathbb{R}^d$ with compact support. The space $C_0(\Omega; \mathbb{R}^d)$ consists of continuous functions vanishing at infinity, i.e:

For every $\epsilon > 0$, there exists a compact $K \subset \Omega$ with $\|f(x)\| \le \epsilon$ for all $x \in \Omega \setminus K$.

Equipped with the supremum norm $\|f\|_{\infty} := \sup_{x \in \Omega} \|f(x)\|$, the space $C_0(\Omega; \mathbb{R}^d)$ is the closure of $C_c(\Omega; \mathbb{R}^d)$.

**DEFINITION 1.1 (Radon measure).** The space of finite $d$-dimensional vector-valued Radon measures on $(\Omega, \mathcal{B}(\Omega))$, denoted by $\mathcal{M}(\Omega)^d$, is the topological dual of $C_0(\Omega; \mathbb{R}^d)$ endowed with the supremum norm. The duality pairing $\langle \cdot, \cdot \rangle$ is defined by

$$ \forall m \in \mathcal{M}(\Omega)^d, \forall f \in C_0(\Omega; \mathbb{R}^d), \quad \langle m, f \rangle := \int_{\Omega} f \, dm. $$

The norm associated with the strong topology in $\mathcal{M}(\Omega)^d$, called the total variation norm, is given by

$$ \forall m \in \mathcal{M}(\Omega)^d, \quad \|m\|_{\text{TV}} := \sup \{ \langle m, f \rangle : f \in C_0(\Omega; \mathbb{R}^d), \|f\|_{\infty} \le 1 \}. $$

Let $(m_h)_h \subset \mathcal{M}(\Omega)^d$ and $m \in \mathcal{M}(\Omega)^d$. We say that $m_h \xrightarrow{\ast} m$ (weak-* convergence) if

$$ \lim_{h \to \infty} \int_{\Omega} f \, dm_h = \int_{\Omega} f \, dm, \quad \text{for every } f \in C_0(\Omega; \mathbb{R}^d). $$

If this convergence holds only for all test functions $f \in C_c(\Omega; \mathbb{R}^d)$, we refer to it as local weak-* convergence, denoted

$$ m_h \xrightarrow{\ast} m \quad \text{locally.} $$

**Funding:** The work of T.A.D. was supported by the French government through the France 2030 investment plan, managed by the ANR (ANR-15-IDEX-01) and the "UCA DS4H" project (ANR-17-EURE-0004). The work of L.B.F. was supported by the French government through the 3IA Côte d'Azur Investments in the Future project, managed by the ANR (ANR-19-P3IA-0002).

<footnote>†Université Côte d'Azur, CNRS, I3S, INRIA, Morpheme project, France (Aneva-Doliciane.Tsafack@univ-cotedazur.fr, blancf@i3s.unice.fr).</footnote>

<footnote>‡Université Côte d'Azur, CNRS, LJAD, France (Gilles.AUBERT@univ-cotedazur.fr)</footnote>
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Remark 1.2 (Weak-* vs. local weak-* convergence). If $m_h \xrightarrow{\ast} m$, then $\sup_h \|m_h\|_{TV} < \infty$ by the Banach-Steinhaus theorem. Conversely, the weak-* convergence of a sequence $(m_h)_h \subset \mathcal{M}(\Omega)^d$ is equivalent to local weak-* convergence together with the uniform bound $\sup_h \|m_h\|_{TV} < \infty$.

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**PROPOSITION 1.3 (Weak-* Compactness of $\mathcal{M}(\Omega)^d$).** Let $(m_h)_h \subset \mathcal{M}(\Omega)^d$ be a sequence of finite Radon measures on a locally compact separable metric space $\Omega$ such that

47

$$ \sup_{h \in \mathbb{N}} \|m_h\|_{TV} < \infty. $$

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Then there exists a subsequence $(m_{h_k})_k$ and a measure $m \in \mathcal{M}(\Omega)^d$ such that $m_{h_k} \xrightarrow{\ast} m$.

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**DEFINITION 1.4 (Total Variation Measure).** Let $m \in \mathcal{M}(\Omega)^d$. The total variation measure of $m$, denoted $|m|$, is defined by

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$$ \forall A \in \mathcal{B}(\Omega), \quad |m|(A) = \sup \{ \langle m, \varphi \rangle : \varphi \in C_c(A, \mathbb{R}^d), \|\varphi\|_{\infty} \le 1 \}. $$

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The measure $|m|$ is a positive Radon measure on $\Omega$, i.e., $|m| \in \mathcal{M}^+(\Omega)$. Moreover, the total variation norm of $m$ satisfies $\|m\|_{TV} = |m|(\Omega)$.

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**PROPOSITION 1.5 (Set-theoretic characterization of the total variation).** Let $m \in \mathcal{M}(\Omega; \mathbb{R}^d)$. For every Borel set $E \subset \Omega$, the total variation of $m$ satisfies

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$$ (1.1) \qquad |m|(E) = \sup \left\{ \sum_{h=1}^{\infty} \|m(E_h)\| : E = \bigcup_{h=1}^{\infty} E_h, E_h \in \mathcal{B}(\Omega) \text{ pairwise disjoint} \right\}. $$

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**DEFINITION 1.6 (Support of a Radon Measure).** Let $m \in \mathcal{M}(\Omega)^d$. The support of $m$, denoted $\text{supp}(m)$, is defined as:

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$$ \text{supp}(m) := \overline{\{x \in \Omega \mid \forall r > 0, |m|(B(x, r)) > 0\}}, $$

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where $B(x, r)$ is the open ball of radius $r$ centered at $x$.

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**DEFINITION 1.7 (Space of distributions).** The space of distributions on $\Omega$, denoted $\mathcal{D}'(\Omega)$, is the topological dual of the space of smooth functions with compact support on $\Omega$, $C_c^\infty(\Omega, \mathbb{R})$. Its closure in the sup-norm satisfies

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$$ \overline{C_c^\infty(\Omega, \mathbb{R})}^{\|\cdot\|_{\infty}} = C_0(\Omega, \mathbb{R}), $$

so that $\mathcal{M}(\Omega)$ is a subset of $\mathcal{D}'(\Omega)$.

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**DEFINITION 1.8 (Divergence of a vector-valued measure).** Let $\mathcal{X} \subset \mathbb{R}^d$ be an open set. Let $m \in \mathcal{M}(\mathcal{X})^d$ be a vector-valued Radon measure on $\mathcal{X}$. The divergence of $m$, denoted $\text{div}(m)$, is defined in the sense of distributions by

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$$ \langle \text{div}(m), \varphi \rangle := - \int_{\mathcal{X}} \nabla \varphi \cdot dm, \quad \forall \varphi \in C_c^\infty(\mathcal{X}, \mathbb{R}). $$

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**DEFINITION 1.9 (The space of divergence-measure vector fields).** Let $\mathcal{X} \subset \mathbb{R}^d$ be an open set. The space of divergence-measure vector fields on $\mathcal{X}$ is defined as

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$$ \mathcal{V}(\mathcal{X}) := \{m \in \mathcal{M}(\mathcal{X})^d \mid \text{div} m \in \mathcal{M}(\mathcal{X})\}, $$

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where the divergence is understood as in definition 1.8. The space $\mathcal{V}(\mathcal{X})$ is a Banach space when endowed with the norm

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$$ \|m\|_{\mathcal{V}(\mathcal{X})} := \|m\|_{TV} + \| \text{div} m \|_{TV}. $$

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**DEFINITION 1.10 (Pushforward Measure).** Let $(X_1, \Sigma_1)$ and $(X_2, \Sigma_2)$ be two measurable spaces, and let $f: (X_1, \Sigma_1) \to (X_2, \Sigma_2)$ be a measurable map. If $\mu$ is a measure on $(X_1, \Sigma_1)$, the pushforward measure $f_*\mu$ of $\mu$ under $f$ is defined as follows:

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$$ f_*\mu(E) = \mu(f^{-1}(E)), \quad \text{for all } E \in \Sigma_2. $$

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In other words, $f_*\mu$ is the measure on $X_2$ that corresponds to the measure $\mu$ on $X_1$, via the map $f$.

Moreover, the pushforward measure satisfies the following change of variable formula:

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$$ \int_{X_2} g \, d(f_*\mu) = \int_{X_1} g \circ f \, d\mu, $$

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for all measurable functions $g: X_2 \to \mathbb{R}$, such that $g \circ f$ is $\mu$-integrable.

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We now turn to definitions and properties of curves and measures supported on curves, which constitute the specific case of interest in this work.
**1.3. Curves and Curve Measures.** Let $d \in \mathbb{N}^*$. Consider a parametrized Lipschitz curve $\gamma: [0, 1] \to \mathbb{R}^d$ and denote by $\Gamma = \gamma([0, 1])$ its support, which is the geometric curve associated with $\gamma$.

**DEFINITION 1.11 (Simple Curve).** We say that $\gamma$ is a simple parametrized curve if its restriction to $[0, 1)$ is injective. It is called an open simple curve if $\gamma(0) \neq \gamma(1)$, and a closed simple curve otherwise.

**DEFINITION 1.12 (Regular parametrized curve).** A parametrized curve $\gamma: [0, 1] \to \mathbb{R}^d$ is said to be regular if it is differentiable at every point and its derivative never vanishes, i.e.,

$$ \dot{\gamma}(t) \neq 0 \quad \text{for all } t \in [0, 1]. $$

**DEFINITION 1.13 (Length of a Parametrized Curve).** The length of $\gamma$ (which is the distance traced by the path $\gamma$), denoted $\ell(\gamma)$, is defined by:

$$ \ell(\gamma) := \int_{0}^{1} \|\dot{\gamma}(t)\| \, dt. $$

The length of $\gamma$ is invariant under reparametrization of the path $\gamma$. However, it is not necessarily the length of the geometric curve $\Gamma$, since $\gamma$ may cover parts of $\Gamma$ multiple times.

**DEFINITION 1.14 (Length of a Geometric Curve).** The length of a geometric curve $\Gamma$ is defined as the one-dimensional Hausdorff measure of the set $\Gamma$, denoted $H^1(\Gamma)$. This coincides with the length $\ell(\gamma)$ when $\gamma$ is a parametrization that traces $\Gamma$ exactly once — for instance, a simple parametrization when $\Gamma$ does not self-intersect.

**THEOREM 1.15 (Area formula).** Let $\gamma: [0, 1] \to \mathbb{R}^d$ be a parametrized curve of finite length, and set

$$ \Gamma := \gamma([0, 1]). $$

Define

$$ N(\gamma, [0, 1], x) := \#\{t \in [0, 1] : \gamma(t) = x\}. $$

Then $N(\gamma, [0, 1], \cdot)$ is Borel measurable, $H^1$-integrable, and finite for $H^1$-almost every $x \in \Gamma$. Moreover, for any Borel function $g: \Gamma \to \mathbb{R}$,

$$ \int_{\Gamma} g(x) N(\gamma, [0, 1], x) dH^1(x) = \int_{0}^{1} g(\gamma(t)) \|\dot{\gamma}(t)\| dt. \quad (1.4) $$

In particular, taking $g \equiv 1$ yields

$$ \ell(\gamma) = \int_{0}^{1} \|\dot{\gamma}(t)\| dt = \int_{\Gamma} N(\gamma, [0, 1], x) dH^1(x). $$

**DEFINITION 1.16 (Measure supported by a curve).** Let $\gamma: [0, 1] \to \mathbb{R}^d$ be a Lipschitz curve. We define the $\mathbb{R}^d$-valued Radon measure $\mu_\gamma \in \mathcal{M}(\mathbb{R}^d; \mathbb{R}^d)$ by

$$ \forall \phi \in C_0(\mathbb{R}^d; \mathbb{R}^d), \quad \langle \mu_\gamma, \phi \rangle = \int_0^1 \phi(\gamma(t)) \cdot \dot{\gamma}(t) dt. $$

The measure $\mu_\gamma$ is supported on the geometric curve $\Gamma := \gamma([0, 1])$ and is referred to as the curve measure associated with $\gamma$.

$\mu_\gamma$ is absolutely continuous with respect to the one-dimensional Hausdorff measure $H^1 \llcorner \Gamma$. More precisely, define

$$ f_\gamma(x) = \sum_{t \in \gamma^{-1}(\{x\})} \frac{\dot{\gamma}(t)}{\|\dot{\gamma}(t)\|} \quad \text{for } x \in \Gamma, \quad f_\gamma(x) = 0 \text{ otherwise.} \quad (1.5) $$

Then

$$ \mu_\gamma = f_\gamma H^1 \llcorner \Gamma, \quad (1.6) $$

and

$$ |\mu_\gamma| = \|f_\gamma\| H^1 \llcorner \Gamma. \quad (1.7) $$
When $\gamma$ is a simple parametrized curve, one has

$$|\mu_{\gamma}| = \mathcal{H}^1 \llcorner \Gamma;$$

and consequently, for all $\varphi \in C_0(\mathbb{R}^d; \mathbb{R})$,

$$\langle |\mu_{\gamma}|, \varphi \rangle = \int_{\Gamma} \varphi(x) \, d\mathcal{H}^1(x)$$

$$= \int_{0}^{1} \varphi(\gamma(t)) \|\dot{\gamma}(t)\| \, dt,$$

by the area formula (1.4). Moreover, the total variation of $\mu_{\gamma}$ satisfies

$$(1.8) \qquad \|\mu_{\gamma}\|_{\text{TV}} = \ell(\gamma) = \mathcal{H}^1(\Gamma).$$

The space of divergence-measure vector fields $\mathcal{V}(\mathbb{R}^d)$ contains curve-supported measures of the form $\mu_{\gamma}$, and its key property is that any of its elements can be decomposed into such curve measures.

**2. The Smirnov-based decomposition theorem with simple regular curves.** The Smirnov decomposition theorem, originally formulated for normal charges (i.e., compactly supported elements of $\mathcal{V}(\mathbb{R}^d)$) in the space of Lipschitz curves of finite length [12], has been refined in several subsequent works. In the two-dimensional case ($d = 2$), Bonicatto et al. [3] established a complete decomposition for arbitrary elements of $\mathcal{V}(\mathbb{R}^2)$ in the space of Lipschitz curves parametrized on $[0, 1]$. More recently, to provide more accessible proofs, Rodríguez-Arenas et al. [9] and independently Irving [7] revisited the Smirnov theorem for all elements of $\mathcal{V}(\mathbb{R}^d)$ in the space of 1-Lipschitz curves parametrized on a fixed interval $[0, 1]$.

In this section, we present a new Smirnov-type decomposition with simple regular curves in dimension 2. We begin by introducing the space of curves involved in our construction.

**DEFINITION 2.1 (Space of parametrized curves $\mathcal{K}^d$).** We define $\mathcal{K}^d$ as the subset of parametrized curves in $\mathcal{C}^3([0, 1]; \mathbb{R}^d)$ satisfying the following uniform bounds:

$$(2.1) \qquad \mathcal{K}^d := \left\{ \gamma \in \mathcal{C}^3([0, 1]; \mathbb{R}^d) \;\middle|\; \begin{array}{l} \|\dot{\gamma}(t)\| \le 1, \quad |\dot{\gamma}_i(t)| \le 2\sqrt{d} \quad \text{for all } i = 1, \dots, d, \\ \|\gamma^{(3)}(t)\| \le L_d \quad \text{for all } t \in [0, 1], \\ \text{and } \gamma^{(3)} \text{ is } M_d\text{-Lipschitz continuous} \end{array} \right\},$$

where $L_d, M_d > 0$ are fixed constants depending on the ambient dimension $d$. We endow $\mathcal{K}^d$ with the uniform $\mathcal{C}^3$-norm:

$$\|\gamma\|_{\mathcal{C}^3} := \sum_{k=0}^{3} \|\gamma^{(k)}\|_{\infty}, \quad \text{where} \quad \|\gamma^{(k)}\|_{\infty} := \sup_{t \in [0,1]} \|\gamma^{(k)}(t)\|.$$

**LEMMA 2.2.** $\mathcal{K}^d$ is locally compact metric space with respect to the induced $\mathcal{C}^3$-norm.

**Proof.** First, $\mathcal{K}^d$ is closed in $\mathcal{C}^3([0, 1]; \mathbb{R}^d)$. Indeed, the uniform Lipschitz bounds on the functions and their derivatives up to order 3 are preserved under $\mathcal{C}^3$-convergence: if a sequence $\{\gamma_k\} \subset \mathcal{K}^d$ converges to $\gamma$ in the $\mathcal{C}^3$-norm, then the Lipschitz inequalities for $\gamma_k$ and its derivatives pass to the limit, ensuring that $\gamma \in \mathcal{K}^d$.

