2.1. Definitions, first properties

Let us now define self-similar fragmentation processes with values in $\mathcal{M}_{\infty}$. For this, let us introduce a family of self-similar fragmentation operators $(\textrm{ssFrag}_\alpha, \alpha\in\mathbb{R})$, defined as follows. For $n\in\mathbb{N}\cup\{\infty\}$, consider a marked partition $x=(\pi,\textbf{v})\in\mathcal{M}_{n}$ and a sequence $\bar{x}^{(\cdot)}$ of càdlàg maps ${\bar{x}^{(k)}\colon[0,\infty)\to\mathcal{M}_{n}}$, satisfying $\bar{x}^{(k)}(0)=(\textbf{1},1)$. Writing $\bar{x}^{(k)}(t)=(\bar\pi^{(k)}(t),\bar {\textbf{v}}^{(k)}(t))$ for all $k\geq 1$ and $t\geq 0$, we define

\begin{equation*}\textrm{ssFrag}_\alpha(x,\bar{x}^{(\cdot)})\,{:\!=}\,(\hat{\pi},\hat{\textbf{v}})\end{equation*}

as the map $[0,\infty)\to\mathcal{M}_{n}$ such that

\begin{gather*}\hat{v}_i(t) = v_i \bar{v}^{(k_i)}(v_i^{\alpha} t), \\i\sim^{\hat{\pi}(t)}j \iff i\sim^{\pi}j \;\text{ and }\;i\sim j\text{ in }\bar{\pi}^{(k_i)}(v_i^{\alpha}t),\end{gather*}

where $k_i$ is defined as the label of the block of $\pi$ containing i (i.e. such that i is in the $k_i$th block of $\pi$). Note that we use the convention $0^\alpha = 0$ for all $\alpha\in\mathbb{R}$, so that thanks to this definition, if a block B of x has mark 0, then the process $\textrm{ssFrag}_\alpha(x,\bar{x}^{(\cdot)})$ is frozen at block B, in the sense that for all $t\geq 0$, B is a block of $\hat{\pi}(t)$, and every $j\in B$ will have $\hat{v}_j(t)=0$. Also, the assumptions on the maps $\bar{x}^{(k)}$ imply that $\textrm{ssFrag}_\alpha(x,\bar{x}^{(\cdot)})$ is càdlàg and satisfies $\textrm{ssFrag}_\alpha(x,\bar{x}^{(\cdot)})(0)=x$.

We can now define the following generalization of self-similar fragmentations.

An ESSF process X will be called non-degenerate if for all $x\in\mathcal{M}^{\star}_{\infty}$, the process has sample paths in $\mathcal{M}^{\star}_{\infty}$, $\mathbb{P}_{x}$-a.s.

As a first result about ESSF processes, let us show a projective Markov property. It is very analogous to [Reference Blancas, Duchamps, Lambert and Siri-Jégousse9, Lemma 3.2] and [Reference Duchamps13, Proposition 2], but we need another statement in the present context.

Lemma 1. Let X be an ESSF process. Then for any $n\in\mathbb{N}$, the process $(X(t)_{|[n]},t\geq 0)$ is Markovian in $\mathcal{M}_{n}$. More precisely, there exists a measurable transition kernel $(p^n_t,t\geq 0)$ on $\mathcal{M}_{n}$ such that for any initial state $x\in\mathcal{M}_{\infty}$,

\begin{equation*} \mathbb{P}_{x}(X(t)_{|[n]}\in \cdot) = p^{n}_t(x_{|[n]},\cdot\,). \end{equation*}

Proof. To show that $p^n$ is well-defined by the formula above, we need to prove that

\begin{equation*} \mathbb{P}_{x}\big(X(t)_{|[n]}\in \cdot\big) = \mathbb{P}_{x'}\big(X(t)_{|[n]}\in \cdot\big) \end{equation*}

for any two initial states $x,x'\in\mathcal{M}_{\infty}$ such that ${x'}\!\!_{|[n]}=x_{|[n]}$.

Consider a probability space $(\Omega,\mathbb{P})$ such that $X^{(\cdot)}$ is a sequence of i.i.d. copies of the ESSF process started from $(\textbf{1},1)$, and let $\alpha\in\mathbb{R}$ be the self-similarity index of X. By the branching property, we have

\begin{align*} & \mathbb{P}_{x}\big(X(t)_{|[n]}\in \cdot\big) = \mathbb{P}\big( \!\textrm{ssFrag}_\alpha(x,X^{(\cdot)})(t) _{|[n]}\in \cdot\big)\\[3pt] \text{and } & \mathbb{P}_{x'}\big(X(t)_{|[n]}\in \cdot\big) = \mathbb{P}\big( \!\textrm{ssFrag}_\alpha(x',X^{(\cdot)})(t) _{|[n]}\in \cdot\big). \end{align*}

It remains to notice that by definition, $\textrm{ssFrag}_\alpha(x,X^{(\cdot)})(t) _{|[n]}$ is in fact a functional which depends only on $x_{|[n]}$ and $X^{(\cdot)}$. Therefore, because ${x'}\!\!_{|[n]}=x_{|[n]}$, we have

\begin{equation*} \textrm{ssFrag}_\alpha(x,X^{(\cdot)})(t) _{|[n]} = \textrm{ssFrag}_\alpha(x',X^{(\cdot)})(t) _{|[n]} \end{equation*}

everywhere on $\Omega$, which implies by the preceding display that

\begin{equation*} \mathbb{P}_{x}\big(X(t)_{|[n]}\in \cdot\big) = \mathbb{P}_{x'}\big(X(t)_{|[n]}\in \cdot\big). \end{equation*}