To show local compactness, consider any $\gamma_0 \in \mathcal{K}^d$ and $r > 0$, and define the set

$$(2.2) \qquad V(\gamma_0) := \{\gamma \in \mathcal{K}^d \mid \|\gamma - \gamma_0\|_{\infty} \le r\}.$$

We claim that $V(\gamma_0)$ is compact in $\mathcal{K}^d$. The relative compactness of $V(\gamma_0)$ follows from the Arzelà-Ascoli theorem [8] applied successively to each derivative up to order 3. Indeed, for each $j = 0, 1, 2, 3$, the family $\{\gamma^{(j)} : \gamma \in V(\gamma_0)\}$ is uniformly bounded and equicontinuous. Uniform boundedness for $j = 0$ follows from $\|\gamma\|_{\infty} \le \|\gamma_0\|_{\infty} + r$, and equicontinuity results from the uniform Lipschitz bounds. Since $V(\gamma_0)$ is closed in $\mathcal{K}^d$, and $\mathcal{K}^d$ is closed in $\mathcal{C}^3([0, 1]; \mathbb{R}^d)$, then $V(\gamma_0)$ is compact. This proves that $\mathcal{K}^d$ is locally compact. $\square$

Since $\mathcal{K}^d$ is a locally compact metric space, every positive finite Borel measure on $\mathcal{K}^d$ is a Radon measure and therefore belongs to the space $M^+(\mathcal{K}^d)$ of positive finite Radon measures [10].

The decomposition theorem relies on a specific construction: a measure on $\mathbb{R}^d$ obtained by integrating simpler measures, each supported on a curve. We formalize this key concept in the following definition.
DEFINITION 2.3 (Measure defined by integration over the space of curves). Let $\nu \in \mathcal{M}^+(\mathcal{K}^d)$ be a finite positive Radon measure on the space of curves $\mathcal{K}^d$.

For each $\gamma \in \mathcal{K}^d$, let $\mu_\gamma \in \mathcal{M}(\mathbb{R}^d; \mathbb{R}^d)$ be the associated curve measure. We define the integrated measure $\mu := \int_{\mathcal{K}^d} \mu_\gamma \, d\nu(\gamma)$ by

$$ \langle \mu, \phi \rangle := \int_{\mathcal{K}^d} \langle \mu_\gamma, \phi \rangle \, d\nu(\gamma), \quad \forall \phi \in C_0(\mathbb{R}^d; \mathbb{R}^d). $$

Note that this is well-defined because for each $\phi$, the map $\gamma \mapsto \langle \mu_\gamma, \phi \rangle := \int_0^1 \phi(\gamma(t)) \cdot \dot{\gamma}(t) \, dt$ is continuous (hence measurable with respect to $\nu$) and bounded. Moreover, the estimate

$$ |\langle \mu, \phi \rangle| \le \int_{\mathcal{K}^d} |\langle \mu_\gamma, \phi \rangle| \, d\nu(\gamma) \le \|\phi\|_{\infty} \int_{\mathcal{K}^d} \|\mu_\gamma\|_{\text{TV}} \, d\nu(\gamma) \le \nu(\mathcal{K}^d) \|\phi\|_{\infty} $$

holds, since $\|\mu_\gamma\|_{\text{TV}} \le 1$ for every $\gamma \in \mathcal{K}^d$ (which follows from $\|\dot{\gamma}\|_{\infty} \le 1$). Consequently, $\mu$ is a bounded linear functional on $C_0(\mathbb{R}^d; \mathbb{R}^d)$ and therefore a finite vector-valued Radon measure on $\mathbb{R}^d$.

The measure $\int_{\mathcal{K}^d} |\mu_\gamma| \, d\nu(\gamma)$ is understood in the same sense, yielding this time a finite scalar Radon measure on $\mathbb{R}^d$.

In the following, we state and prove a Smirnov-type decomposition theorem in the space of curves $\mathcal{K}^d$ introduced earlier, for divergence-free vector-valued measures in dimension $d$.

THEOREM 2.4 (Decomposition of divergence-free measure fields into curves). Let $\mu \in \mathcal{V}(\mathbb{R}^d)$ be a vector-valued Radon measure with $\text{div}(\mu) = 0$. Then there exists a positive finite Radon measure $\nu \in \mathcal{M}^+(\mathcal{K}^d)$ such that

$$ \mu = \int_{\mathcal{K}^d} \mu_\gamma \, d\nu(\gamma), $$

and

$$ |\mu| = \int_{\mathcal{K}^d} |\mu_\gamma| \, d\nu(\gamma). $$

Furthermore, the following properties hold:

$$ \|\mu\|_{\text{TV}} = \nu(\mathcal{K}^d), \quad \|\mu_\gamma\|_{\text{TV}} = \ell(\gamma) = 1 \quad \text{for } \nu\text{-almost every } \gamma \in \mathcal{K}^d. $$

Proof. The proof follows the same strategy as in [12]. In particular, it relies on the proof approach presented by [9], which is based on the Liouville theorem.

We begin by introducing the kernel function used in the decomposition procedure to regularize a measure and produce a smooth vector field. Let $h: \mathbb{R}^d \to \mathbb{R}$ be a smooth, strictly positive, and integrable function defined by

$$ h(z) = \exp \left( -\sqrt{\|z\|^2 + 1} \right), \quad \forall z \in \mathbb{R}^d. $$

We then define the normalized kernel $k: \mathbb{R}^d \to \mathbb{R}$ by

$$ k(z) = \frac{h(z)}{\int_{\mathbb{R}^d} h(x) \, dx}. $$

The function $k$ belongs to $C^\infty(\mathbb{R}^d)$, is strictly positive, integrable, and satisfies $\int_{\mathbb{R}^d} k(z) \, dz = 1$.

We now recall the following version of Liouville's theorem proved in [9], which is a key ingredient in the decomposition.

THEOREM 2.5 (Liouville's Theorem). Let $F \in C^1(\mathbb{R}^d; \mathbb{R}^d)$ be a linearly bounded vector field. Then any locally finite Borel measure $\rho$ satisfying

$$ \text{div}(F \cdot \rho) = 0 $$

is invariant under the flow $u$ generated by $F$, i.e., the pushforward measure

$$ \rho_t = u(t, \cdot)_{\#} \rho $$

is independent of $t$, for all $t \in \mathbb{R}$.
Let us define:

$$206 \qquad (2.8)$$

$$F = \frac{\mu * k}{|\mu| * k}$$

Since $F \in C^\infty(\mathbb{R}^d)$ and satisfies $\|F\|_\infty \le 1$, it is bounded. Therefore, $F$ generates a well-defined flow $u$ defined by: $u: \mathbb{R} \times \mathbb{R}^d \to \mathbb{R}^d$, defined by $u(t, x) = \gamma_x(t)$, where $\gamma_x$ is the solution of the following ordinary differential equation:

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$$\left\{ \begin{array}{l} \displaystyle \frac{d\gamma_x(t)}{dt} = F(\gamma_x(t)), \\ \gamma_x(0) = x. \end{array} \right.$$

As $F \in C^\infty(\mathbb{R}^d)^d$, $u$ belongs to $C^\infty(\mathbb{R} \times \mathbb{R}^d)$.

Let

$$213 \qquad (2.9)$$

$$\rho = (|\mu| * k)dx \quad \text{and} \quad \tilde{\mu} = \mu * k.$$

The measure $\rho$ is a finite Borel measure on $\mathbb{R}^d$ since

$$\begin{align} \rho(\mathbb{R}^d) &= \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} k(x-y) d|\mu|(y) \right) dx \nonumber \\ &= \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} k(x-y) dx \right) d|\mu|(y) \nonumber \\ &= |\mu|(\mathbb{R}^d) \quad \text{since } \int_{\mathbb{R}^d} k(x-y) dx = 1 \nonumber \\ &= \|\mu\|_{\text{TV}} \tag{2.10} \end{align}$$

and

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$$\begin{align*} \operatorname{div}(F \cdot \rho) &= \operatorname{div}((\mu * k) dx) \\ &= (\operatorname{div}(\mu) * k) dx \\ &= 0 \quad \text{since } \operatorname{div}(\mu) = 0. \end{align*}$$

Therefore, $\rho$ is invariant under the flow $u$ according to Theorem 2.5. That is, the pushforward measure $\rho_t = u(t, \cdot)_{\#} \rho$ is independent of $t$, meaning that: $\forall \varphi \in \mathcal{D}(\mathbb{R}^d)$,

$$222 \qquad (2.11) \qquad \int_{\mathbb{R}^d} \varphi \, d\rho_t := \int_{\mathbb{R}^d} \varphi(u(t,x)) \, d\rho(x) = \int_{\mathbb{R}^d} \varphi(u(0,x)) \, d\rho(x) = \int_{\mathbb{R}^d} \varphi(x) \, d\rho(x).$$

For all $x$, we denote by $\mu_{\gamma_x}$ a curve measure supported by the curve $\gamma_x = u(\cdot, x)$. We have

$$224 \qquad (2.12) \qquad \int_{\mathbb{R}^d} \langle \mu_{\gamma_x}, \varphi \rangle \, d\rho(x) = \langle \tilde{\mu}, \varphi \rangle, \quad \forall \varphi \in \mathcal{D}(\mathbb{R}^d)^d.$$

Indeed: Let $\varphi \in \mathcal{D}(\mathbb{R}^d)^d$.

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Since $F_i \cdot \varphi_i \in \mathcal{D}(\mathbb{R}^d)$, for $i \in \{1, 2, \dots, n\}$, then according to the equation (2.11), we have

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$$\int_0^1 \int_{\mathbb{R}^d} \varphi_i(u(t,x)) F_i(u(t,x)) \, d\rho(x) \, dt = \int_0^1 \int_{\mathbb{R}^d} \varphi_i(x) F_i(x) \, dt \, d\rho(x) , \forall i.$$
So,

$$
\begin{align*}
\int_{\mathbb{R}^d} \langle \mu_{\gamma_x}, \varphi \rangle \, d\rho(x) &= \sum_{i=1}^n \int_{\mathbb{R}^d} \varphi_i(x) F_i(x) \, d\rho(x) \\
&= \int_{\mathbb{R}^d} \varphi(x) \cdot F(x) \, d\rho(x) \\
&= \int_{\mathbb{R}^d} \varphi(x) \cdot \frac{\mu * k}{|\mu| * k} \cdot (|\mu| * k) \, dx \\
&= \int_{\mathbb{R}^d} \varphi(x) \cdot (\mu * k)(x) \, dx \\
&= \langle \tilde{\mu}, \varphi \rangle.
\end{align*}
$$

LEMMA 2.6. The map G defined by

$$
\begin{array}{rcl}
G : \mathbb{R}^d & \to & \mathcal{K}^d \\
x & \mapsto & \gamma_x : [0, 1] \to \mathbb{R}^d, \quad t \mapsto u(t, x)
\end{array}
$$

is well defined and continuous from $\mathbb{R}^d$ (with its standard norm topology) to $\mathcal{K}^d$ (equipped with the topology induced by the $C^3$-norm). The proof is provided in Appendix A.

Since $\rho$ is a positive finite Radon measure on $\mathbb{R}^d$ and $G: \mathbb{R}^d \to \mathcal{K}^d$ is continuous, the pushforward
measure $\nu_0 := G_{\#}\rho$ is well defined and is also a positive finite Radon measure on $\mathcal{K}^d$. Moreover,

$$
\tilde{\mu} = \int_{\mathcal{K}^d} \mu_{\gamma} \, d\nu_0(\gamma).
$$

Indeed, for any test function $\varphi \in \mathcal{D}(\mathbb{R}^d)$, define the map

$$
\begin{array}{rcl}
N_{\varphi} : \mathcal{K}^d & \to & \mathbb{R} \\
\gamma & \mapsto & \langle \mu_{\gamma}, \varphi \rangle := \int_0^1 \varphi(\gamma(t)) \cdot \dot{\gamma}(t) \, dt.
\end{array}
$$

This map is continuous, and applying the change of variable formula (1.3), we obtain:

$$
\begin{align*}
\int_{\mathcal{K}^d} N_\varphi(\gamma) \, d(G_{\#} \rho)(\gamma) &= \int_{\mathbb{R}^d} (N_\varphi \circ G)(x) \, d\rho(x) \\
&= \int_{\mathbb{R}^d} \langle \mu_{\gamma_x}, \varphi \rangle \, d\rho(x) \\
&= \langle \tilde{\mu}, \varphi \rangle, \quad \text{by equation (2.12).}
\end{align*}
$$

Therefore, for all $\varphi \in \mathcal{D}(\mathbb{R}^d)$,

$$
\int_{\mathcal{K}^d} \langle \mu_{\gamma}, \varphi \rangle \, d\nu_0(\gamma) = \langle \tilde{\mu}, \varphi \rangle,
$$

which proves equation (2.14). Moreover,

$$
\nu_0(\mathcal{K}^d) = \rho(G^{-1}(\mathcal{K}^d)) = \rho(\mathbb{R}^d) = \|\mu\|_{\text{TV}},
$$

where the last equality follows from equation (2.10).

For all $\epsilon > 0$ sufficiently small, define

$$
k_{\epsilon}(x) = \frac{1}{\epsilon^d} k\left(\frac{x}{\epsilon}\right).
$$

Like $k$, the function $k_{\epsilon}$ belongs to $C^{\infty}(\mathbb{R}^d)$, is strictly positive, and satisfies

$$
\int_{\mathbb{R}^d} k_{\epsilon}(x) \, dx = 1.
$$

Therefore, as in the case of $k$ discussed above, we have

$$
\tilde{\mu}_{\epsilon} := \mu * k_{\epsilon} = \int_{\mathcal{K}^d} \mu_{\gamma} \, d\nu_{\epsilon}(\gamma),
$$
where $\nu_{\epsilon} = (G_{\epsilon})_{\#} \rho_{\epsilon}$, $\rho_{\epsilon} = (|\mu| * k_{\epsilon}) dx$, and $\nu_{\epsilon}(\mathcal{K}^d) = \|\mu\|_{\text{TV}}$.

The sequence $(\nu_{\epsilon})_{\epsilon>0}$ is bounded in $\mathcal{M}^{+}(\mathcal{K}^d)$ since $\nu_{\epsilon}(\mathcal{K}^d) = \|\mu\|_{\text{TV}}$, $\forall \epsilon > 0$. Therefore, by the weak*-compactness of the space of Radon measure, we can extract a subsequence $(\nu_{\epsilon_h})_{\epsilon_h>0}$ that weak*-converges to a Radon measure $\nu \in \mathcal{M}^{+}(\mathcal{K}^d)$. That is

$$ (2.19) \qquad \forall \Phi \in C_0(\mathcal{K}^d), \quad \int_{\mathcal{K}^d} \Phi \, d\nu_{\epsilon_h} \xrightarrow{\epsilon_h \to 0} \int_{\mathcal{K}^d} \Phi \, d\nu. $$

For all $\varphi \in \mathcal{D}(\mathbb{R}^d)$, the continuous map $N_{\varphi}$ defined above belongs to $C_0(\mathcal{K}^d)$. Indeed, let $\delta > 0$. Since $\varphi$ has compact support, there exists $R_{\delta} > 0$ such that

$$ \|x\| \ge R_{\delta} \Rightarrow |\varphi(x)| \le \delta. $$

The set

$$ E_{\delta} = \{ \gamma \in \mathcal{K}^d : \|\gamma(0)\| \le R_{\delta} + 1 \} \text{ is a compact subset of } \mathcal{K}^d $$

and for all $\gamma \in \mathcal{K}^d \setminus E_{\delta}$, we have $\|\gamma(0)\| > R_{\delta} + 1$. Since $\|\gamma(t) - \gamma(0)\| \le |t| \le 1$, it follows that

$$ \|\gamma(t)\| \ge \|\gamma(0)\| - 1 > R_{\delta}, \quad \forall t \in [0, 1], \text{ so that } |\varphi(\gamma(t))| \le \delta \text{ for all } t. $$

Hence,

$$ |N_{\varphi}(\gamma)| = \left| \int_{0}^{1} \varphi(\gamma(t)) \cdot \dot{\gamma}(t) \, dt \right| \le \int_{0}^{1} |\varphi(\gamma(t))| \cdot \|\dot{\gamma}(t)\| \, dt \le \delta, $$

which shows that $N_{\varphi} \in C_0(\mathcal{K}^d)$.

So, for all $\varphi \in \mathcal{D}(\mathbb{R}^d)^d$, we have by equation (2.19),

$$ \int_{\mathcal{K}^d} N_{\varphi}(\gamma) \, d\nu_{\epsilon_h}(\gamma) \xrightarrow{\epsilon_h \to 0} \int_{\mathcal{K}^d} N_{\varphi}(\gamma) \, d\nu(\gamma). $$

That is $\int_{\mathcal{K}^d} \langle \mu_{\gamma}, \varphi \rangle \, d\nu_{\epsilon_h}(\gamma) \xrightarrow{\epsilon_h \to 0} \int_{\mathcal{K}^d} \langle \mu_{\gamma}, \varphi \rangle \, d\nu(\gamma)$ and thus

$$ \int_{\mathcal{K}^d} \mu_{\gamma} \, d\nu_{\epsilon_h}(\gamma) \stackrel{*}{\longrightarrow} \int_{\mathcal{K}^d} \mu_{\gamma} \, d\nu(\gamma) \text{ as } \epsilon_h \to 0. $$

On the other hand, the sequence $\tilde{\mu}_{\epsilon_h} := \mu * k_{\epsilon_h}$ converges to $\mu$ in the weak-* topology as $\epsilon_h \to 0$. Hence, by uniqueness of the weak-* limit,

$$ \mu = \int_{\mathcal{K}^d} \mu_{\gamma} \, d\nu(\gamma), $$

proving equation (2.3).