This shows that $p^n$ is well-defined. It remains to note that this is indeed a measurable transition kernel, i.e. to show that for any Borel set $A\subset\mathcal{M}_{n}$, the map $(x,t)\in\mathcal{M}_{n}\times[0,\infty)\mapsto p^n_t(x,A)$ is measurable. This is the case because the map $(x,t)\in\mathcal{M}_{\infty}\times[0,\infty)\mapsto \mathbb{P}_{x}(X(t)\in A)$ is measurable by definition of X as a strong Markov process, and we can write

\begin{equation*} p^n_t(x,A) = \mathbb{P}_{\varphi_n(x)}(X(t)\in A) \end{equation*}

for any well-chosen measurable map $\varphi_n\,{:}\,\mathcal{M}_{n}\to\mathcal{M}_{\infty}$ such that $\varphi_n(x)_{|[n]}=x$, for instance such that $\varphi_n(x)$ is the unique such marked partition with a unique infinite block $\{n+1,n+2,\ldots\}$ with mark 0.

The previous lemma shows that given an ESSF process X, one can define its law started from any $x_0\in\mathcal{M}_{n}$, for any $n\in \mathbb{N}$, as the law of the restriction $X_{|[n]}$ of the initial process started from any $x\in\mathcal{M}_{\infty}$ such that $x_{|[n]}=x_0$.

As a result, the restriction $X_{|[1]}$ of an ESSF process $X=(\Pi,\textbf{V})$ to $\mathcal{M}_{1}=\mathcal{P}_{1}\times S^1$ is a Markov process. Since the space $\mathcal{P}_{1}$ is a singleton, the lemma implies that the real-valued process $V_1=(V_1(t),t\geq 0)$ is a Markov process in $S^1$, and note that by exchangeability, the process $V_i$ has the same marginal distribution for all $i\geq 1$. Further, Definition 2 implies that it is an a.s. càdlàg strong Markov process satisfying a self-similar property; more precisely, for $v\geq 0$ let $P_v$ denote the distribution of $V_1$ started at v on the Skorokhod space of càdlàg maps $[0,\infty)\to S^1$, and let V denote the canonical process on that space. Then

\begin{equation*}(V(t),\,t\geq 0) \text{ under } P_v \overset{(d)}{=} (vV(v^\alpha t),\,t\geq 0) \text{ under }P_1,\end{equation*}

where $\alpha$ is the self-similarity index of X. In other words, V is a positive self-similar Markov process (pssMp). Note that in the literature, the index of self-similarity of a pssMp refers in general to $-\alpha$ [Reference Pardo and Rivero21] or $-1/\alpha$ when $\alpha\neq 0$, e.g. in [Reference Lamperti20], where Lamperti calls this the order of the process rather than the index. Here we use the convention found in the self-similar fragmentation literature, e.g. [Reference Berestycki2, Reference Bertoin3, Reference Krell, Donati-Martin, émery, Rouault, Stricker and Springer19]. Let us summarize in a proposition some properties of V that can be deduced from the well-developed theory of self-similar Markov processes. First, if $X=(\Pi,\textbf{V})$ is an ESSF process, for each $i\geq 1$ define $\zeta_i \,{:\!=}\, \inf\{t\geq 0,\, V_i(t)=0\}$, and for $t\in[0,\zeta_i]$,

\begin{equation*}\varphi_i(t) \,{:\!=}\, \int_{0}^{t}V_i(t)^{\alpha}\,\textrm{d} s.\end{equation*}

Note that $\varphi_i$ is continuous and increasing. We define its right-continuous inverse $\tau_i(t)$, for $t\in[0,\infty)$, by

\begin{equation*}\tau_i(t) \,{:\!=}\, \begin{cases}\varphi_i^{-1}(t) &\text{if }t < \varphi_i(\zeta_i),\\ \\[-7pt] \infty &\text{if }t\geq \varphi_i(\zeta_i).\end{cases}\end{equation*}

We need a convention for infinite times, so we let $V_i(\infty)\equiv 0$, so that $V_i(\tau_i(t))$ is always defined. Also, note that the definition of the Frag operator implies that a.s., $\Pi$ has nonincreasing sample paths for the finer-than partial order—$\pi$ is finer than $\pi'$ if the blocks of $\pi'$ can be written as unions of blocks of $\pi$. Since, a.s. for all $n\geq 1$, $\Pi(t)_{|[n]}$ is nonincreasing in a finite set, it is eventually constant. This implies that $\Pi(t)$ converges a.s. when $t\to\infty$, and we may denote its limit by $\Pi(\infty)$. Let us now state the proposition.

Proposition 2. Let $X=(\Pi,\textbf{V})$ be an $\alpha$-ESSF process and $i\geq 1$, and define

\begin{align*} \zeta_i &\,{:\!=}\, \inf\{t\geq 0,\, V_i(t)=0\},\\[3pt] \varphi_i(t) &\,{:\!=}\, \int_{0}^{t}V_i(t)^{\alpha}\,\textrm{d} s, \qquad t\in [0,\zeta_i], \\[3pt] \tau_i(t) &\,{:\!=}\, \varphi_i^{-1}(t),\qquad t\geq 0. \end{align*}

Then the following properties hold:

• Either $\zeta_i<\infty$ $\mathbb{P}_{(\textbf{1},1)}$-a.s., or $\zeta_i=\infty$ $\mathbb{P}_{(\textbf{1},1)}$-a.s.