Let $E \subset \mathbb{R}^d$ be an open set. By definition, we have

$$ |\mu|(E) = \sup \{ \langle \mu, \varphi \rangle \mid \varphi \in C_c(E, \mathbb{R}^d), \|\varphi\|_{\infty} \le 1 \}. $$

For any $\varphi \in C_c(E, \mathbb{R}^d)$ such that $\|\varphi\|_{\infty} \le 1$, we have:

$$ \begin{align*} \langle \mu, \varphi \rangle &= \int_{\mathcal{K}^d} \langle \mu_{\gamma}, \varphi \rangle \, d\nu(\gamma) \\ &\le \int_{\mathcal{K}^d} \sup \{ \langle \mu_{\gamma}, \psi \rangle \mid \psi \in C_c(E, \mathbb{R}^d), \|\psi\|_{\infty} \le 1 \} \, d\nu(\gamma) \\ &= \int_{\mathcal{K}^d} |\mu_{\gamma}|(E) \, d\nu(\gamma). \end{align*} $$

Taking the supremum over such $\varphi$ yields

$$ |\mu|(E) \le \int_{\mathcal{K}^d} |\mu_{\gamma}|(E) \, d\nu(\gamma). $$

Using the regularity of Radon measures, the inequality extends to all Borel subsets of $\mathbb{R}^d$. Indeed:
Let $A \subset \mathbb{R}^d$ be a closed set. Then $A$ can be written as the countable intersection of open sets:

$$A = \bigcap_{m=1}^{\infty} A_m, \quad \text{where } A_m = \{x \in \mathbb{R}^d \mid d(x, A) < 1/m\}.$$

Since the sequence $(A_m)$ is decreasing and $|\mu|$ is a finite positive measure, we have

$$|\mu|(A) = \lim_{m \to \infty} |\mu|(A_m).$$

Similarly, since $|\mu_\gamma|$ is also a finite positive measure for every $\gamma \in \mathcal{K}^d$, we have

$$|\mu_\gamma|(A) = \lim_{m \to \infty} |\mu_\gamma|(A_m).$$

Therefore, we obtain

$$|\mu|(A) = \lim_{m \to \infty} |\mu|(A_m) \le \lim_{m \to \infty} \int_{\mathcal{K}^d} |\mu_\gamma|(A_m) d\nu(\gamma) = \int_{\mathcal{K}^d} |\mu_\gamma|(A) d\nu(\gamma),$$

where the last equality follows from the monotone convergence theorem.

Now, let $B \subset \mathbb{R}^d$ be an arbitrary Borel set. By the inner regularity of $|\mu|$, we have

$$|\mu|(B) = \sup \{|\mu|(C) \mid C \subset B, C \text{ compact}\}.$$

For every such compact (and hence closed) set $C \subset B$, the inequality above gives

$$|\mu|(C) \le \int_{\mathcal{K}^d} |\mu_\gamma|(C) d\nu(\gamma) \le \int_{\mathcal{K}^d} |\mu_\gamma|(B) d\nu(\gamma).$$

Taking the supremum over all compact subsets $C \subset B$, we conclude that

$$|\mu|(B) \le \int_{\mathcal{K}^d} |\mu_\gamma|(B) d\nu(\gamma).$$

For $E = \mathbb{R}^d$, this implies:

$$\|\mu\|_{\text{TV}} \le \int_{\mathcal{K}^d} \|\mu_\gamma\|_{\text{TV}} d\nu(\gamma).$$

Since $\|\mu_\gamma\|_{\text{TV}} \le \ell(\gamma) \le 1$, and

$$\nu(\mathcal{K}^d) \le \liminf_{\epsilon_h \to 0} \nu_{\epsilon_h}(\mathcal{K}^d) = \|\mu\|_{\text{TV}},$$

We obtain the chain of inequalities

$$\|\mu\|_{\text{TV}} \le \int_{\mathcal{K}^d} \|\mu_\gamma\|_{\text{TV}} d\nu(\gamma) \le \int_{\mathcal{K}^d} \ell(\gamma) d\nu(\gamma) \le \int_{\mathcal{K}^d} 1 d\nu = \nu(\mathcal{K}^d) \le \|\mu\|_{\text{TV}},$$

which implies that

$$\|\mu\|_{\text{TV}} = \int_{\mathcal{K}^d} \|\mu_\gamma\|_{\text{TV}} d\nu(\gamma),$$

(2.20)

and

$$\nu(\mathcal{K}^d) = \|\mu\|_{\text{TV}}, \quad \text{and} \quad \|\mu_\gamma\|_{\text{TV}} = \ell(\gamma) = 1 \quad \text{for } \nu\text{-almost every } \gamma \in \mathcal{K}^d,$$

proving equation (2.5).

Since for any Borel set $B \subset \mathbb{R}^d$,

$$|\mu|(B) \le \int_{\mathcal{K}^d} |\mu_\gamma|(B) d\nu(\gamma), \quad \text{and} \quad |\mu|(\mathbb{R}^d) = \int_{\mathcal{K}^d} |\mu_\gamma|(\mathbb{R}^d) d\nu(\gamma),$$

we conclude that

$$|\mu| = \int_{\mathcal{K}^d} |\mu_\gamma| d\nu(\gamma),$$

thus proving equation (2.4). This concludes the proof of Theorem 2.4. □
Remark 2.7. For any $\gamma \in \mathcal{K}^d$, the condition $\ell(\gamma) = 1$ implies

$$ \int_0^1 \|\dot{\gamma}(t)\| dt = 1. $$

Since $\|\dot{\gamma}(t)\| \le 1$ for all $t \in [0, 1]$, it follows that $\|\dot{\gamma}(t)\| = 1$ for almost every $t \in [0, 1]$. By continuity of the map $t \mapsto \|\dot{\gamma}(t)\|$, this equality holds for all $t \in [0, 1]$. Therefore, $\gamma$ is a regular curve parametrized by arc length.

Consequently, in Theorem 2.4, $\nu$-almost every $\gamma \in \mathcal{K}^d$ is regular and parametrized by arc length.

**COROLLARY 2.8.** Let $d \in \{2, 3\}$. Then, in the decomposition given by Theorem 2.4, $\nu$-almost every $\gamma \in \mathcal{K}^d$ is injective.

**Proof.** By Remark 2.7, $\nu$-almost every $\gamma \in \mathcal{K}^d$ is a regular $C^3$ curve parametrized by arc length. Moreover, its curvature $\kappa_\gamma(t) = \|\ddot{\gamma}(t)\|$ satisfies

$$ \kappa_{\gamma}(t) \le 2d \le 2\pi = \frac{2\pi}{\ell(\gamma)}, \quad \text{since } \ell(\gamma) = 1. $$

Therefore, by [5, Corollary 3.1], the curve $\gamma$ is injective. $\square$

The following theorem extends Theorem 2.4 to divergence-measure fields with nonzero divergence.

**THEOREM 2.9 (Decomposition into curves of divergence measure fields).** We define the space of curves

$$ (2.21) \qquad \mathcal{K}_{\text{proj}}^d := \left\{ \gamma \in C^3([0, 1]; \mathbb{R}^d) \;\middle|\; \begin{array}{l} \|\dot{\gamma}(t)\| \le 1, \\ |\ddot{\gamma}_i(t)| \le 2\sqrt{d+1} \quad \text{for all } i = 1, \dots, d, \\ \|\gamma^{(3)}(t)\| \le L_{d+1} \quad \text{for all } t \in [0, 1], \\ \gamma^{(3)} \text{ is } M_{d+1}\text{-Lipschitz continuous} \end{array} \right\}. $$

For any $\mu \in \mathcal{V}(\mathbb{R}^d)$, there exists a positive finite Radon measure $\nu \in \mathcal{M}^+(\mathcal{K}_{\text{proj}}^d)$ such that

$$ (2.22) \qquad \mu = \int_{\mathcal{K}_{\text{proj}}^d} \mu_{\gamma} \, d\nu(\gamma), $$

and

$$ (2.23) \qquad |\mu| = \int_{\mathcal{K}_{\text{proj}}^d} |\mu_{\gamma}| \, d\nu(\gamma). $$

Moreover,

$$ (2.24) \qquad \nu(\mathcal{K}_{\text{proj}}^d) \le 2 \|\mu\|_{\mathcal{V}(\mathbb{R}^d)}. $$

Furthermore,

(2.25) $\nu$-almost every $\gamma$ is either constant or a regular curve with constant speed,

and, in particular,

(2.26) if $d=2$, then $\nu$-almost every non-constant $\gamma$ is injective.

**Proof.** We follow the strategy originally introduced by Smirnov [12] and further developed in [9, 7]. The key idea is to reduce the problem to the divergence-free case by introducing an auxiliary measure, apply the previous decomposition result (Theorem 2.4), and then derive a decomposition of the original measure within our framework.

Define a $(d+1)$-dimensional vector-valued Radon measure on $\mathbb{R}^{d+1}$ by

$$ (2.27) \qquad m := (\mu \otimes (\delta_0 - \delta_1), -\operatorname{div} \mu \otimes \mathcal{L}^{1\perp[0,1]}). $$

Here, $\otimes$ denotes the product of Radon measures on the product space $\mathbb{R}^d \times \mathbb{R}$.
For any test function $\tilde{\varphi}(x, x_{d+1}) = (\varphi(x, x_{d+1}), \varphi_{d+1}(x, x_{d+1})) \in \mathcal{D}(\mathbb{R}^{d+1})^{d+1}$, the duality pairing reads

$$ 
\begin{aligned}
\langle m, \tilde{\varphi} \rangle &= \langle \mu, \langle \delta_0 - \delta_1, \varphi(x, \cdot) \rangle \rangle + \langle -\operatorname{div} \mu, \langle \mathcal{L}^{1\perp[0,1]}, \varphi_{d+1}(x, \cdot) \rangle \rangle; \\
&= \int_{\mathbb{R}^d} (\varphi(x, 0) - \varphi(x, 1)) \, d\mu(x) - \int_{\mathbb{R}^d} \left( \int_0^1 \varphi_{d+1}(x, t) \, dt \right) \, d(\operatorname{div} \mu)(x).
\end{aligned}
$$

Note that $\operatorname{div}(m) = 0$. Indeed, for any $\phi \in \mathcal{D}(\mathbb{R}^{d+1})$, we have:

$$ 
\begin{aligned}
\langle \operatorname{div}(m), \phi \rangle &= -\langle m, \nabla \phi \rangle; \\
&= -\int_{\mathbb{R}^d} (\nabla_x \phi(x, 0) - \nabla_x \phi(x, 1)) \, d\mu(x) + \int_{\mathbb{R}^d} \int_0^1 \partial_{x_{d+1}} \phi(x, t) \, dt \, d(\operatorname{div} \mu)(x); \\
&= -\int_{\mathbb{R}^d} (\nabla_x \phi(x, 0) - \nabla_x \phi(x, 1)) \, d\mu(x) + \int_{\mathbb{R}^d} (\phi(x, 1) - \phi(x, 0)) \, d(\operatorname{div} \mu)(x); \\
&= \int_{\mathbb{R}^d} (\phi(x, 0) - \phi(x, 1)) \, d(\operatorname{div} \mu)(x) + \int_{\mathbb{R}^d} (\phi(x, 1) - \phi(x, 0)) \, d(\operatorname{div} \mu)(x); \\
&= 0.
\end{aligned}
$$

Applying Theorem 2.4, we obtain the existence of a measure $\tilde{\nu} \in \mathcal{M}^+(\mathcal{K}^{d+1})$ such that

$$ (2.28) \qquad m = \int_{\mathcal{K}^{d+1}} \mu_{\tilde{\gamma}} \, d\tilde{\nu}(\tilde{\gamma}), $$

$$ (2.29) \qquad |m| = \int_{\mathcal{K}^{d+1}} |\mu_{\tilde{\gamma}}| \, d\tilde{\nu}(\tilde{\gamma}). $$

$$ (2.30) \qquad \|m\|_{\mathrm{TV}} = \tilde{\nu}(\mathcal{K}^{d+1}), \quad \|\mu_{\tilde{\gamma}}\|_{\mathrm{TV}} = \ell(\tilde{\gamma}) = 1 \quad \text{for } \tilde{\nu}\text{-almost every } \tilde{\gamma} \in \mathcal{K}^{d+1}. $$

Note that, from equation (2.29), the support of $m$ satisfies

$$ \mathrm{supp}(m) = \bigcup_{\tilde{\gamma} \in \mathrm{supp}(\tilde{\nu})} \tilde{\gamma}([0, 1]) \subset \mathbb{R}^d \times [0, 1], $$

which will be rigorously justified by Lemma 2.13 below. Consequently, for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{K}^{d+1}$,

$$ \tilde{\gamma}([0, 1]) \subset \mathbb{R}^d \times [0, 1]. $$

The restriction of the vector-valued measure $m \in \mathbb{R}^{d+1}$ to the hyperplane $\mathbb{R}^d \times \{0\}$ coincides with the original measure $\mu$ in the first $d$ components, while the $(d+1)$-th component vanishes, that is,

$$ (2.31) \qquad m_{\perp}(\mathbb{R}^d \times \{0\}) = (\mu, 0) $$

Indeed, for any Borel set $A \subset \mathbb{R}^d$, we have

$$ 
\begin{aligned}
m(A \times \{0\}) &= ((\mu \otimes (\delta_0 - \delta_1))(A \times \{0\}), (-\operatorname{div} \mu \otimes \mathcal{L}^{1\perp[0,1]})(A \times \{0\})) \\
&= (\mu(A), 0).
\end{aligned}
$$

Hence, the idea is to extract from the curves $\tilde{\gamma}$ in the support of $m$ the portions that lie in the hyperplane $\mathbb{R}^d \times \{0\}$. These portions will form the curves that appear in the decomposition of the original measure $\mu$.

To this end, we define

$$ \mathcal{M} := \{ \tilde{\gamma} \in \mathcal{K}^{d+1} \mid \tilde{\gamma}([0, 1]) \cap (\mathbb{R}^d \times \{0\}) \neq \emptyset \}. $$
Remark 2.10. The set $\mathcal{M}$ is not empty. Assume, for contradiction, that $\mathcal{M} = \emptyset$. Then every curve $\tilde{\gamma} \in \mathcal{K}^{d+1}$ satisfies

$$ \tilde{\gamma}([0, 1]) \cap (\mathbb{R}^d \times \{0\}) = \emptyset, $$

so that, by definition of $|\mu_{\tilde{\gamma}}|$ (equation (1.7)),

$$ |\mu_{\tilde{\gamma}}|(\mathbb{R}^d \times \{0\}) = \int_{\tilde{\gamma}([0,1])} \mathbb{1}_{\mathbb{R}^d \times \{0\}}(x) \|f_{\tilde{\gamma}}(x)\| d\mathcal{H}^1(x) = 0. $$

Using the decomposition (2.29) of $|m|$, it follows that

$$ |m|(\mathbb{R}^d \times \{0\}) = \int_{\mathcal{K}^{d+1}} |\mu_{\tilde{\gamma}}|(\mathbb{R}^d \times \{0\}) d\tilde{\nu}(\tilde{\gamma}) = 0. $$

However, by the set-theoretic definition (1.1) of the total variation,

$$ \begin{aligned} |m|(\mathbb{R}^d \times \{0\}) &= \sup \left\{ \sum_h \|m(A_h \times \{0\})\| : A_h \in \mathcal{B}(\mathbb{R}^d) \text{ pairwise disjoint, } \bigcup_h A_h = \mathbb{R}^d \right\}; \\ &= \sup \left\{ \sum_h \|\mu(A_h)\| : A_h \in \mathcal{B}(\mathbb{R}^d) \text{ pairwise disjoint, } \bigcup_h A_h = \mathbb{R}^d \right\}, \\ &\qquad \text{since by equation (2.31), } m(A_h \times \{0\}) = (\mu(A_h), 0); \\ &= |\mu|(\mathbb{R}^d) > 0, \end{aligned} $$

since $\mu$ is non-zero. This contradiction shows that $\mathcal{M} \neq \emptyset$.

Having identified the curves that intersect the hyperplane $\mathbb{R}^d \times \{0\}$ through $\mathcal{M}$, we now focus on the precise structure of the preimage of this hyperplane under each curve that appears in the decomposition of $m$.