• Either $\varphi_i(\zeta_i)<\infty$ $\mathbb{P}_{(\textbf{1},1)}$-a.s., or $\varphi_i(\zeta_i)=\infty$ $\mathbb{P}_{(\textbf{1},1)}$-a.s.

• In the case $\zeta_i < \infty$, either $V_i$ reaches 0 continuously $\mathbb{P}_{(\textbf{1},1)}$-a.s., or $V_i$ eventually jumps to 0 $\mathbb{P}_{(\textbf{1},1)}$-a.s.

• $\varphi_i(\zeta_i) = \infty$ iff $V_i$ reaches 0 continuously.

• The process $\xi_i \,{:\!=}\, \log(V_i\circ\tau_i)$, i.e. defined by

  \begin{equation*} \xi_i(t) = \log V_i(\tau_i(t)), \qquad 0\leq t < \varphi_i(\zeta_i), \end{equation*}
  is a (killed in the case $\varphi_i(\zeta_i) < \infty$) Lévy process called the inverse Lamperti transform of $V_i$.

Proof. These are classical results on pssMp; we refer to [Reference Lamperti20] for proofs.

This proposition tells us that it is natural to consider the time-changed processes $V_i\circ \tau_i$ for $n\geq 1$, which behave as exponentials of Lévy processes. However, there is no unique time-change that could make the whole process X behave nicely. Instead, we have to rely on stopping lines, which are tools generalizing stopping times in the context of branching Markov processes (see e.g. [Reference Chauvin10] for their use in branching Brownian motion, or [Reference Bertoin3, Reference Bertoin5] in the context of fragmentations).

2.2. Stopping lines, changing the index of self-similarity

First let us define some filtrations associated with an ESSF process $X=(\Pi,\textbf{V})$. To this end, let us endow the power set $2^{\mathbb{N}}\,{:\!=}\,\{A\subset\mathbb{N}\}$ with the topology generated by the metric $d(A,B) \,{:\!=}\, \big(\!\sup\{n\in\mathbb{N}, A\cap[n] =B\cap[n]\}\big)^{-1}$, which makes $2^{\mathbb{N}}$ a compact space. Now for $i\in \mathbb{N}$, let us define the block process $(B_i(t),t\geq 0)$ as the $2^{\mathbb{N}}$-valued càdlàg process such that for all $t\geq 0$, $B_i(t)$ is the block of $\Pi(t)$ containing i; that is,

\begin{equation*}B_i(t) = \{j\in\mathbb{N}, \,i\sim^{\Pi(t)}j \}.\end{equation*}

Now we can define a sequence of natural filtrations associated with X by

\begin{equation*}\mathcal{G}_i=(\mathcal{G}_i(t),\,t\geq 0) \quad \text{ with }\mathcal{G}_i(t) = \sigma\big(B_i(s),V_i(s),\; s\in[0,t]\big), \qquad i\geq 1, t\geq 0.\end{equation*}

Since (ii) entails that $i\sim^{\Pi(L_i)} j$ is an equivalence relation, its equivalence classes form a well-defined partition of $\mathbb{N}$, which we denote by $\Pi(L)$ with a slight abuse of notation. Also, with $\textbf{V}(L)$ denoting the vector $(V_i(L_i),i\geq 1)$, it is clear that $X(L)\,{:\!=}\,(\Pi(L),\textbf{V}(L))$ is a well-defined (random) element of $\mathcal{M}_{\infty}$.

Remark 3. A fixed time $t\geq 0$ can be seen as a stopping line (an L for which $L_i\equiv t$ for all $i\geq 1$), and it is easily checked that for a stopping line L, one can define $L+t$ and $L\wedge t$ by

\begin{equation*} (L+t)_i = L_i+t \qquad \text{ and }(L\wedge t)_i = L_i \wedge t, \end{equation*}

which are again stopping lines. Thus for a stopping line L we will be able to consider the processes $X(L+\cdot) \,{:\!=}\, (X(L+t),\,t\geq 0)$ and $X(L\wedge\cdot) \,{:\!=}\, (X(L\wedge t),\,t\geq 0)$. Since it will be useful, we define the following $\sigma$-algebra:

\begin{equation*} \mathcal{G}_L \,{:\!=}\, \sigma\big(X(L\wedge t),\,t\geq 0\big). \end{equation*}

We can now state the Markov property for stopping lines, which is analogous to what can be found in [5, Lemma 3.14].

Proposition 3. (Stopping line Markov property) Let X be an $\alpha$-ESSF, and L a stopping line. Then conditional on $\mathcal{G}_L$, the following equality in distribution holds:

(3) \begin{equation} X(L+\cdot) \overset{(d)}{=} \textrm{ssFrag}_\alpha\!\big(X(L),X^{(\cdot)}\big), \end{equation}

where $X^{(\cdot)}$ is an independent, i.i.d. sequence of copies of the process started from $(\textbf{1},1)$.

Proof. See Appendix A.2.