**LEMMA 2.11** (Preimage of the hyperplane). For $\tilde{\nu}$-almost every curve $\tilde{\gamma} = (\gamma, \gamma_{d+1}) \in \mathcal{M}$, the preimage of the hyperplane

$$ \tilde{\gamma}^{-1}(\mathbb{R}^d \times \{0\}) $$

is the closed interval

$$ [a(\tilde{\gamma}), b(\tilde{\gamma})] \subset [0, 1], $$

where

$$ a(\tilde{\gamma}) := \inf\{t \in [0, 1] \mid \tilde{\gamma}(t) \in \mathbb{R}^d \times \{0\}\}, \quad b(\tilde{\gamma}) := \sup\{t \in [0, 1] \mid \tilde{\gamma}(t) \in \mathbb{R}^d \times \{0\}\}. $$

Note that if $a(\tilde{\gamma}) = b(\tilde{\gamma})$, then the preimage consists of a single point. If $a(\tilde{\gamma}) \neq b(\tilde{\gamma})$, then $\tilde{\gamma}$ cannot be constant on $(a(\tilde{\gamma}), b(\tilde{\gamma}))$, since being constant would imply $\tilde{\gamma} \equiv 0$, contradicting the fact that $\|\tilde{\gamma}\| = 1$.

**Proof of Lemma 2.11.** To prove this, it is necessary to understand the behavior of such curves when they are not in the hyperplane $\mathbb{R}^d \times \{0\}$, and in particular when they pass through

$$ H := \mathbb{R}^d \times (0, 1). $$

Note that the restriction of $m$ to $H$ is

$$ m \llcorner H = (0_{\mathbb{R}^d}, -\operatorname{div} \mu \otimes \mathcal{L}^{1 \llcorner [0,1]}). $$

Indeed, consider any test function

$$ \tilde{\varphi}(x, x_{d+1}) = (\varphi(x, x_{d+1}), \varphi_{d+1}(x, x_{d+1})) \in \mathcal{D}(H)^{d+1}, $$

i.e., smooth with compact support in $H$. The duality pairing with the measure $m$ restricted to $H$ reads

$$ \begin{aligned} \langle m \llcorner H, \tilde{\varphi} \rangle &= \underbrace{\int_H \varphi(x, 0) - \varphi(x, 1) d\mu(x)}_{=0 \text{ since } \varphi \text{ has compact support in } H} + \int_H \varphi_{d+1}(x, x_{d+1}) d(-\operatorname{div} \mu \otimes \mathcal{L}^{1 \llcorner [0,1]}) \\ &= - \int_H \varphi_{d+1}(x, x_{d+1}) d(\operatorname{div} \mu \otimes \mathcal{L}^{1 \llcorner [0,1]}), \\ &= \langle -(\operatorname{div} \mu \otimes \mathcal{L}^{1 \llcorner [0,1]}), \varphi_{d+1} \rangle \\ &= \langle (0_{\mathbb{R}^d}, -(\operatorname{div} \mu \otimes \mathcal{L}^{1 \llcorner [0,1]})), \tilde{\varphi} \rangle. \end{aligned} $$
Thus, we can write

$$ (2.32) \qquad m_{\perp}H = e_{d+1}(-\operatorname{div} \mu \otimes \mathcal{L}_{\perp[0,1]}), $$

where $e_{d+1}$ denotes the $(d+1)$-th canonical basis vector in $\mathbb{R}^{d+1}$.

Using the polar decomposition of a vector-valued measure, we can write

$$ -\operatorname{div} \mu = g |\operatorname{div} \mu|, $$

where $g \in L^1(\mathbb{R}^d, |\operatorname{div} \mu|)$ satisfies $|g(x)| = 1$ for all $x \in \mathbb{R}^d$.

We then define the sets

$$ E^{-} := \{x \in \mathbb{R}^{d} \mid g(x) = -1\}, \quad E^{+} := \{x \in \mathbb{R}^{d} \mid g(x) = 1\}. $$

**LEMMA 2.12.** For $\tilde{\nu}$-almost every curve $\tilde{\gamma} = (\gamma, \gamma_{d+1}) \in \mathcal{K}^{d+1}$, and for $\mathcal{L}^1$-almost every $t \in [0, 1]$, we have

$$ \tilde{\gamma}(t) \in E^{+} \times (0, 1) \implies \dot{\gamma}_{d+1}(t) = 1, \quad \tilde{\gamma}(t) \in E^{-} \times (0, 1) \implies \dot{\gamma}_{d+1}(t) = -1. $$

**Proof of the Lemma 2.12.** Note that, from equation (2.32) and since $g = 1$ on $E^+$, we have

$$ \begin{aligned} m_{\perp}H(E^{+} \times (0, 1)) &= e_{d+1} (|\operatorname{div} \mu| \otimes \mathcal{L}_{\perp}^{1}(0, 1))(E^{+} \times (0, 1)) \\ &= e_{d+1} |\operatorname{div} \mu|(E^{+}) \mathcal{L}_{\perp}^{1}(0, 1)((0, 1)). \end{aligned} $$

Since $\mathcal{L}_{\perp}^{1}(0, 1)((0, 1)) = 1$, we conclude that

$$ m_{\perp}H(E^{+} \times (0, 1)) = e_{d+1} |\operatorname{div} \mu|(E^{+}). $$

On the other hand, from the decomposition of $m$ given in equation (2.28), we have

$$ \begin{aligned} m_{\perp}H(E^{+} \times (0, 1)) &= m(E^{+} \times (0, 1)) \\ &= \int_{\mathcal{K}^{d+1}} \mu_{\tilde{\gamma}}(E^{+} \times (0, 1)) d\tilde{\nu}(\tilde{\gamma}). \end{aligned} $$

Combining equations (2.33) and (2.34), we obtain

$$ (2.35) \qquad \int_{\mathcal{K}^{d+1}} \mu_{\tilde{\gamma}}(E^{+} \times (0, 1)) d\tilde{\nu}(\tilde{\gamma}) = e_{d+1} |\operatorname{div} \mu|(E^{+}). $$

Moreover, from equation (2.32), the total variation measure of the restriction $m_{\perp}H$ is

$$ |m_{\perp}H| = | - \operatorname{div} \mu \otimes \mathcal{L}_{\perp}^{1}[0, 1] | = |\operatorname{div} \mu| \otimes \mathcal{L}_{\perp}^{1}[0, 1], $$

so that

$$ |m_{\perp}H|(E^{+} \times (0, 1)) = |\operatorname{div} \mu|(E^{+}). $$

On the other hand, from equation (2.29), we have

$$ \begin{aligned} |m_{\perp}H|(E^{+} \times (0, 1)) &= |m|(E^{+} \times (0, 1)) \\ &= \int_{\mathcal{K}^{d+1}} |\mu_{\tilde{\gamma}}|(E^{+} \times (0, 1)) d\tilde{\nu}(\tilde{\gamma}). \end{aligned} $$

Combining equations (2.36) and (2.37), we obtain

$$ (2.38) \qquad \int_{\mathcal{K}^{d+1}} |\mu_{\tilde{\gamma}}|(E^{+} \times (0, 1)) d\tilde{\nu}(\tilde{\gamma}) = |\operatorname{div} \mu|(E^{+}). $$

From equations (2.35) and (2.38), we obtain

$$ (2.39) \qquad \int_{\mathcal{K}^{d+1}} \mu_{\tilde{\gamma}}(E^{+} \times (0, 1)) d\tilde{\nu}(\tilde{\gamma}) = e_{d+1} \int_{\mathcal{K}^{d+1}} |\mu_{\tilde{\gamma}}|(E^{+} \times (0, 1)) d\tilde{\nu}(\tilde{\gamma}) $$
Recall that, by the definition of $|\mu_{\tilde{\gamma}}|$ (equation (1.7)),

$$|\mu_{\tilde{\gamma}}|(E^+ \times (0,1)) = \int_{\tilde{\Gamma} \cap (E^+ \times (0,1))} \left\| \sum_{t \in \tilde{\gamma}^{-1}(\{x\})} \frac{\dot{\tilde{\gamma}}(t)}{\|\dot{\tilde{\gamma}}(t)\|} \right\| d\mathcal{H}^1(x).$$

Since $\tilde{\gamma}$ is parametrized at unit speed and $\|\mu_{\tilde{\gamma}}\|_{\text{TV}} = \ell(\tilde{\gamma})$, by [2, Lemma A.2], for $\mathcal{H}^1$-a.e. $x \in \tilde{\Gamma}$, $\dot{\tilde{\gamma}}(t)$ is independent of $t \in \tilde{\gamma}^{-1}(x)$.

Let $N(\tilde{\gamma}, [0, 1], x)$ denote the cardinality of the preimage $\tilde{\gamma}^{-1}(x)$. Then we obtain

$$ (2.40) \qquad |\mu_{\tilde{\gamma}}|(E^+ \times (0,1)) = \int_{\tilde{\Gamma} \cap (E^+ \times (0,1))} N(\tilde{\gamma}, [0, 1], x) d\mathcal{H}^1(x). $$

By the Area formula (1.4), we can push this integral back to the parameter $t$:

$$ \begin{aligned} |\mu_{\tilde{\gamma}}|(E^+ \times (0,1)) &= \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) \|\dot{\tilde{\gamma}}(t)\| dt \\ &= \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) dt, \quad \text{since } \tilde{\gamma} \text{ is unit-speed.} \end{aligned} $$

Furthermore, for the vector-valued measure itself, we have

$$ \mu_{\tilde{\gamma}}(E^+ \times (0,1)) = \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) \dot{\tilde{\gamma}}(t) dt. $$

It then follows from equation (2.39) that

$$ (2.41) \qquad \int_{\mathcal{K}^{d+1}} \left[ \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) \dot{\tilde{\gamma}}(t) dt - e_{d+1} \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) dt \right] d\tilde{\nu}(\tilde{\gamma}) = 0. $$

Considering the $(d+1)$-th component, equation (2.41) becomes

$$ (2.42) \qquad \int_{\mathcal{K}^{d+1}} \left[ \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) \dot{\tilde{\gamma}}_{d+1}(t) dt - \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) dt \right] d\tilde{\nu}(\tilde{\gamma}) = 0. $$

Since $\dot{\tilde{\gamma}}_{d+1}(t) \le 1$ for all $t \in [0, 1]$, the integrand in equation (2.42) is non-positive for every $\tilde{\gamma}$:

$$ \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) \dot{\tilde{\gamma}}_{d+1}(t) dt - \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) dt \le 0. $$

Since the integral over $\mathcal{K}^{d+1}$ vanishes, this can only happen if the integrand itself is zero for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{K}^{d+1}$:

$$ (2.43) \qquad \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) \dot{\tilde{\gamma}}_{d+1}(t) dt = \int_0^1 \mathbb{1}_{E^+ \times (0,1)}(\tilde{\gamma}(t)) dt. $$

Therefore, for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{K}^{d+1}$, we must have

$$ \dot{\tilde{\gamma}}_{d+1}(t) = 1 \quad \text{for } \mathcal{L}^1\text{-almost every } t \in [0, 1] \text{ such that } \tilde{\gamma}(t) \in E^+ \times (0, 1). $$

Applying the same arguments to the set $E^- \times (0, 1)$, we obtain that, for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{K}^{d+1}$,

$$ \dot{\tilde{\gamma}}_{d+1}(t) = -1 \quad \text{for } \mathcal{L}^1\text{-almost every } t \in [0, 1] \text{ such that } \tilde{\gamma}(t) \in E^- \times (0, 1), $$

which concludes the proof of Lemma 2.12. □

Concerning the behavior of the curves in $\mathbb{R}^d \times \{1\}$, suppose that there exists $t^* \in [0, 1]$ such that $\tilde{\gamma}_{d+1}(t^*) = 1$. Since $\tilde{\gamma}_{d+1}(t^*) \le 1$ for all $t$, this point corresponds to a maximum. If $t^* = 0$, then $a(\tilde{\gamma}) > 0$, and if $t^* = 1$, then $b(\tilde{\gamma}) < 1$. Finally, if $t^* \in (0, 1)$, Fermat's theorem implies that $\dot{\tilde{\gamma}}_{d+1}(t^*) = 0$.

By continuity of $\gamma_{d+1}$ and by the definitions of $a(\tilde{\gamma})$ and $b(\tilde{\gamma})$, we have

$$ \gamma_{d+1}(a(\tilde{\gamma})) = \gamma_{d+1}(b(\tilde{\gamma})) = 0. $$
Suppose that there exists a time $t_0 \in (a(\tilde{\gamma}), b(\tilde{\gamma}))$ such that

$$ \gamma_{d+1}(t_0) \neq 0. $$

Since the curve $\tilde{\gamma}([0, 1])$ is contained in $\mathbb{R}^d \times [0, 1]$, it follows that either

$$ \gamma_{d+1}(t_0) \in (0, 1), \quad \text{or} \quad \gamma_{d+1}(t_0) = 1. $$

We first show that $\gamma_{d+1}(t_0) \neq 1$.

Since $\gamma_{d+1}$ is absolutely continuous and $\gamma_{d+1}(a(\tilde{\gamma})) = 0$, we can write

$$ \gamma_{d+1}(t_0) = \int_{a(\tilde{\gamma})}^{t_0} \dot{\gamma}_{d+1}(s) \, ds. $$

Moreover, since $|\dot{\gamma}_{d+1}(s)| \le 1$ for all $s$, we obtain

$$ \gamma_{d+1}(t_0) \le \int_{a(\tilde{\gamma})}^{t_0} 1 \, ds = t_0 - a(\tilde{\gamma}) < 1. $$

Hence, we conclude that $\gamma_{d+1}(t_0) \in (0, 1)$, so that $\tilde{\gamma}(t_0) \in H$.

Thus

$$ t_0 \in (a(\tilde{\gamma}), b(\tilde{\gamma})) \cap \tilde{\gamma}^{-1}(H), $$

which is an open subset of $[0, 1]$.

This open set can be decomposed into disjoint open intervals, its connected components. Let $J = (a_1(\tilde{\gamma}), b_1(\tilde{\gamma}))$ be the connected component containing $t_0$.

By definition, $J = (a_1(\tilde{\gamma}), b_1(\tilde{\gamma}))$ is maximal in the sense that it cannot be extended to a larger interval while remaining contained in $(a(\tilde{\gamma}), b(\tilde{\gamma})) \cap \tilde{\gamma}^{-1}(H)$.

Consequently, at each endpoint one of the following situations occurs:

$$ a_1(\tilde{\gamma}) = a(\tilde{\gamma}) \quad \text{or} \quad \tilde{\gamma}(a_1(\tilde{\gamma})) \notin H, $$

and

$$ b_1(\tilde{\gamma}) = b(\tilde{\gamma}) \quad \text{or} \quad \tilde{\gamma}(b_1(\tilde{\gamma})) \notin H. $$

Recall that, by definition of $a(\tilde{\gamma})$ and $b(\tilde{\gamma})$,

$$ \gamma_{d+1}(a(\tilde{\gamma})) = \gamma_{d+1}(b(\tilde{\gamma})) = 0. $$

We claim that

$$ (2.44) \qquad \gamma_{d+1}(a_1(\tilde{\gamma})) = \gamma_{d+1}(b_1(\tilde{\gamma})) = 0. $$

Indeed, if $a_1(\tilde{\gamma}) = a(\tilde{\gamma})$, the conclusion is immediate. Otherwise, $\tilde{\gamma}(a_1(\tilde{\gamma})) \notin H$, hence

$$ \gamma_{d+1}(a_1(\tilde{\gamma})) \in \{0, 1\}. $$

Since $\gamma_{d+1}$ is absolutely continuous and $|\dot{\gamma}_{d+1}| \le 1$, we have

$$ \gamma_{d+1}(a_1(\tilde{\gamma})) = \int_{a(\tilde{\gamma})}^{a_1(\tilde{\gamma})} \dot{\gamma}_{d+1}(s) \, ds \le a_1(\tilde{\gamma}) - a(\tilde{\gamma}) < 1, $$

since $(a_1(\tilde{\gamma}), b_1(\tilde{\gamma}))$ is strictly contained in $(a(\tilde{\gamma}), b(\tilde{\gamma})) \subset [0, 1]$. Therefore,

$$ \gamma_{d+1}(a_1(\tilde{\gamma})) = 0. $$

The argument for $b_1(\tilde{\gamma})$ is analogous. If $b_1(\tilde{\gamma}) = b(\tilde{\gamma})$ there is nothing to prove; otherwise, $\tilde{\gamma}(b_1(\tilde{\gamma})) \notin H$ and thus $\gamma_{d+1}(b_1(\tilde{\gamma})) \in \{0, 1\}$. Moreover,

$$ \gamma_{d+1}(b_1(\tilde{\gamma})) = - \int_{b_1(\tilde{\gamma})}^{b(\tilde{\gamma})} \dot{\gamma}_{d+1}(s) \, ds \le b(\tilde{\gamma}) - b_1(\tilde{\gamma}) < 1, $$

which implies $\gamma_{d+1}(b_1(\tilde{\gamma})) = 0$, and thus proves the claim (2.44).
Since for all $s \in J$ we have $\tilde{\gamma}(s) \in H$, it follows from Lemma 2.12 that

$$ \dot{\gamma}_{d+1}(s) = \pm 1 \quad \text{for } \mathcal{L}^1\text{-almost every } s \in J. $$

Moreover, since $\|\dot{\gamma}(s)\| = 1$ for $\mathcal{L}^1$-almost every $s$ and $\dot{\gamma}_{d+1}(s) = \pm 1$, we necessarily have

$$ \dot{\gamma}(s) = 0 \quad \text{for } \mathcal{L}^1\text{-almost every } s \in J. $$

By continuity of $\dot{\gamma}$, we deduce that

$$ \dot{\gamma}(s) = 0 \quad \text{for all } s \in J, \quad \text{and then } \dot{\gamma}_{d+1}(s) = \pm 1 \quad \text{for all } s \in J. $$

Consequently, the function $\dot{\gamma}_{d+1}$ has a constant sign on $J$. Indeed, if it were to change sign, continuity of $\dot{\gamma}_{d+1}$ would imply the existence of $s \in J$ such that $\dot{\gamma}_{d+1}(s) = 0$, which is impossible.