The next step in the analysis of ESSF processes is to bring the index of self-similarity to 0. This will be done via the random time changes $(\tau_i(t), i\geq 1, t\geq 0)$ defined above by

\begin{equation*}\tau_i(t) \,{:\!=}\, \varphi_i^{-1}(t), \quad \text{ where } \quad \varphi_i(u) \,{:\!=}\, \int_{0}^{u}V_i(s)^{\alpha} \,\textrm{d} s, \quad u\geq 0.\end{equation*}

These time changes enable us to turn an $\alpha$-ESSF into a homogeneous ESSF. The following proposition makes this claim more precise.

Proposition 4. Let $X=(\Pi,\textbf{V})$ be an $\alpha$-ESSF, with $\alpha\in\mathbb{R}$, and let $\beta\in\mathbb{R}$. For $i\geq 1$, define the random map $\tau^{\beta}_i$ as the right-continuous inverse of the map $u\mapsto \int_{0}^{u}V_i(s)^{\beta} \,\textrm{d} s$; that is,

\begin{equation*} \forall t\geq 0,\qquad\tau^{\beta}_i(t) = \left (\int_{0}^{\cdot}V_i(s)^{\beta} \,\textrm{d} s\right )^{-1}(t) \;=\;\inf \bigg\{u\geq 0,\,\int_{0}^{u}V_i(s)^{\beta} \,\textrm{d} s=t\bigg\}, \end{equation*}

with the convention $\inf \varnothing=\infty$. Then for each $t\geq 0$, $\tau^{\beta}(t)$ is a stopping line, and the process $X\circ\tau^{\beta} \,{:\!=}\, (X(\tau^{\beta}(t)),\,t\geq 0)$ is an $(\alpha-\beta)$-ESSF. Furthermore, if X is non-degenerate, then $X\circ\tau^{\beta}$ is also non-degenerate.

Proof. See Appendix A.3.

By bringing the index of self-similarity to 0 we can transform any ESSF into a homogeneous process. It follows from the definition that if $X=(\Pi, \textbf{V})$ is a 0-ESSF, then $\Pi$ is simply a classical homogeneous fragmentation process (possibly stopped at a random line if the marks can jump to 0), and the process of marks $\textbf{V}$ can be seen as a secondary process, evolving jointly and branching simultaneously with $\Pi$. As in the classical theory, it is necessary to study the homogeneous case to describe and give results on the general self-similar case. We study these 0-ESSF processes further in Section 3, but first let us point out how self-similar growth-fragmentation processes can be understood as special cases of ESSF processes.

Link with self-similar growth-fragmentation processes.

Consider an ESSF $X=(\Pi,\textbf{V})$ and define the associated point-measure valued process $\tilde{\textbf{V}}$ by

\begin{equation*}\tilde{\textbf{V}}(t) \,{:\!=}\, \sum_{k\geq 1}\delta_{\tilde{V}_k(t)},\end{equation*}

where $\tilde{V}_k(t)$ denotes the mark associated with the kth block of X(t). Assume that the fragmentation events are binary and conservative in the sense that for all $t\geq 0$ such that $\Pi(t-) \neq \Pi(t)$,

• there is a unique block $B_k$ of $\Pi(t-)$ such that $\Pi(t)_{|B_k} \neq \Pi(t-)_{|B_k} = \textbf{1}_{B_k}$, and this block can be written as the union of two blocks of $\Pi(t)$, say $B_{k_1}$ and $B_{k_2}$;

• $\tilde{V}_k(t-)=\tilde{V}_{k_1}(t)+\tilde{V}_{k_2}(t)$ holds, where $k,k_1,k_2$ are as above.

Then it follows from the definition of ESSF processes that $\tilde{\textbf{V}}$ is a self-similar growth-fragmentation process as introduced by Bertoin [Reference Bertoin7]. In fact, any self-similar growth-fragmentation process can be built as the process of marks of a well-chosen ESSF. We will make this claim precise after giving a characterization of ESSF processes, in Section 3.3.

3.1. Decomposition of ESSF processes

Let us consider here a homogeneous 0-ESSF process $X=(\Pi,\textbf{V})$, started from $(\textbf{1},1)$. We know by Lemma 1 that it satisfies a projective Markov property, i.e. for all $n\in\mathbb{N}$, $X_{|[n]}$ defines a Markov process with values in $\mathcal{P}_{n}$. Let $n\in\mathbb{N}$ be fixed, and define the stopping time

\begin{equation*}T_n \,{:\!=}\, \inf\{t\geq 0,\, \Pi(t)_{|[n]} \neq \textbf{1}_n \text{ or } V_1(t) = 0\},\end{equation*}

as well as the killed process

\begin{equation*}\tilde{\xi}_n \,{:\!=}\, (\log V_1(t), \,0\leq t < T_n).\end{equation*}

Note that homogeneity implies that the pair $(\tilde{\xi}_n-\log v,T_n)$ has the same distribution under every $\mathbb{P}_{(\textbf{1},v)}$ for all $v\in S^1\setminus\{0\}$. Therefore for $t\geq 0$, conditional on $\{T_n>t\}$, the Markov property applied at time t shows that $(\tilde{\xi}_n(t+\cdot)-\tilde{\xi}_n(t),T_n-t)$ has the same distribution as $(\tilde{\xi}_n,T_n)$ under $\mathbb{P}_{(\textbf{1},1)}$. This shows that the killed process $\tilde{\xi}_n$ is distributed as

\begin{equation*}\tilde{\xi}_n\overset{(d)}{=}(\xi_n(t),\,0\leq t < T'_n),\end{equation*}

where $\xi_n$ is a Lévy process and $T'_n$ is an independent exponential random variable. Up to enriching our probability space, whenever we need to, we will assume that the whole (non-killed) Lévy process $\xi_n=(\xi_n(t),t\geq 0)$ is defined in our space and that the above equality in distribution holds a.s., that is,

\begin{equation*} \forall n\geq 1, \qquad \tilde{\xi}_n = (\log V_1(t),\,0\leq t < T_n) = (\xi_n(t),\,0\leq t < T_n) \qquad \text{a.s.} \end{equation*}

Note that this implies that if $T_n<\infty$, then $V_1(T_n-) = \exp(\xi_n(T_n))> 0$. Now for $n\in \mathbb{N}$ such that $T_n<\infty$ a.s., consider $D_n$, the dislocation (or freezing) at time $T_n$, defined by

\begin{equation*}D_n \,{:\!=}\, (\Pi(T_n),\textbf{V}(T_n)/V_1(T_n-))_{|[n]} \in\mathcal{M}_{n},\end{equation*}

where the division $\textbf{V}(T_n)/V_1(T_n-)$ is to be understood coordinatewise. Equivalently, $D_n$ is the unique random marked partition such that

\begin{equation*}X(T_n)_{|[n]} = \textrm{Frag}\big(X(T_n-)_{|[n]},D_n\big),\end{equation*}

with a slight abuse of notation in this case since $X(T_n-)_{|[n]}$ has only one block ($D_n$ is not a sequence, but additional terms are useless to define a fragmentation of a single block).

Note that this implies that $D_n$ has the same distribution under every $\mathbb{P}_{(\textbf{1},v)}$ for all $v\in S^1\setminus\{0\}$. Thus, for any bounded measurable maps $g\,{:}\,\mathbb{R}\to\mathbb{R}$, $h\,{:}\,\mathcal{M}_{n}\to\mathbb{R}$ and $t\geq 0$, applying the Markov property at time $t\geq 0$, one gets

\begin{equation*}\mathbb{E}_{(\textbf{1},1)}[g(\xi_n(t))\unicode{x1D7D9}_{T_n>t} h(D_n)] = \mathbb{E}_{(\textbf{1},1)}[g(\xi_n(t))\unicode{x1D7D9}_{T_n>t}] \, \mathbb{E}_{(\textbf{1},1)}[h(D_n)],\end{equation*}

which shows that the killed Lévy process $(\xi_n,T_n)$ and the marked partition $D_n$ are independent. Let us define $\mathcal{D}_n$ as the law of $D_n$, and notice also that exchangeability of $X_{|[n]}$ implies that $\mathcal{D}_n$ is an exchangeable probability measure on $\mathcal{M}_{n}$.

Since $(\xi_n,T_n)$ is a killed Lévy process, one can define uniquely $d_n\in\mathbb{R}$, $\beta_n\geq 0$, $J_n\geq 0$, and $\lambda_n$ a measure on $\mathbb{R}\setminus\{0\}$ satisfying $\int 1\wedge y^2 \, \lambda_n(\textrm{d} y) < \infty$, such that

• the process $\xi_n$ is a Lévy process with characteristic exponent

  \begin{equation*} \psi_n(\theta) \,{:\!=}\, \log\mathbb{E}[\textrm{e}^{i\theta\xi_n(1)}] = id_n\theta -\frac{\beta_n}{2}\theta^2 + \int_{\mathbb{R}}\left (\textrm{e}^{i\theta y} - 1 - i\theta y \unicode{x1D7D9}_{\lvert {y} \rvert\leq 1}\right )\,\lambda_n(\textrm{d} y); \end{equation*}

• $\xi_n$ is killed at rate $J_n = 1/\mathbb{E}[T_n]$, which may be 0 if $T_n = \infty$ a.s.

Remark 4. Note that knowing $(\psi_n, J_n, \mathcal{D}_n)$ for $n\in\mathbb{N}$ is enough to reconstruct the process X. Indeed, starting from $(\textbf{1},v)$, the process $X_{|[n]}$ up to time $T_n$ has distribution equal to that of

\begin{equation*} Y_n \,{:\!=}\, \big((\textbf{1}_n, v\textrm{e}^{\xi_n(t)}),\, 0\leq t < T_n\big), \end{equation*}

and at time $T_n$ jumps to $(\Pi, v\textrm{e}^{\xi_n(T_n-)}\textbf{V})$, where $(\Pi,\textbf{V})$ is independently drawn according to $\mathcal{D}_n$.

By the branching property, one only needs to iterate this construction at each jump time, independently for each marked block, to get the whole process $X_{|[n]}$. By Kolmogorov’s extension theorem—since $(X_{|[m]})_{|[n]}=X_{|[n]}$ for each $n\leq m$—these distributions characterize the distribution of X.

Let us now state our main result, which identifies the form that these characteristics can take.