Without loss of generality, assume that

$$ \dot{\gamma}_{d+1}(t_0) = 1. $$

Then

$$ \dot{\gamma}_{d+1}(s) = 1 \quad \text{for all } s \in J, $$

and therefore $\gamma_{d+1}$ is strictly increasing on $J$.

However, since $\gamma_{d+1}$ is strictly increasing on $J = (a_1(\tilde{\gamma}), b_1(\tilde{\gamma}))$ and, by (2.44),

$$ \gamma_{d+1}(a_1(\tilde{\gamma})) = \gamma_{d+1}(b_1(\tilde{\gamma})) = 0, $$

we reach a contradiction, as a strictly increasing function cannot take the same value at both endpoints of an interval.

Therefore, the assumption that there exists $t_0 \in (a(\tilde{\gamma}), b(\tilde{\gamma}))$ such that $\gamma_{d+1}(t_0) \neq 0$ is false, concluding the proof of Lemma 2.11. □

Consider the projection map

$$ \mathcal{P} : \mathcal{M} \to \mathcal{K}_{\text{proj}}^d, \quad \tilde{\gamma} = (\gamma, \gamma_{d+1}) \mapsto [t \mapsto \gamma(a(\tilde{\gamma}) + (b(\tilde{\gamma}) - a(\tilde{\gamma}))t)], \quad \forall t \in [0, 1]. $$

This map is well-defined. Indeed, let $\tilde{\gamma} \in \mathcal{M}$, and set $\beta := \mathcal{P}(\tilde{\gamma})$. Since $\tilde{\gamma} \in C^3$, the curve $\beta$ is also of class $C^3$. For any integer $k \in \{1, 2, 3\}$ and all $t \in [0, 1]$, we have

$$ (2.45) \qquad \beta^{(k)}(t) = (b(\tilde{\gamma}) - a(\tilde{\gamma}))^k \gamma^{(k)}(s(t)), \quad \text{where } s(t) := a(\tilde{\gamma}) + (b(\tilde{\gamma}) - a(\tilde{\gamma}))t \in [a(\tilde{\gamma}), b(\tilde{\gamma})]. $$

Using $b(\tilde{\gamma}) - a(\tilde{\gamma}) \le 1$ and the uniform bounds satisfied by $\tilde{\gamma}$ in $\mathcal{K}^{d+1}$, we deduce

$$ (2.46) \qquad \|\dot{\beta}(t)\| \le 1, \quad |\ddot{\beta}_i(t)| \le 2\sqrt{d+1}, \quad \forall i \in \{1, \dots, d\}, \quad \|\beta^{(3)}(t)\| \le L_{d+1}. $$

Finally, since $\tilde{\gamma}^{(3)}$ is $M_{d+1}$-Lipschitz, the same holds for $\gamma^{(3)}$, and therefore

$$ \|\beta^{(4)}(t)\| = (b(\tilde{\gamma}) - a(\tilde{\gamma}))^4 \|\gamma^{(4)}(s(t))\| \le M_{d+1}, \quad \text{for almost every } t \in [0, 1]. $$

It follows that $\beta^{(3)}$ is $M_{d+1}$-Lipschitz continuous. Consequently, we have

$$ \beta \in \mathcal{K}_{\text{proj}}^d. $$

The projection map $\mathcal{P}$ is Borel measurable [7] when $\mathcal{M}$ and $\mathcal{K}_{\text{proj}}^d$ are equipped with the $C^3$-norm topology.

We then define the pushforward measure

$$ (2.47) \qquad \nu := \mathcal{P}_{\#}(\tilde{\nu}_{\perp} \cdot \mathcal{M}), $$

which is a positive Borel measure on $\mathcal{K}_{\text{proj}}^d$.

Moreover, we have

$$ \nu(\mathcal{K}_{\text{proj}}^d) \le \tilde{\nu}(\mathcal{K}^{d+1}) = \|m\|_{\text{TV}} < \infty, $$

so that $\nu$ is a finite positive Borel measure and hence, $\nu \in \mathcal{M}^+(\mathcal{K}_{\text{proj}}^d)$.
Let $\varphi \in \mathcal{D}(\mathbb{R}^d)^d$, following the idea of [7], we consider the function $\tilde{\varphi} \in \mathcal{B}(\mathbb{R}^{d+1})^{d+1}$ defined by

$$ \tilde{\varphi}(x, x_{d+1}) = \begin{cases} (\varphi(x), 0) & \text{if } x_{d+1} = 0, \\ 0 & \text{if } x_{d+1} \neq 0. \end{cases} $$

For $\tilde{\nu}$-almost every $\tilde{\gamma} = (\gamma, \gamma_{d+1}) \in \mathcal{M}$, we have:

$$ \begin{aligned} \langle \mu_{\mathcal{P}(\tilde{\gamma})}, \varphi \rangle &= \int_0^1 \frac{d}{dt} \mathcal{P}(\tilde{\gamma})(t) \cdot \varphi(\mathcal{P}(\tilde{\gamma})(t)) dt \\ &= \int_{a(\tilde{\gamma})}^{b(\tilde{\gamma})} \dot{\gamma}(t) \cdot \varphi(\gamma(t)) dt \quad (\text{by the change of variable from } [0, 1] \text{ to } [a(\tilde{\gamma}), b(\tilde{\gamma})]) \\ &= \int_0^1 \dot{\tilde{\gamma}}(t) \cdot \tilde{\varphi}(\tilde{\gamma}(t)) dt \quad (\text{by definition of } \tilde{\varphi}, \text{ since } \gamma_{d+1} \equiv 0 \text{ on } [a(\tilde{\gamma}), b(\tilde{\gamma})] \text{ and } \gamma_{d+1} \neq 0 \text{ outside}) \\ &= \langle \mu_{\tilde{\gamma}}, \tilde{\varphi} \rangle. \end{aligned} $$

Note that by definition of $\tilde{\varphi}$,

$$ (2.48) \qquad \int_{\mathbb{R}^{d+1}} \tilde{\varphi} \cdot dm = \int_{\mathbb{R}^d} \varphi \cdot d\mu. $$

Observe that for every $\tilde{\gamma} \in \mathcal{K}^{d+1} \setminus \mathcal{M}$, we have $\langle \mu_{\tilde{\gamma}}, \tilde{\varphi} \rangle = 0$, by the definitions of $\mathcal{M}$ and $\tilde{\varphi}$. Therefore, we deduce:

$$ \begin{aligned} \int_{\mathbb{R}^{d+1}} \tilde{\varphi} \cdot dm &= \int_{\mathcal{K}^{d+1}} \langle \mu_{\tilde{\gamma}}, \tilde{\varphi} \rangle d\tilde{\nu}(\tilde{\gamma}) && (\text{by equation (2.28)}) \\ &= \int_{\mathcal{M}} \langle \mu_{\tilde{\gamma}}, \tilde{\varphi} \rangle d\tilde{\nu}(\tilde{\gamma}) && (\text{since } \langle \mu_{\tilde{\gamma}}, \tilde{\varphi} \rangle = 0 \text{ outside } \mathcal{M}) \\ &= \int_{\mathcal{M}} \int_{a(\tilde{\gamma})}^{b(\tilde{\gamma})} \dot{\gamma}(t) \cdot \varphi(\gamma(t)) dt d\tilde{\nu}(\tilde{\gamma}) \\ &= \int_{\mathcal{M}} \langle \mu_{\mathcal{P}(\tilde{\gamma})}, \varphi \rangle d\tilde{\nu}(\tilde{\gamma}) \\ &= \int_{\mathcal{K}_{\text{proj}}^d} \langle \mu_{\gamma}, \varphi \rangle d\nu(\gamma) && (\text{by change of variables under } \mathcal{P}). \end{aligned} $$

By equation (2.48), we then obtain

$$ \int_{\mathbb{R}^d} \varphi \cdot d\mu = \int_{\mathcal{K}_{\text{proj}}^d} \langle \mu_{\gamma}, \varphi \rangle d\nu(\gamma) \quad \text{for all } \varphi \in \mathcal{D}(\mathbb{R}^d)^d, $$

which implies

$$ \mu = \int_{\mathcal{K}_{\text{proj}}^d} \mu_{\gamma} d\nu(\gamma), $$

thus proving equation (2.24).

Similarly, let $\psi \in \mathcal{D}(\mathbb{R}^d)$ and define $\tilde{\psi} \in \mathcal{B}(\mathbb{R}^{d+1})$ by

$$ \tilde{\psi}(x, x_{d+1}) = \begin{cases} \psi(x), & \text{if } x_{d+1} = 0, \\ 0, & \text{if } x_{d+1} \neq 0. \end{cases} $$

By definition of $\tilde{\psi}$ and equation (2.31), we have

$$ (2.49) \qquad \int_{\mathbb{R}^{d+1}} \tilde{\psi} \, d|m| = \int_{\mathbb{R}^d} \psi \, d|\mu|. $$

On the other hand, using the decomposition (2.29) of $|m|$, we can write

$$ \int_{\mathbb{R}^{d+1}} \tilde{\psi} \, d|m| = \int_{\mathcal{K}_{d+1}} \langle |\mu_{\tilde{\gamma}}|, \tilde{\psi} \rangle \, d\tilde{\nu}(\tilde{\gamma}). $$
For $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{K}^{d+1}$, the curve is parametrized at unit speed and $\|\mu_{\tilde{\gamma}}\|_{\text{TV}} = \ell(\tilde{\gamma})$, so that the Area formula gives

$$ \langle |\mu_{\tilde{\gamma}}|, \tilde{\psi} \rangle = \int_{0}^{1} \tilde{\psi}(\tilde{\gamma}(t)) \|\dot{\tilde{\gamma}}(t)\| dt. $$

In particular, for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{K}^{d+1} \setminus \mathcal{M}$, we have

$$ \langle |\mu_{\tilde{\gamma}}|, \tilde{\psi} \rangle = 0, $$

since $\tilde{\gamma}([0, 1]) \cap (\mathbb{R}^d \times \{0\}) = \emptyset$ and $\text{supp}(\tilde{\psi}) \subset \mathbb{R}^d \times \{0\}$.

For $\tilde{\nu}$-almost every $\tilde{\gamma} = (\gamma, \gamma_{d+1}) \in \mathcal{M}$, we have

$$ \begin{aligned} \langle |\mu_{\tilde{\gamma}}|, \tilde{\psi} \rangle &= \int_{0}^{1} \tilde{\psi}(\tilde{\gamma}(t)) \|\dot{\tilde{\gamma}}(t)\| dt \\ &= \int_{a(\tilde{\gamma})}^{b(\tilde{\gamma})} \tilde{\psi}(\tilde{\gamma}(t)) \|\dot{\tilde{\gamma}}(t)\| dt \quad (\text{since } \tilde{\psi}(\tilde{\gamma}(t)) = 0 \text{ outside } [a(\tilde{\gamma}), b(\tilde{\gamma})]) \\ &= \int_{a(\tilde{\gamma})}^{b(\tilde{\gamma})} \psi(\gamma(t)) \|\dot{\gamma}(t)\| dt \quad (\text{since } \tilde{\psi}(x, 0) = \psi(x) \text{ and } \gamma_{d+1}(t) = 0 \text{ on } [a(\tilde{\gamma}), b(\tilde{\gamma})]) \\ &= \langle |\mu_{\mathcal{P}(\tilde{\gamma})}|, \psi \rangle. \end{aligned} $$

As a consequence, we obtain

$$ \begin{aligned} \int_{\mathbb{R}^d} \psi \, d|\mu| &= \int_{\mathcal{M}} \langle |\mu_{\mathcal{P}(\tilde{\gamma})}|, \psi \rangle \, d\tilde{\nu}(\tilde{\gamma}) \\ &= \int_{\mathcal{K}^{d+1}} \langle |\mu_{\mathcal{P}(\tilde{\gamma})}|, \psi \rangle \, d(\tilde{\nu} \llcorner \mathcal{M})(\tilde{\gamma}) \\ &= \int_{\mathcal{K}_{\text{proj}}^d} \langle |\mu_{\gamma}|, \psi \rangle \, d\nu(\gamma) \quad (\text{by the change of variable under } \mathcal{P}). \end{aligned} $$

Hence, we conclude

$$ |\mu| = \int_{\mathcal{K}_{\text{proj}}^d} |\mu_{\gamma}| \, d\nu(\gamma), $$

as stated in equation (2.23).

Moreover, we have

$$ \nu(\mathcal{K}_{\text{proj}}^d) = \tilde{\nu}(\mathcal{M}) \le \tilde{\nu}(\mathcal{K}^{d+1}) = \|m\|_{\text{TV}} \le 2\|\mu\|_{V(\mathbb{R}^d)}, $$

where the last inequality follows from the definition of $m$, thus establishing equation (2.24).

From equation (2.45), we have

$$ \left\| \frac{d}{dt} \mathcal{P}(\tilde{\gamma})(t) \right\| = |b(\tilde{\gamma}) - a(\tilde{\gamma})| \cdot \|\dot{\gamma}(s(t))\|. $$

Moreover, for $\tilde{\nu}$-almost every $\tilde{\gamma} = (\gamma, \gamma_{d+1}) \in \mathcal{M}$, we have $\|\dot{\tilde{\gamma}}(s(t))\| = 1$ and $\gamma_{d+1}(s(t)) = 0$ for all $t \in [0, 1]$. This implies that $\|\dot{\gamma}(s(t))\| = 1$ for all $t$, and thus

$$ \left\| \frac{d}{dt} \mathcal{P}(\tilde{\gamma})(t) \right\| = |b(\tilde{\gamma}) - a(\tilde{\gamma})|. $$

Therefore, for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{M}$, the curve $\mathcal{P}(\tilde{\gamma})$ has constant speed less than or equal to 1. Since the set

$$ R := \{ \gamma \in \mathcal{K}_{\text{proj}}^d \mid \exists c \in [0, 1], \forall t \in [0, 1], \|\dot{\gamma}(t)\| = c \} $$

is measurable, and

$$ \nu(R^c) = (\tilde{\nu} \llcorner \mathcal{M})(\mathcal{P}^{-1}(R^c)) = \tilde{\nu}(\mathcal{M} \cap \mathcal{P}^{-1}(R^c)) = \tilde{\nu}(\{\tilde{\gamma} \in \mathcal{M} \mid \mathcal{P}(\tilde{\gamma}) \in R^c\}) = 0, $$

it follows that $\nu$-a.e $\gamma \in \mathcal{K}_{\text{proj}}^d$ has constant speed. Since $\nu(R^c) = 0$ and $R$ is closed (proof in appendix B), it follows that the support of $\nu$ is contained in $R$. Therefore, every curve in the support of $\nu$ is either a constant curve or a regular curve with constant speed as stated in (2.25).
Let $d = 2$.

Define the subset $S$ by

$$ (2.50) \qquad S = \left\{ \gamma \in \mathcal{K}_{\text{proj}}^2 : \|\dot{\gamma}(t)\| \equiv c \ge 0, \text{ and if } c > 0, \sup_{t \in [0,1]} \kappa_{\gamma}(t) \le \frac{2\pi}{c} \right\}, $$

where $\kappa_\gamma(t)$ is the curvature of $\gamma$. Here, the length $\ell(\gamma)$ is equal to $c$, since the speed of $\gamma$ is constant on the parametrized interval $[0, 1]$. Remark that $S$ is a closed subset of $\mathcal{K}_{\text{proj}}^2$ (proof in appendix B).

Note that for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{M}$, we have $\mathcal{P}(\tilde{\gamma}) \in S$. Indeed, for $\tilde{\nu}$-a.e. $\tilde{\gamma} = (\gamma, \gamma_3) \in \mathcal{M}$, it holds that $\gamma_3 \equiv 0$ on $[a(\tilde{\gamma}), b(\tilde{\gamma})]$, and $\mathcal{P}(\tilde{\gamma}) = \gamma_{|[a(\tilde{\gamma}), b(\tilde{\gamma})]}$.

If $a(\tilde{\gamma}) = b(\tilde{\gamma})$, then $\mathcal{P}(\tilde{\gamma})$ reduces to a constant curve. Otherwise, denoting $\beta := \mathcal{P}(\tilde{\gamma})$, the curvature of $\beta$ is given by

$$ \kappa_{\beta}(t) = \frac{|\dot{\beta}(t) \wedge \ddot{\beta}(t)|}{\|\dot{\beta}(t)\|^3} \le \frac{\|\dot{\beta}(t)\| \|\ddot{\beta}(t)\|}{\|\dot{\beta}(t)\|^3} = \frac{\|\ddot{\beta}(t)\|}{\|\dot{\beta}(t)\|^2}. $$

As shown previously, we have

$$ \|\dot{\beta}(t)\| = b(\tilde{\gamma}) - a(\tilde{\gamma}), \quad \|\ddot{\beta}(t)\| = (b(\tilde{\gamma}) - a(\tilde{\gamma}))^2 \|\ddot{\gamma}(s(t))\|. $$

Since $\|\ddot{\gamma}(s(t))\| \le 2\sqrt{3}\sqrt{2} \le 2\pi$, it follows that

$$ \kappa_{\beta}(t) \le 2\pi \quad \text{for all } t \in [0, 1]. $$

Moreover, $\ell(\beta) = b(\tilde{\gamma}) - a(\tilde{\gamma}) \le 1$, which implies $2\pi \le \frac{2\pi}{\ell(\beta)}$. Thus, $\beta \in S$.

Since $\mathcal{P}(\tilde{\gamma}) \in S$ for $\tilde{\nu}$-almost every $\tilde{\gamma} \in \mathcal{M}$, we have

$$ \nu(S^c) = (\tilde{\nu} \llcorner \mathcal{M})(\mathcal{P}^{-1}(S^c)) = \tilde{\nu}(\mathcal{M} \cap \mathcal{P}^{-1}(S^c)) = \tilde{\nu}(\{\tilde{\gamma} \in \mathcal{M} \mid \mathcal{P}(\tilde{\gamma}) \in S^c\}) = 0. $$

As $S$ is closed, it follows that

$$ \text{supp}(\nu) \subset S. $$

Therefore, every curve in the support of $\nu$ is either constant or a regular injective curve with constant speed, where injectivity follows from [5, Corollary 3.1]. This establishes equation (2.26) and concludes the proof of Theorem 2.9. $\square$

**LEMMA 2.13.** For a measure $\mu \in \mathcal{V}(\mathbb{R}^d)$, if there exists a Radon measure $\nu \in \mathcal{M}^+(\mathcal{K}_{\text{proj}}^d)$ such that equation (2.23) is satisfied, then

$$ (2.51) \qquad \text{supp}(\mu) = \overline{\bigcup_{\gamma \in \text{supp}(\nu)} \gamma([0, 1])}. $$

*Proof.* Let

$$ K := \overline{\bigcup_{\gamma \in \text{supp}(\nu)} \gamma([0, 1])}. $$

Since $K$ is closed and $\text{supp}(\mu_\gamma) \subset \gamma([0, 1]) \subset K$ for all $\gamma \in \text{supp}(\nu)$, we have $\|\mu_\gamma\|(K^c) = 0$. Therefore,

$$ \|\mu\|(K^c) = \int_{\mathcal{K}_{\text{proj}}^n} \|\mu_\gamma\|(K^c) d\nu(\gamma) = 0. $$

Hence $\text{supp}(\mu) \subset K$.

Conversely, if $x \in \text{supp}(\mu)$, then for every $r > 0$, $|\mu|(B(x, r)) > 0$. Since

$$ |\mu|(B(x, r)) = \int_{\mathcal{K}_{\text{proj}}^n} |\mu_\gamma|(B(x, r)) d\nu(\gamma), $$

there exists $\gamma_0 \in \text{supp}(\nu)$ with $|\mu_{\gamma_0}|(B(x, r)) > 0$, which implies $\gamma_0([0, 1]) \cap B(x, r) \ne \emptyset$. Thus every ball $B(x, r)$ intersects $\bigcup_{\gamma \in \text{supp}(\nu)} \gamma([0, 1])$, and we deduce that $x \in K$. $\square$

**LEMMA 2.14.** The decomposition given in Theorem 2.9 is not unique. However, for a measure $\mu \in \mathcal{V}(\mathbb{R}^d)$, if there exist two Radon measures $\nu_1, \nu_2 \in \mathcal{M}^+(\mathcal{K}_{\text{proj}}^d)$ that both satisfy the decomposition of $\mu$ given in Theorem 2.9, then

$$ (2.52) \qquad \text{supp}(\mu) = \overline{\bigcup_{\gamma \in \text{supp}(\nu_1)} \gamma([0, 1])} = \overline{\bigcup_{\gamma \in \text{supp}(\nu_2)} \gamma([0, 1])}. $$
Proof. For the non-uniqueness of the decomposition, fix a simple, regular curve with constant speed $\gamma_0 \in \mathcal{K}_{\text{proj}}^d$ and set $\mu = \mu_{\gamma_0}$. The measure $\nu_1 := \delta_{\gamma_0}$ satisfies

$$\mu = \int_{\mathcal{K}_{\text{proj}}^d} \mu_{\gamma} \, d\nu_1(\gamma), \quad |\mu| = \int_{\mathcal{K}_{\text{proj}}^d} |\mu_{\gamma}| \, d\nu_1(\gamma).$$

Let $0 < \alpha < 1$ and consider the two subpaths of $\gamma_0$ obtained by restricting it to $[0, \alpha]$ and $[\alpha, 1]$:

$$\gamma_1(t) := \gamma_0(\alpha t), \quad \gamma_2(t) := \gamma_0(\alpha + (1-\alpha)t), \quad t \in [0, 1].$$

Define $\nu_2 := \delta_{\gamma_1} + \delta_{\gamma_2}$. Then

$$\mu = \mu_{\gamma_0} = \mu_{\gamma_1} + \mu_{\gamma_2} = \int_{\mathcal{K}_{\text{proj}}^d} \mu_{\gamma} \, d\nu_2(\gamma),$$

and

$$\begin{aligned} |\mu| &= |\mu_{\gamma_0}| = |\mu_{\gamma_1}| + |\mu_{\gamma_2}| \quad \text{since } \gamma_1((0,1)) \text{ and } \gamma_2((0,1)) \text{ are disjoint,} \\ &= \int_{\mathcal{K}_{\text{proj}}^d} |\mu_{\gamma}| \, d\nu_2(\gamma). \end{aligned}$$

Thus, both $\nu_1$ and $\nu_2$ yield valid decompositions of $m$, but $\nu_1 \neq \nu_2$, which proves non-uniqueness. Here, $\delta_{\gamma} \in \mathcal{M}^{+}(\mathcal{K}_{\text{proj}}^d)$ denotes the Dirac measure on $\gamma$. The equality in (2.52) follows directly from Lemma 2.13. $\square$

We recall that in the case $d=2$, the set $S \subset \mathcal{K}_{\text{proj}}^2$ of curves introduced above is closed. As $\mathcal{K}_{\text{proj}}^2$ is locally compact, it follows that $S$ is also locally compact. Consequently, any positive finite Borel measure on $S$ is a Radon measure, i.e., belongs to $\mathcal{M}^+(S)$. Moreover, any measure $\mu \in \mathcal{M}^+(\mathcal{K}_{\text{proj}}^2)$ with support in $S$ can be identified with its restriction $\mu_S \in \mathcal{M}^+(S)$. Therefore, Theorem 2.9 for $d=2$ yields the following corollary.

COROLLARY 2.15 (Decomposition theorem for $d=2$). For every $\mu \in \mathcal{V}(\mathbb{R}^2)$, there exists a positive finite Radon measure $\nu \in \mathcal{M}^+(S)$ such that

$$\mu = \int_S \mu_{\gamma} \, d\nu(\gamma), \quad |\mu| = \int_S |\mu_{\gamma}| \, d\nu(\gamma),$$

(2.53)

and

$$\nu(S) \leq 2 \|\mu\|_{\mathcal{V}(\mathbb{R}^2)}.$$

(2.54)

Note that each element of $S$ is either a constant curve or a regular injective curve with constant speed.

The extension of this decomposition result to vector-valued measures defined on open subsets of $\mathbb{R}^2$ are given in the following corollary.

COROLLARY 2.16 (Decomposition theorem on open subsets of $\mathbb{R}^2$). Let $\mathcal{X} \subset \mathbb{R}^2$ be an open set, and let $\mu \in \mathcal{V}(\mathcal{X})$. Then there exists a positive $\sigma$-finite Borel measure $\nu$ on $S$ such that

$$\mu = \int_S \mu_{\gamma} \, d\nu(\gamma), \quad |\mu| = \int_S |\mu_{\gamma}| \, d\nu(\gamma),$$

(2.55)

and

$$\int_S \|\mu_{\gamma}\|_{\text{TV}} \, d\nu(\gamma) = |\mu|(\mathcal{X}) < \infty.$$

(2.56)

Moreover, $\nu$-a.e. $\gamma \in S$ satisfies $\text{supp}(\gamma) \subseteq \mathcal{X}$.

Proof. The proof of this corollary is given in [7, Theorem 2.17]. The main ingredients come from [11, Theorem 2.14], which provides sufficient conditions ensuring that the trace of a divergence-measure field on the boundary of a subset can be represented by a measure, and from [4], which shows that these conditions are satisfied for a countable decreasing family of open sets $(\mathcal{X}_{\delta_k})_{k \geq 1}$ strictly contained in $\mathcal{X}$, such that $\mathcal{X} = \bigcup_{k=1}^{\infty} \mathcal{X}_{\delta_k}$ and

$$|\mu|(\partial \mathcal{X}_{\delta_k}) = |\text{div } \mu|(\partial \mathcal{X}_{\delta_k}) = 0 \quad \text{for all } k \geq 1.$$

(2.57)
The idea in [7] is to construct the sequence of pairwise disjoint sets

$$A_k := \mathcal{X}_{\delta_k} \setminus \mathcal{X}_{\delta_{k-1}}, \quad \mathcal{X}_{\delta_0} := \emptyset.$$

For each $k$, the trace of $\mu$ on the boundary of $A_k$ is a measure, so that the truncated measure $\mu \llcorner A_k$ belongs to $\mathcal{V}(\mathcal{X})$ and has compact support in $\mathcal{X}$. Hence, it can be naturally extended to an element of $\mathcal{V}(\mathbb{R}^2)$ by setting it equal to 0 outside $\mathcal{X}$.

Applying Theorem 2.15 to each measure $\mu \llcorner A_k$, one obtains a finite positive Radon measure $\nu_k \in \mathcal{M}^+(S)$ associated with its decomposition. The positive $\sigma$-finite Borel measure $\nu$ satisfying the statement of the corollary is then defined by

$$\nu := \sum_{k=1}^{\infty} \nu_k,$$

where equation (2.56) holds as a consequence of (2.57). Since for every $k$, $\nu_k$-almost every curve $\gamma$ satisfies $\text{supp}(\gamma) \subset \bar{A}_k \subset \mathcal{X}$, we have $\nu_k(\{\gamma : \text{supp}(\gamma) \subset \mathcal{X}\}) = \nu_k(S)$ for each $k$. Consequently,

$$\nu(\{\gamma : \text{supp}(\gamma) \subset \mathcal{X}\}) = \sum_k \nu_k(\{\gamma : \text{supp}(\gamma) \subset \mathcal{X}\}) = \sum_k \nu_k(S) = \nu(S).$$

Therefore, $\nu$-almost every curve $\gamma$ satisfies $\text{supp}(\gamma) \subset \mathcal{X}$.

We emphasize that there is no guarantee that $\nu$ is a Radon measure (i.e., finite on compact subsets of $S$). Nevertheless, $\nu$ is a $\sigma$-finite Borel measure. Indeed, we can decompose $S$ as

$$S = \bigcup_{k=1}^{\infty} \text{supp}(\nu_k) \cup \left(S \setminus \bigcup_{k=1}^{\infty} \text{supp}(\nu_k)\right),$$

and for each $k \ge 1$,

$$\begin{aligned} \nu(\text{supp}(\nu_k)) &= \sum_{j=1}^{\infty} \nu_j(\text{supp}(\nu_k)) \\ &= \nu_{k-1}(\text{supp}(\nu_k)) + \nu_k(\text{supp}(\nu_k)) + \nu_{k+1}(\text{supp}(\nu_k)) < \infty, \end{aligned}$$

since each $\nu_j$ is finite and the supports of non-consecutive $\nu_j$ are disjoint. Moreover, $\nu$ vanishes on the complement $S \setminus \bigcup_k \text{supp}(\nu_k)$. Hence, $S$ is covered by a countable union of sets of finite $\nu$-measure, showing that $\nu$ is $\sigma$-finite. $\square$

Note that extending a measure $\mu \in \mathcal{V}(\mathcal{X})$ to $\mathcal{V}(\mathbb{R}^2)$ is, in general, a subtle issue. Indeed, a naive extension of $\mu$ by zero outside $\mathcal{X}$ does not guarantee that the trace on $\partial\mathcal{X}$ is representable by a measure, and consequently, the divergence may fail to be a finite Radon measure. The following corollary provides a construction that overcomes this difficulty.

**COROLLARY 2.17** (Extension of divergence-measure fields to $\mathbb{R}^2$). Let $\mu \in \mathcal{V}(\mathcal{X})$ be a divergence-measure field on an open set $\mathcal{X} \subset \mathbb{R}^2$. Then, using the measure $\nu$ constructed in Corollary 2.16, we define the vector-valued distribution on $\mathbb{R}^2$

$$\tilde{\mu} := \int_S \mu_{\gamma} \, d\nu(\gamma), \quad \text{defined by} \quad \langle \tilde{\mu}, \varphi \rangle = \int_S \langle \mu_{\gamma}, \varphi \rangle \, d\nu(\gamma), \quad \forall \varphi \in \mathcal{C}_c^{\infty}(\mathbb{R}^2)^2.$$

This distribution defines an element of $\mathcal{V}(\mathbb{R}^2)$ such that

$$(2.58) \qquad \tilde{\mu}|_{\mathcal{X}} = \mu, \quad \text{div}(\tilde{\mu})|_{\mathcal{X}} = \text{div}(\mu), \quad \|\tilde{\mu}\|_{\text{TV}} = |\mu|(\mathcal{X}), \quad |\text{div}(\tilde{\mu})|_{\text{TV}} = |\text{div}(\mu)|(\mathcal{X}) < \infty.$$

In particular, $\tilde{\mu}$ is a divergence-measure field on $\mathbb{R}^2$ extending $\mu$ by zero outside $\mathcal{X}$.

**Proof.** For every test function $\varphi \in \mathcal{C}_c^{\infty}(\mathbb{R}^2)^2$, we have

$$|\langle \tilde{\mu}, \varphi \rangle| = \left| \int_S \langle \mu_{\gamma}, \varphi \rangle \, d\nu(\gamma) \right| \le \int_S |\langle \mu_{\gamma}, \varphi \rangle| \, d\nu(\gamma) \le \|\varphi\|_{\infty} \int_S \|\mu_{\gamma}\|_{\text{TV}} \, d\nu(\gamma).$$

Hence, $|\langle \tilde{\mu}, \varphi \rangle| \le c \|\varphi\|_{\infty}$, where $c := \int_S \|\mu_{\gamma}\|_{\text{TV}} \, d\nu(\gamma) = |\mu|(\mathcal{X}) < \infty$ by (2.56). Therefore, $\tilde{\mu}$ is a finite Radon measure on $\mathbb{R}^2$ such that $\|\tilde{\mu}\|_{\text{TV}} \le |\mu|(\mathcal{X})$. Moreover, by (2.55), we have

$$|\mu|(\mathcal{X}) = \int_S |\mu_{\gamma}|(\mathcal{X}) \, d\nu(\gamma) = \int_S \|\mu_{\gamma}\|_{\text{TV}} \, d\nu(\gamma) \ge \|\tilde{\mu}\|_{\text{TV}}.$$
733 Hence, equality holds: $||\tilde{\mu}||_{TV} = |\mu|(\mathcal{X})$.