Theorem 1. Let X be a non-degenerate 0-ESSF, and for each n, write $(\psi_n, J_n, \mathcal{D}_n)$ for the characteristics describing the law of $X_{|[n]}$. Then there is a unique quadruple $(c,d,\beta,\Lambda)$, where $c,\beta\geq 0$, $d\in\mathbb{R}$, and $\Lambda$ is a measure on $\mathscr{Z}^\downarrow\setminus\{(\textbf{1},1)\}$, which necessarily satisfies

(4) \begin{equation} \int_{\mathscr{Z}^\downarrow}\big(1-s_1\unicode{x1D7D9}_{v_1>0} + (\log v_1)^2 \wedge 1\big)\,\Lambda(\textrm{d} \textbf{z}) < \infty, \end{equation}

such that for all $n\in\mathbb{N}$,

1. (i) $\displaystyle\psi_n(\theta) = id\theta -\frac{\beta}{2}\theta^2 + \int_{\mathscr{Z}^\downarrow}\bigg(\sum_{\substack{j\geq 1\\ v_j>0}}(s_j)^n\big(\textrm{e}^{i\theta\log v_j} - 1\big) - i\theta\log v_1 \unicode{x1D7D9}_{\lvert {\log v_1} \rvert\leq 1}\bigg)\,\Lambda(\textrm{d} \textbf{z});$

2. (ii) $J_n = nc + \displaystyle \int_{\mathscr{Z}^\downarrow}\Big (1-\sum_{\substack{i\geq 1\\ v_i>0}}(s_i)^n\Big )\,\Lambda(\textrm{d} \textbf{z})$;

3. (iii) if $J_n>0$, then

   \begin{equation*}\mathcal{D}_n=\frac{1}{J_n}\Big(\sum_{i=1}^{n}c\delta_{\mathfrak{e}^{n}_i} + \int_{\mathscr{Z}^{\downarrow}}\varrho^{n}_{\textbf{z}}\big(\cdot\cap\{\pi\neq\textbf{1}_n\text{ or }(\pi,\textbf{v})=(\textbf{1}_n,0)\}\big)\,\Lambda(\textrm{d}\textbf{z})\Big),\end{equation*}
   where $\varrho^{n}_{\textbf{z}}$ is the paintbox process defined in Section 1.2, and $\delta_{\mathfrak{e}^{n}_i}$ denotes the Dirac point measure on $\mathfrak{e}^{n}_i$, the marked partition defined as
   \begin{equation*} \mathfrak{e}^{n}_i \,{:\!=}\, \Big(\big\{[n]\setminus\{i\},\{i\}\big\},\, (1,\ldots,1,\underbrace{0}_{i-\text{th index}},1,\ldots,1) \Big). \end{equation*}

Conversely if $c,\beta \geq 0$, $d\in\mathbb{R}$, and $\Lambda$ is a measure on $\mathscr{Z}^\downarrow\setminus\{(\textbf{1},1)\}$ satisfying (4), then there exists a 0-ESSF with characteristics as above.

Proof. See Appendix A.4.

Combining Theorem 1 with Proposition 4, we get the following characterization of all ESSF processes.

Let us point out that the condition (4) is surprisingly nonrestrictive. There are no integrability assumptions concerning the marks of the smallest blocks (with labels greater than 1). Consequently, the point measure

\begin{equation*}\sum_k\delta_{\tilde{V}_k(t)},\end{equation*}

where $\tilde{V}_k(t)$ denotes the mark of the kth block in X(t), might assign infinite mass to any interval $(a,b)\subset [0,\infty)$ for any $t>0$. Indeed it suffices for instance that $\Lambda(\textrm{d} \textbf{z})$ be of the form

\begin{equation*}\int_{\mathscr{Z}^\downarrow}\prod_{i\geq 1}F_i(s_i,v_i)\,\Lambda(\textrm{d} \textbf{z}) = \int_{(0,1)\times S^{1}} F_1(s_1,v_1)\,\mathbb{E}\bigg[\prod_{i\geq 2}F_i\Big(\frac{1-s_1}{2^{i-1}},Z_i\Big)\bigg]\,\nu(\textrm{d} z_1),\end{equation*}

where $\nu$ is a measure on $(0,1)\times S^{1}$ with infinite mass and satisfying

\begin{equation*}\int_{(0,1)\times S^{1}}\big(1-s_1\unicode{x1D7D9}_{v_1>0} + (\log v_1)^2 \wedge 1\big)\,\nu(\textrm{d} z_1) < \infty,\end{equation*}

and $Z_2,Z_3,\ldots$ are i.i.d. Exp(1) random variables. On the other hand, if one assumes an integrability condition such as

\begin{equation*}\int_{\mathscr{Z}^\downarrow}\big(\sum_{i\geq 1} v_i^{\theta}\big) - 1 -\theta\log v_1 \unicode{x1D7D9}_{\lvert {\log v_1} \rvert\leq 1}\,\Lambda(\textrm{d} \textbf{z}) < \infty\end{equation*}

for some $\theta\in\mathbb{R}$, then one observes a process of point measures

\begin{equation*}(\sum_k\delta_{\tilde{V}_k(t)},t\geq 0)\end{equation*}

that is nice in the sense that for all $t\geq 0$, $\mathbb{E}\sum_k\tilde{V}_k(t)^{\theta}<\infty$. This is the object of the next section.