734 We have the divergence of $\tilde{\mu}$ in the sense of distributions: for all $\phi \in C_c^\infty(\mathbb{R}^2)$,

$$
\begin{align*}
\langle \mathrm{div}(\tilde{\mu}), \phi \rangle &= - \int_S \langle \mu_\gamma, \nabla \phi \rangle \, \mathrm{d}\nu(\gamma) \\
&= - \sum_{k=1}^\infty \int_S \langle \mu_\gamma, \nabla \phi \rangle \, \mathrm{d}\nu_k(\gamma) \quad \text{since } \gamma \mapsto \langle \mu_\gamma, \nabla \phi \rangle \text{ is } \nu\text{-integrable} \\
&= \sum_{k=1}^\infty \int_S (\phi(\gamma(0)) - \phi(\gamma(1))) \, \mathrm{d}\nu_k(\gamma) \\
&= \sum_{k=1}^\infty \int_S \langle \delta_{\gamma(0)}, \phi \rangle \, \mathrm{d}\nu_k(\gamma) - \int_S \langle \delta_{\gamma(1)}, \phi \rangle \, \mathrm{d}\nu_k(\gamma).
\end{align*}
$$

739 Hence, in the sense of distributions,

$$
740 \qquad \mathrm{div}(\tilde{\mu}) = \sum_{k=1}^{\infty} \int_S \delta_{\gamma(0)} \, \mathrm{d}\nu_k(\gamma) - \int_S \delta_{\gamma(1)} \, \mathrm{d}\nu_k(\gamma).
$$

741 For each $k \ge 1$, $\int_S \delta_{\gamma(0)} \, d\nu_k(\gamma)$ and $\int_S \delta_{\gamma(1)} \, d\nu_k(\gamma)$ are finite Radon measures, being the pushforwards of
742 the finite Radon measure $\nu_k$ under the continuous maps $\gamma \mapsto \gamma(0)$ and $\gamma \mapsto \gamma(1)$, respectively. Therefore,
743 $m_k = \int_S \delta_{\gamma(0)} \, d\nu_k(\gamma) - \int_S \delta_{\gamma(1)} \, d\nu_k(\gamma)$ is a finite signed Radon measure, and the series

$$
744 \qquad \sum_{k=1}^{\infty} \left( \int_S \delta_{\gamma(0)} \, d\nu_k(\gamma) - \int_S \delta_{\gamma(1)} \, d\nu_k(\gamma) \right)
$$

745 defines a well-defined $\sigma$-finite signed Borel measure, which equals $\mathrm{div}(\tilde{\mu})$.

746 Since $\mathbb{R}^2 \setminus \mathcal{X}$ is closed, it is a Borel set. By definition of the total variation of a signed measure, we
747 have

$$
748 \qquad \begin{aligned}
& |\mathrm{div}(\tilde{\mu})|(\mathbb{R}^2 \setminus \mathcal{X}) \\
& = \sup \left\{ \sum_{h=0}^{\infty} \left| \mathrm{div}(\tilde{\mu})(E_h) \right| : E_h \in \mathcal{B}(\mathbb{R}^2) \text{ pairwise disjoint, } \mathbb{R}^2 \setminus \mathcal{X} = \bigcup_h E_h \right\}.
\end{aligned}
$$

750 However, for any Borel set $E \subset \mathbb{R}^2 \setminus \mathcal{X}$, we have

$$
751 \qquad \mathrm{div}(\tilde{\mu})(E) = \sum_{k=1}^{\infty} \int_{S} (\delta_{\gamma(0)}(E) - \delta_{\gamma(1)}(E)) \, \mathrm{d}\nu_{k}(\gamma) = 0,
$$

752 since for all $k \ge 1$, $\nu_k$-a.e. $\gamma \in S$ has support contained in $\bar{A}_k \subset \mathcal{X}$. It follows that

$$
753 \qquad (2.59) \qquad \qquad |\mathrm{div}(\tilde{\mu})|(\mathbb{R}^2 \setminus \mathcal{X}) = 0.
$$

754 Note that $\tilde{\mu} = \mu$ on $\mathcal{X}$, so $\mathrm{div}(\tilde{\mu})|_{\mathcal{X}} = \mathrm{div}(\mu)$. From equation (2.59), it follows that

$$
755 \qquad |\mathrm{div}(\tilde{\mu})|(\mathbb{R}^2) = |\mathrm{div}(\tilde{\mu})|(\mathcal{X}) = |\mathrm{div}(\mu)|(\mathcal{X}) < \infty.
$$

756 Hence, $\mathrm{div}(\tilde{\mu})$ is a finite signed Radon measure on $\mathbb{R}^2$, and the corollary is proved. $\square$

757 The reader may be interested in further properties of the measure $\nu \in \mathcal{M}(S)$ governing the decom-
758 position of a given measure $m \in \mathcal{V}(\mathbb{R}^2)$. This motivates the following proposition.

759 PROPOSITION 2.18. Let $\nu$ be a positive Borel measure on the space of curves $S$. Then

$$
760 \qquad \left| \int_S \mu_\gamma \mathrm{d}\nu(\gamma) \right| = \int_S |\mu_\gamma| \mathrm{d}\nu(\gamma)
$$

761 if and only if the tangent alignment condition ( t.a.c ) holds: for $\mathcal{H}^1$-a.e. $x \in \mathbb{R}^2$, there exists a unit
762 vector $u(x) \in \mathbb{R}^2$ such that for $\nu$-a.e. $\gamma \in S(x) := \{\gamma \in S : x \in \gamma([0, 1])\}$, the tangent vector satisfies

$$
763 \qquad \dot{\gamma}(\gamma^{-1}(x)) = \alpha(\gamma, x) u(x), \quad \text{with } \alpha(\gamma, x) \ge 0.
$$
In other words, for $\mathcal{H}^1$-almost every $x \in \mathbb{R}^2$, if $\gamma_1, \gamma_2 \in S(x)$, then their tangent vectors $\dot{\gamma}_1(\gamma_1^{-1}(x))$ and $\dot{\gamma}_2(\gamma_2^{-1}(x))$ are collinear and point in the same direction.

Geometrically, this means that wherever two curves in the support of $\nu$ meet on a set of positive Hausdorff measure, their tangent vectors along the intersection are collinear and oriented in the same direction.

**Proof.** We start by observing that, for each curve $\gamma \in S$, the vector-valued Radon measure $\mu_\gamma$ is absolutely continuous with respect to the 1-dimensional Hausdorff measure $\mathcal{H}^1$:

$$ (2.60) \qquad \mu_{\gamma} = f_{\gamma} \mathcal{H}^1, \quad \text{where} \quad f_{\gamma} : \mathbb{R}^2 \to \mathbb{R}^2, \quad f_{\gamma}(x) = \begin{cases} \dot{\gamma}(\gamma^{-1}(x)) & \text{if } x \in \Gamma := \gamma([0, 1]), \\ 0 & \text{otherwise.} \end{cases} $$

Then, for any $B \in \mathcal{B}(\mathbb{R}^2)$, we have

$$ \begin{aligned} \left( \int_S \mu_\gamma \, d\nu(\gamma) \right) (B) &= \int_S \mu_\gamma(B) \, d\nu(\gamma) \\ &= \int_S \int_B f_\gamma(x) \, d\mathcal{H}^1(x) \, d\nu(\gamma) \\ &= \int_B \int_S f_\gamma(x) \, d\nu(\gamma) \, d\mathcal{H}^1(x) \quad (\text{by Fubini's theorem}) \\ &= \int_B \int_{S(x)} f_\gamma(x) \, d\nu(\gamma) \, d\mathcal{H}^1(x), \end{aligned} $$

where $S(x) := \{\gamma \in S : x \in \gamma([0, 1])\}$.

It follows that the measure

$$ m := \int_S \mu_\gamma \, d\nu(\gamma) $$

is absolutely continuous with respect to $\mathcal{H}^1$, so that we can write

$$ (2.61) \qquad m = g \mathcal{H}^1, \quad g : \mathbb{R}^2 \to \mathbb{R}^2, \quad g(x) := \int_{S(x)} f_\gamma(x) \, d\nu(\gamma). $$

Therefore,

$$ (2.62) \qquad \left| \int_S \mu_\gamma \, d\nu(\gamma) \right| = \|g\| \mathcal{H}^1, \quad \text{i.e.} \quad \left| \int_S \mu_\gamma \, d\nu(\gamma) \right| (B) = \int_B \|g(x)\| \, d\mathcal{H}^1(x) \quad \forall B \in \mathcal{B}(\mathbb{R}^2). $$

(=$\Leftarrow$) Assume now that the tangent alignment condition holds.

Let $B \in \mathcal{B}(\mathbb{R}^2)$. From equation (2.62), we have

$$ \begin{aligned} \left| \int_S \mu_\gamma \, d\nu(\gamma) \right| (B) &= \int_B \left| \int_{S(x)} f_\gamma(x) \, d\nu(\gamma) \right| \, d\mathcal{H}^1(x) \\ &= \int_B \left| \int_{S(x)} \alpha(\gamma, x) u(x) \, d\nu(\gamma) \right| \, d\mathcal{H}^1(x) \\ &= \int_B \|u(x)\| \left| \int_{S(x)} \alpha(\gamma, x) \, d\nu(\gamma) \right| \, d\mathcal{H}^1(x) \\ &= \int_B \int_{S(x)} \alpha(\gamma, x) \, d\nu(\gamma) \, d\mathcal{H}^1(x), \end{aligned} $$

since $\|u(x)\| = 1$ and the integral $\int_{S(x)} \alpha(\gamma, x) \, d\nu(\gamma)$ is positive.
791 On the other hand, we have

$$
\begin{align*}
\left( \int_S |\mu_\gamma| \, d\nu(\gamma) \right) (B) &= \int_S |\mu_\gamma|(B) \, d\nu(\gamma) \\
&= \int_S \int_B \|f_\gamma(x)\| \, d\mathcal{H}^1(x) \, d\nu(\gamma) && \text{(by equation (2.60))} \\
&= \int_B \int_{S(x)} \|f_\gamma(x)\| \, d\nu(\gamma) \, d\mathcal{H}^1(x) \\
&= \int_B \int_{S(x)} \|\alpha(\gamma, x) u(x)\| \, d\nu(\gamma) \, d\mathcal{H}^1(x) && \text{(by hypothesis)} \\
&= \int_B \int_{S(x)} \alpha(\gamma, x) \, d\nu(\gamma) \, d\mathcal{H}^1(x) && \text{(since } \alpha(\gamma, x) \ge 0 \text{ and } \|u(x)\| = 1).
\end{align*}
$$

797 Comparing the two expressions, we see that for every Borel set $B \subset \mathbb{R}^2$,

$$
798 \quad \left| \int_S \mu_\gamma \, d\nu(\gamma) \right| (B) = \left( \int_S |\mu_\gamma| \, d\nu(\gamma) \right) (B).
$$

799 Hence, the desired implication follows.

800 (⇒) Assume that

$$
801 \quad \left| \int_S \mu_\gamma \, d\nu(\gamma) \right| = \int_S |\mu_\gamma| \, d\nu(\gamma).
$$

802 Then

$$
803 \quad (2.63) \qquad \int_{\mathbb{R}^2} \|g(x)\| \, d\mathcal{H}^1(x) = \int_{\mathbb{R}^2} \int_{S(x)} \|f_\gamma(x)\| \, d\nu(\gamma) \, d\mathcal{H}^1(x),
$$

804 which implies

$$
805 \quad (2.64) \qquad \text{for } \mathcal{H}^1\text{-a.e. } x \in \mathbb{R}^2, \quad \|g(x)\| = \int_{S(x)} \|f_\gamma(x)\| \, d\nu(\gamma).
$$

806 Fix such an $x$ with $\int_{S(x)} \|f_\gamma(x)\| \, d\nu(\gamma) \neq 0$ (the case where it is 0 is trivial). Then

$$
807 \quad (2.65) \quad \int_{S(x)} \left( \|f_{\gamma}(x)\| - f_{\gamma}(x) \cdot \frac{\int_{S(x)} f_{\gamma}(x) \, d\nu(\gamma)}{\int_{S(x)} \|f_{\gamma}(x)\| \, d\nu(\gamma)} \right) \, d\nu(\gamma) = 0.
$$

808 Moreover, for each $\gamma \in S(x)$, we have

$$
809 \quad f_{\gamma}(x) \cdot \frac{\int_{S(x)} f_{\gamma}(x) \, d\nu(\gamma)}{\int_{S(x)} \|f_{\gamma}(x)\| \, d\nu(\gamma)} \le \left\| f_{\gamma}(x) \cdot \frac{\int_{S(x)} f_{\gamma}(x) \, d\nu(\gamma)}{\int_{S(x)} \|f_{\gamma}(x)\| \, d\nu(\gamma)} \right\| \le \|f_{\gamma}(x)\|.
$$

810 Hence,

$$
811 \quad \|f_{\gamma}(x)\| - f_{\gamma}(x) \cdot \frac{\int_{S(x)} f_{\gamma}(x) \, d\nu(\gamma)}{\int_{S(x)} \|f_{\gamma}(x)\| \, d\nu(\gamma)} \ge 0.
$$

812 Since the integral with respect to $\nu$ of this nonnegative quantity over $S(x)$ is zero from (2.65), we deduce
813 that

$$
814 \quad \|f_{\gamma}(x)\| = f_{\gamma}(x) \cdot \frac{\int_{S(x)} f_{\gamma}(x) \, d\nu(\gamma)}{\int_{S(x)} \|f_{\gamma}(x)\| \, d\nu(\gamma)} \quad \text{for } \nu\text{-a.e. } \gamma \in S(x).
$$

815 This equality implies that

$$
816 \quad \cos \left( \frac{\int_{S(x)} f_{\gamma}(x) \, d\nu(\gamma)}{\int_{S(x)} \|f_{\gamma}(x)\| \, d\nu(\gamma)} \right) = 1, \quad \text{for } \nu\text{-a.e. } \gamma \in S(x).
$$

817 Defining

$$
818 \quad u(x) := \frac{\int_{S(x)} f_{\gamma}(x) \, d\nu(\gamma)}{\int_{S(x)} \|f_{\gamma}(x)\| \, d\nu(\gamma)},
$$

819 we have $\|u(x)\| = 1$, and there exists $\alpha(\gamma, x) \ge 0$ such that $f_\gamma(x) = \alpha(\gamma, x) u(x)$, which concludes the
820 proof. $\square$
Remark that the measure $\nu \in \mathcal{M}^+(S)$ constructed in the decomposition Corollary 2.15 for a given measure $\mu \in \mathcal{V}(\mathbb{R}^2)$ satisfies

$$ \left| \int_S \mu_{\gamma} d\nu(\gamma) \right| = \int_S |\mu_{\gamma}| d\nu(\gamma), $$

we deduce from proposition 2.18 that wherever two curves in the support of $\nu$ meet on a set of positive Hausdorff measure, their tangent vectors along the intersection are collinear and consistently oriented. The Figure 1 illustrates this condition.

<figure><img src="image_2.png" /><figcaption>FIG. 1. This figure illustrates the tangent alignment condition for two curves $\gamma_1$ (red) and $\gamma_2$ (blue) in the support of $\nu$. The curves coincide on a set of positive Hausdorff length near the point $x$. At such intersection points, their tangent vectors must be positively collinear. The densely dotted and loosely dotted arrows indicate two different possible orientations of the tangent vectors at $x$ for $\gamma_1$ (red) and $\gamma_2$ (blue), respectively.</figcaption></figure>

**PROPOSITION 2.19.** The set

$$ \mathcal{O} = \{ \nu \in \mathcal{M}^+(S) : \left| \int_S \mu_{\gamma} d\nu(\gamma) \right| = \int_S |\mu_{\gamma}| d\nu(\gamma) \} $$

is not closed with respect to the weak*-topology in $\mathcal{M}^+(S)$.

*Proof.* Let us consider two parametrized curves $\gamma_1$ and $\gamma_2$ in $S$, whose geometric supports are represented in Figure 2 in red and blue, respectively. We define

$$ \nu = \delta_{\gamma_1} + \delta_{\gamma_2} \in \mathcal{M}^+(S). $$

For each curve, we consider sequences of approximating curves $\{\gamma_1^n\}$ and $\{\gamma_2^n\}$ in $S$ such that

$$ \gamma_1^n \to \gamma_1 \quad \text{and} \quad \gamma_2^n \to \gamma_2 \quad \text{in the norm topology defined in } S $$

The corresponding geometric supports are illustrated in Figure 2 by the red and blue dotted curves, respectively. We then set

$$ \nu_n = \delta_{\gamma_1^n} + \delta_{\gamma_2^n} \in \mathcal{M}^+(S). $$

For every $n$, the measure $\nu_n$ belongs to $\mathcal{O}$, since the supports of $\gamma_1^n$ and $\gamma_2^n$ are disjoint, so the tangent alignment condition (t.a.c.) is satisfied.

Moreover, $(\nu_n)$ converges to $\nu$ in the weak*-topology of $\mathcal{M}(S)$: indeed, for any $\Phi \in C_0(S)$,

$$ \langle \nu_n, \Phi \rangle = \Phi(\gamma_1^n) + \Phi(\gamma_2^n) \to \Phi(\gamma_1) + \Phi(\gamma_2) = \langle \nu, \Phi \rangle \quad \text{as } n \to \infty. $$

However, $\nu \notin \mathcal{O}$, because the supports of $\gamma_1$ and $\gamma_2$ intersect on a set of positive $\mathcal{H}^1$-measure $(\mathcal{H}^1(\Gamma_1 \cap \Gamma_2) > 0)$, and at points of intersection the tangent vectors of $\gamma_1$ and $\gamma_2$ are collinear but oriented in opposite directions. Hence, the set $\mathcal{O}$ is not closed under weak*-convergence. □
<figure><img src="image_3.png" /><figcaption>FIG. 2. Description of the counter example for the non-closedness of O.</figcaption></figure>

## Appendix A. Proof of Lemma 2.6.

*Proof.* Let $x \in \mathbb{R}^d$.

(a) $\gamma_x = u(\cdot, x) \in C^\infty([0, 1]; \mathbb{R}^d) \subset C^3([0, 1]; \mathbb{R}^d)$.

(b) For all $t \in [0, 1]$, we have

$$ \dot{\gamma}_x(t) = F(\gamma_x(t)), \quad \text{so that} \quad \|\dot{\gamma}_x(t)\| = \|F(\gamma_x(t))\| \le 1. $$

(c) For each $i \in \{1, \dots, d\}$, let $F_i$ denote the $i$-th component of the vector field defined by $F_i(z) = \frac{(\mu_i * k)(z)}{(|\mu| * k)(z)}$, where $\mu_i$ denotes the $i$-th component of the vector-valued measure $\mu$.

For each $i$, we compute the second derivative:

$$ \ddot{\gamma}_x^i(t) = \frac{d}{dt} F_i(\gamma_x(t)) = \nabla F_i(\gamma_x(t)) \cdot \dot{\gamma}_x(t), $$

hence

$$ |\dot{\gamma}_x^i(t)| \le \|\nabla F_i(\gamma_x(t))\| \cdot \|\dot{\gamma}_x(t)\| \le \|\nabla F_i(\gamma_x(t))\|. $$

For each $j \in \{1, \dots, d\}$, we compute:

$$ \frac{\partial F_i}{\partial z_j}(z) = \frac{(\mu_i * \frac{\partial k}{\partial z_j})(z)(|\mu| * k)(z) - (\mu_i * k)(z)(|\mu| * \frac{\partial k}{\partial z_j})(z)}{(|\mu| * k)^2(z)}. $$

Taking absolute values and using $|\mu_i| \le |\mu|$, we obtain

$$ \left| \frac{\partial F_i}{\partial z_j}(z) \right| \le 2 \cdot \frac{\int_{\mathbb{R}^d} \left| \frac{\partial k}{\partial z_j}(z-y) \right| d|\mu|(y)}{(|\mu| * k)(z)}. $$

Next, since

$$ \left| \frac{\partial k}{\partial z_j}(z) \right| = \left| \frac{-z_j}{\sqrt{1 + \|z\|^2}} \cdot k(z) \right| \le \frac{|z_j|}{\sqrt{1 + \|z\|^2}} \cdot k(z), $$

we conclude

$$ \begin{aligned} \left| \frac{\partial F_i}{\partial z_j}(z) \right| &\le 2 \cdot \sup_{y \in \mathbb{R}^d} \frac{|z_j - y_j|}{\sqrt{1 + \|z - y\|^2}} \\ &\le 2, \end{aligned} $$
since $|z_j - y_j| \le \|z - y\| \le \sqrt{1 + \|z - y\|^2}$.

Therefore, for all $i \in \{1, \dots, d\}$ and all $t \in [0, 1]$, we have

$$ (A.1) \qquad \left| \ddot{\gamma}_x^i(t) \right| \le \sqrt{\sum_{j=1}^d \left| \frac{\partial F_i}{\partial z_j}(z) \right|^2} \le 2\sqrt{d}. $$

(d) We have:

$$ \begin{aligned} \frac{d^3}{dt^3} \gamma_x^i(t) &= \frac{d}{dt} \left[ \nabla F_i(\gamma_x(t)) \cdot \dot{\gamma}_x(t) \right] \\ &= \frac{d}{dt} \left[ \sum_{j=1}^d \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \cdot \dot{\gamma}_x^j(t) \right] \\ &= \sum_{j=1}^d \frac{d}{dt} \left( \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \cdot \dot{\gamma}_x^j(t) \right) \\ &= \sum_{j=1}^d \left[ \left( \nabla \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \cdot \dot{\gamma}_x(t) \right) \cdot \dot{\gamma}_x^j(t) + \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \cdot \ddot{\gamma}_x^j(t) \right] \end{aligned} $$

Hence,

$$ \begin{aligned} \left| \frac{d^3}{dt^3} \gamma_x^i(t) \right| &\le \sum_{j=1}^d \left[ \sup_{z \in \mathbb{R}^d} \left\| \nabla \left( \frac{\partial F_i}{\partial z_j}(z) \right) \right\| \cdot \left| \dot{\gamma}_x^j(t) \right| + \left| \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \right| \cdot \left| \ddot{\gamma}_x^j(t) \right| \right] \\ &\le \sum_{j=1}^d \left[ \sup_{z \in \mathbb{R}^d} \left( \sqrt{\sum_{l=1}^d \left| \frac{\partial^2 F_i}{\partial z_l \partial z_j}(z) \right|^2} \right) + 4\sqrt{d} \right], \end{aligned} $$

since $|\dot{\gamma}_x^j(t)| \le 1$ and $|\ddot{\gamma}_x^j(t)| \le 2\sqrt{d}$ for all $j \in \{1, \dots, d\}$, and $\left| \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \right| \le 2$ as previously established.

Note that:

$$ \begin{aligned} \frac{\partial^2 F_i}{\partial z_l \partial z_j}(z) &= \frac{\partial}{\partial z_l} \left[ \frac{(\mu_i * \frac{\partial k}{\partial z_j})(z)(|\mu|*k)(z) - (\mu_i * k)(z)(|\mu| * \frac{\partial k}{\partial z_j})(z)}{(|\mu|*k)^2(z)} \right] \\ &= \frac{1}{(|\mu|*k)^2} \left[ (\mu_i * \frac{\partial^2 k}{\partial z_l \partial z_j})(|\mu|*k) + (\mu_i * \frac{\partial k}{\partial z_j})(|\mu| * \frac{\partial k}{\partial z_l}) \right. \\ &\qquad \left. - (|\mu| * \frac{\partial^2 k}{\partial z_l \partial z_j})(\mu_i * k) - (|\mu| * \frac{\partial k}{\partial z_j})(\mu_i * \frac{\partial k}{\partial z_l}) \right] \\ &\qquad - 2 \cdot \frac{1}{(|\mu|*k)^4} \cdot (|\mu|*k)(|\mu| * \frac{\partial k}{\partial z_l}) \\ &\qquad \cdot \left[ (\mu_i * \frac{\partial k}{\partial z_j})(|\mu|*k) - (\mu_i * k)(|\mu| * \frac{\partial k}{\partial z_j}) \right]. \end{aligned} $$

We compute the partial derivatives of the kernel $k$ in order to bound the second-order derivatives of $F_i$. For all $j, l \in \{1, \dots, d\}$, we have:

$$ \frac{\partial k}{\partial z_j}(z) = - \frac{z_j}{\sqrt{1 + \|z\|^2}} \cdot k(z), \quad \text{so} \quad \left| \frac{\partial k}{\partial z_j}(z) \right| \le k(z). $$

Next, we compute the second derivative:

$$ \frac{\partial^2 k}{\partial z_j^2}(z) = \left[ \frac{z_j^2}{(1 + \|z\|^2)^{3/2}} - \frac{1}{\sqrt{1 + \|z\|^2}} + \frac{z_j^2}{1 + \|z\|^2} \right] \cdot k(z). $$

This implies:

$$ \left| \frac{\partial^2 k}{\partial z_j^2}(z) \right| \le 3k(z). $$
Similarly, for $j \neq l$, we compute the mixed derivatives:

$$ \frac{\partial^2 k}{\partial z_l \partial z_j}(z) = \left[ \frac{z_j z_l}{(1 + \|z\|^2)^{3/2}} + \frac{z_j z_l}{1 + \|z\|^2} \right] \cdot k(z), $$

which gives the bound:

$$ \left| \frac{\partial^2 k}{\partial z_l \partial z_j}(z) \right| \leq 2k(z). $$

These bounds allow us to control the second derivatives of $F_i$, and we obtain the uniform bound:

$$ \left| \frac{\partial^2 F_i}{\partial z_l \partial z_j}(z) \right| \leq 12, \quad \text{for all } z \in \mathbb{R}^d \text{ and all } j, l \in \{1, \dots, d\}. $$

So that:

$$ \begin{aligned} \left| \frac{d^3}{dt^3} \gamma_x^i(t) \right| & \leq \sum_{j=1}^{d} \left[ \sup_{z \in \mathbb{R}^d} \left( \sqrt{\sum_{l=1}^{d} \left| \frac{\partial^2 F_i}{\partial z_l \partial z_j}(z) \right|^2} \right) + 4\sqrt{d} \right] \\ & \leq \sum_{j=1}^{d} \left[ \sqrt{12^2 \cdot d} + 4\sqrt{d} \right] = 16d\sqrt{d}. \end{aligned} $$

This implies that:

$$ \|\gamma_x^{(3)}(t)\| \leq 16d^2, \quad \forall t \in [0, 1]. $$

Therefore, the constant $L_d$ appearing in the definition of the space of curves $\mathcal{K}^d$ in (2.1) should be chosen greater than or equal to $16d^2$.

(e) To show that $\gamma_x^{(3)}$ is $M_d$-Lipschitz, it suffices to prove that the fourth derivative $\gamma_x^{(4)}$ is uniformly bounded by $M_d$, since $\gamma_x \in C^\infty$.

We know that for all $i \in \{1, \dots, d\}$, we have:

$$ \frac{d^4}{dt^4} \gamma_x^i(t) = \frac{d}{dt} \left[ \sum_{j=1}^{d} \left( \left( \nabla \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \cdot \dot{\gamma}_x(t) \right) \cdot \dot{\gamma}_x^j(t) + \frac{\partial F_i}{\partial z_j}(\gamma_x(t)) \cdot \ddot{\gamma}_x^j(t) \right) \right]. $$

As in the previous case, bounding this derivative ultimately requires controlling the third-order derivatives $\frac{\partial^3 F_i}{\partial z_m \partial z_l \partial z_j}(z)$, for all $j, l, m \in \{1, \dots, d\}$. This can be achieved thanks to the fact that the kernel $k$ satisfies:

$$ \left| \frac{\partial^3 k}{\partial z_m \partial z_l \partial z_j}(z) \right| \leq C k(z), $$

for some constant $C > 0$ independent of $z \in \mathbb{R}^d$. Consequently, one obtains a uniform bound on $\|\gamma_x^{(4)}(t)\|$, independent of $x \in \mathbb{R}^d$. The constant $M_d$ appearing in the definition of the space $\mathcal{K}^d$ in (2.1) should therefore be chosen greater than or equal to this uniform bound.

In conclusion, for all $x \in \mathbb{R}^d$, we have $\gamma_x \in \mathcal{K}^d$, so the map $G$ is well-defined.

To prove that $G : \mathbb{R}^d \to \mathcal{K}^d$ is continuous, it suffices to show that it is sequentially continuous, since $\mathbb{R}^d$ is a metric space.

Let $(x_n)_{n \in \mathbb{N}} \subset \mathbb{R}^d$ be a sequence converging to $x_0 \in \mathbb{R}^d$. For each $t \in [0, 1]$ and each $k \in \{0, 1, 2, 3\}$, we have

$$ \gamma_{x_n}^{(k)}(t) = \partial_t^k u(t, x_n) \xrightarrow{n \to \infty} \partial_t^k u(t, x_0) = \gamma_{x_0}^{(k)}(t), $$

by continuity of the map $x \mapsto \partial_t^k u(t, x)$, which holds since $u(t, \cdot) \in C^\infty(\mathbb{R}^d)$. Thus, for each $k \in \{0, 1, 2, 3\}$, the sequence $\gamma_{x_n}^{(k)}$ converges pointwise to $\gamma_{x_0}^{(k)}$ on $[0, 1]$. Moreover, since the curves $\gamma_{x_n}^{(k)}$ are uniformly equicontinuous (being uniformly Lipschitz) on the compact interval $[0, 1]$, we deduce that $\gamma_{x_n}^{(k)} \to \gamma_{x_0}^{(k)}$ uniformly on $[0, 1]$ for each $k \in \{0, 1, 2, 3\}$. Hence, $\gamma_{x_n} \to \gamma_{x_0}$ in the $C^3$-norm, proving the continuity of $G$. $\square$

## Appendix B. Proof of the Closedness of the set $S$ and $R$.

*Proof.* Let $(\gamma_k) \subset R$ be a sequence converging to some $\gamma \in \mathcal{K}_{\text{proj}}^n$ in the $C^3$-topology. For each $k$, there exists a constant $c_k \in [0, 1]$ such that $\|\dot{\gamma}_k(t)\| = c_k$ for all $t \in [0, 1]$. Since $(c_k) \subset [0, 1]$ is bounded, we can extract a subsequence that we still denote $(c_k)$ such that $c_k \to c \in [0, 1]$. Then for all $t \in [0, 1]$, we have $\|\dot{\gamma}(t)\| = \lim_{k \to \infty} \|\dot{\gamma}_k(t)\| = \lim_{k \to \infty} c_k = c$, which shows that $\gamma \in R$. Hence, $R$ is closed. $\square$
Proof. Let $(\gamma_k)_{k \ge 1} \subset S$ be a sequence converging to some $\gamma \in K_{\text{proj}}^2$ in the $C^3$-topology.

**Case 1:** Infinitely many $\gamma_k$ are constant. Assume there exists a subsequence $(\gamma_{k_j})_{j \ge 1}$ such that $\|\dot{\gamma}_{k_j}(t)\| = 0$ for all $j$. Since $\gamma_{k_j} \to \gamma$ in $C^3$, we have $\dot{\gamma}_{k_j} \to \dot{\gamma}$ uniformly on $[0, 1]$. Thus $\dot{\gamma} = 0$, meaning $\gamma$ is constant, hence $\gamma \in N$.

**Case 2:** Infinitely many $\gamma_k$ have positive constant speed. In this case, we can relabel the sequence (if necessary) so that there exists $k_0 \in \mathbb{N}$ such that

$$ \|\dot{\gamma}_k(t)\| = c_k > 0 \quad \text{for all } k \ge k_0. $$

By uniform convergence of $\dot{\gamma}_k$ to $\dot{\gamma}$, we have

$$ \|\dot{\gamma}(t)\| = \lim_{k \to \infty} \|\dot{\gamma}_k(t)\| = \lim_{k \to \infty} c_k =: c \ge 0. $$

(a) If $c = 0$: Then $\dot{\gamma} = 0$, so $\gamma$ is constant and thus $\gamma \in N$.

(b) If $c > 0$: The curve $\gamma$ is regular with constant speed $c > 0$. Its Curvature is given by:

$$ \kappa_{\gamma}(t) := \frac{|\dot{\gamma}(t) \wedge \ddot{\gamma}(t)|}{\|\dot{\gamma}(t)\|^3} = \frac{|\dot{\gamma}(t) \wedge \ddot{\gamma}(t)|}{c^3}, \quad t \in [0, 1]. $$

Similarly, for each $k \ge k_0$,

$$ \kappa_{\gamma_k}(t) = \frac{|\dot{\gamma}_k(t) \wedge \ddot{\gamma}_k(t)|}{c_k^3}. $$

Because $\gamma_k \to \gamma$ in $C^3$, both $\dot{\gamma}_k \to \dot{\gamma}$ and $\ddot{\gamma}_k \to \ddot{\gamma}$ uniformly on $[0, 1]$. Since there all all bounded, we get,

$$ \dot{\gamma}_k(t) \wedge \ddot{\gamma}_k(t) \to \dot{\gamma}(t) \wedge \ddot{\gamma}(t) \text{ uniformly on } [0, 1]. $$

Moreover, since $c_k \to c$ in $(0, 1]$ with $c_k, c > 0$, we obtain

$$ \kappa_{\gamma_k}(t) \to \kappa_{\gamma}(t) \quad \text{for all } t \in [0, 1]. $$

On the other hand, we also have

$$ \frac{2\pi}{c_k} \to \frac{2\pi}{c}. $$

Passing to the limit in the inequality

$$ \kappa_{\gamma_k}(t) \le \frac{2\pi}{c_k} \quad \text{for all } t \in [0, 1], $$

we obtain

$$ \kappa_{\gamma}(t) \le \frac{2\pi}{c} \quad \text{for all } t \in [0, 1]. \quad \square $$

Therefore, $\gamma \in S$ and thus $S$ is closed.

**Acknowledgments.** T.A.D. would like to thank Bastien Laville for pointing out an inconsistency in the previous version of the manuscript, which has been corrected in the present version.

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