3.2. Absorption in finite time

Consider here a non-degenerate $\alpha$-ESSF with characteristics $(c,d,\beta,\Lambda)$, started from $(\textbf{1},1)$. We are interested in the case where the pssMp $V_1$ reaches 0 in finite time, and if $T_i$ denotes the hitting time of 0 by the process $V_i$, we aim at giving a sufficient condition for which

\begin{equation*}\sup_{i\in\mathbb{N}} T_i < \infty \qquad \text{a.s.}\end{equation*}

If this holds, then at this random time the process is frozen in a state $X(\infty)=(\Pi(\infty),\textbf{V}(\infty))$ where $\textbf{V}(\infty)=\textbf{0}=(0,0,\ldots)$, and we say the process X is absorbed in finite time. We first put aside a trivial case and assume that

\begin{equation*}c>0 \quad \text{ or }\quad \int_{\mathscr{Z}^\downarrow}(1-s_1)\,\Lambda(\textrm{d} \textbf{z}) > 0,\end{equation*}

since otherwise $\Pi$ would be a.s. constant, equal to the coarsest partition $\{\mathbb{N}\}$. Recall that a classical self-similar fragmentation which is absorbed in finite time (this is always true when $\alpha < 0$ [4, Proposition 2]) is always totally fragmented in the limit, in the sense that $\Pi(\infty)$ is the partition of $\mathbb{N}$ into singletons. Clearly in our case, because of the possible freezing of blocks at dislocation events, $\Pi(\infty)$ is a.s. totally fragmented if and only if

\begin{equation*}\forall i\geq 1, \,s_i>0\implies v_i>0 \qquad \Lambda\text{-almost everywhere ($\Lambda$-a.e.)\ on }\mathscr{Z}^\downarrow.\end{equation*}

A stronger property than absorption in finite time is the following: we say X has finite total length if

\begin{equation*} \int_{0}^{\infty} \# X(t) \, \textrm{d} t <\infty \quad \text{a.s.},\end{equation*}

where $\#x$ denotes the number of blocks with positive mark in the marked partition x. One can interpret this quantity as the total length of the tree describing the genealogy of blocks in the fragmentation, hence the name. Note that this implies that for a fixed time $t\geq 0$, $\# X(t)$ is a.s. finite, which is well known [4, Proposition 2] in the classical self-similar fragmentation case for $\alpha < -1$.

In this section our aim is to provide sufficient conditions for ESSF processes to be absorbed in finite time and to have finite total length. The following result extends that of the classical setting and makes use of natural martingales—the so-called additive martingales—appearing in the homogeneous case. In order to be able to state it, we need a couple of additional definitions. For a marked partition $x=(\pi,\textbf{v})\in\mathcal{M}_{n}$ with $n\in\mathbb{N}\cup\{\infty\}$, and $\theta\in\mathbb{R}$, let us write

(5) \begin{equation} S_\theta(x) \,{:\!=}\, \sum_{k\in\mathbb{N}}\tilde{v}_k^{\theta},\end{equation}

where $\tilde{v}_k$ denotes the mark associated with the kth block of x. Let us also introduce $\kappa\,{:}\, \mathbb{R}\to(-\infty,\infty]$, defined by

(6) \begin{equation} \kappa(\theta) \,{:\!=}\, d\theta + \frac{\beta}{2}\theta^2 + \int_{\mathscr{Z}^\downarrow}\Big(\big(\sum_{i\geq 1} v_i^{\theta}\big) -1 -\theta \log v_1 \unicode{x1D7D9}_{\lvert {\log v_1} \rvert \leq 1} \Big)\,\Lambda(\textrm{d} \textbf{z}).\end{equation}

Note that the integral in the last display is well-defined with values in $(-\infty,\infty]$, since

\begin{align*}1 +\theta \log v_1 \unicode{x1D7D9}_{\lvert {\log v_1} \rvert \leq 1} - \sum_{i\geq 1} v_i^{\theta} &\leq \big(1+\theta\log v_1\unicode{x1D7D9}_{\lvert {\log v_1} \rvert \leq 1}-v_1^{\theta} \big)_{+}\\&\leq C \big((\log v_1^2)\wedge 1\big),\end{align*}

where C is a positive constant which depends on $\theta$, so the negative part of the integrand in the definition of $\kappa$ is $\Lambda$-integrable.

Proposition 5. Let X be a non-degenerate 0-ESSF with characteristics $(c,d,\beta,\Lambda)$ started from $(\textbf{1},1)$. For all $\theta\in\mathbb{R}$ and $t\geq 0$,

\begin{equation*} \mathbb{E}[S_\theta(X(t))] = \textrm{e}^{t\kappa(\theta)}, \end{equation*}

with $S_\theta$ and $\kappa(\theta)$ respectively defined as in (5) and (6). These quantities may be infinite, meaning that if $\kappa(\theta)=\infty$, then $\mathbb{E}[S_\theta(X(t))]=\infty$ for all $t>0$. If there is $\theta\in\mathbb{R}$ such that $\kappa(\theta) < \infty$, then the process

\begin{equation*} \big(\textrm{e}^{-t\kappa(\theta)}S_\theta(X(t)),\; t\geq 0\big) \end{equation*}

is a martingale. If there is $\theta\neq 0$ such that $\kappa(\theta) < 0$, then for any $\alpha\in\mathbb{R}$,

• if $-\alpha/\theta > 0$, the $\alpha$-ESSF with characteristics $(c,d,\beta,\Lambda)$ is absorbed in finite time;

• if $-\alpha/\theta \geq 1$, the $\alpha$-ESSF with characteristics $(c,d,\beta,\Lambda)$ has finite total length.

Proof. See Appendix A.5.

Remark 6. For a classical self-similar fragmentation with erosion coefficient $c\geq 0$ and dislocation measure $\nu$, we have

\begin{equation*} \kappa(\theta) = -c\theta + \int_{\mathscr{S}^\downarrow}\Big(\sum_{i\geq 1} s_i^{\theta} - 1 \Big)\, \nu(\textrm{d} \textbf{s}). \end{equation*}

Since $\sum_i s_i \leq 1$ $\nu$-a.e., for all $\theta > 1$ we have $\kappa(\theta) < 0$, so we recover absorption in finite time for any $\alpha < 0$ and finite total length for any $\alpha < -1$.

Then the point process recording the positions of the logarithm of marks

\begin{equation*} \sum_{k\in\mathbb{N}}\delta_{\log \tilde{V}_k(t)},\qquad t\geq 0, \end{equation*}

is a classical binary branching Brownian motion with drift d. One gets a cumulant function

\begin{equation*} \kappa(\theta) = d\theta +\frac{\theta^2}{2} + 1. \end{equation*}

This polynomial in $\theta$ takes negative values if and only if $d^2 - 2 > 0$, and we essentially recover the well-known fact that if $d>\sqrt{2}$, the lowest particle of a branching Brownian motion with drift d goes to $+\infty$.

3.3. Self-similar growth-fragmentation processes as ESSF

Growth-fragmentation processes as introduced by Bertoin [Reference Bertoin7] can be thought of as modeling cell systems, i.e. families of cells which grow (their sizes vary with time in a Markovian way) and reproduce independently of one another. To be more specific, the size of a single cell is a Markov process $Y=(Y(t),t\geq 0)$ with values in $[0,\infty)$ with càdlàg paths and no positive jumps, called the cell process. At each negative jump of the process (a time t such that $\lvert {\Delta Y(t)} \rvert = Y(t-)-Y(t) > 0$), a new cell is born in the system with size $\lvert {\Delta Y(t)} \rvert$. The growth-fragmentation process $\textbf{X} = (\textbf{X}(t), t \geq 0)$ is the point-measure-valued process recording the size of all cells alive at time t.

This process is said to be self-similar if it can be built from a self-similar cell process Y, say with index $\alpha$ and characteristics $(d, \beta, \lambda, k)$, with $k,\beta \geq 0$, $d\in \mathbb{R}$, and $\lambda$ a Lévy measure on $(-\infty, 0)$. This means that Y is the Lamperti transform of a Lévy process $\xi$ satisfying $\mathbb{E} \textrm{e}^{q\xi_1}=\exp(\varphi(q))$ for all $q > 0$, with

\begin{equation*}\varphi(q) = -k + \frac{\beta}{2} q^2 + dq + \int_{(-\infty, 0)}\!\!\big(\textrm{e}^{yq}-1-qy\unicode{x1D7D9}_{y>-1}\big)\,\lambda(\textrm{d} q).\end{equation*}

Note that different cell processes may yield the same growth-fragmentation (see Shi [Reference Shi23]), and in fact the distribution of a growth-fragmentation is characterized uniquely by the index $\alpha$ and the cumulant function

\begin{equation*}\kappa(q) \,{:\!=}\, \varphi(q)+\int_{(-\infty, 0)}\!\!(1-\textrm{e}^{y})^q\,\lambda(\textrm{d} q),\end{equation*}

which takes finite values for $q\geq 2$. Furthermore, this map $\kappa$ has the following interpretation: if $\alpha=0$, then $\mathbb{E} \sum_{x\in\textbf{X}(t)}x^q = \textrm{e}^{t\kappa(q)}$. A consequence of [Reference Shi23] is that $\kappa$ is unchanged (and therefore the growth-fragmentation is the same) if we replace $\lambda$ by $\lambda_{|[-\log 2, 0)} + \bar{\lambda}$ (and modify the drift term d accordingly), where $\bar{\lambda}$ is the measure on $[-\log 2, 0)$ defined as the push-forward of $\lambda_{|(-\infty, -\log 2)}$ by $y \mapsto \log(1-\textrm{e}^{y})$. Therefore we assume that $\lambda$ is a measure on $[-\log 2, 0)$, which intuitively means that the cell process Y always keeps track of the biggest cell in a division, i.e. that $Y(t)\geq Y(t-)/2 \geq \lvert {\Delta Y(t)} \rvert$ a.s. for all $t\geq 0$.

Let us now build a version of this growth-fragmentation process $\textbf{X}$ as the process of marks in an ESSF. Unfortunately there seems to be no natural candidate for this, since many ESSFs may have the same process of marks. Consider a map $Z\,{:}\,[-\log 2, 0)\to\mathscr{Z}^\downarrow$ and write $Z(y)=(s_i(y), v_i(y), i\geq 1)$. We assume that $v_1(y) = \textrm{e}^{y}$, $v_2(y) = 1-\textrm{e}^{y}$, $s_2(y) = 1-s_1(y)$, and $s_1$ is such that

\begin{equation*}\int_{[-\log 2, 0)} 1 - s_1(y)\,\lambda(\textrm{d} y)<\infty.\end{equation*}

For instance $s_1(y) = \textrm{e}^{-y^2}$ or $s_1(y)=\textrm{e}^{y}(1-y)$ both satisfy this condition, but of course there is no natural choice here. Now define the push-forward $\Lambda\,{:\!=}\,\lambda\circ Z^{-1} + k\delta_{(\textbf{1},0)}$, which satisfies by construction

\begin{equation*}\int_{\mathscr{Z}^\downarrow} \big(1-s_1\unicode{x1D7D9}_{v_1 > 0} +(\log v_1)^2 \wedge 1\big) \,\Lambda(\textrm{d} \textbf{z}) < \infty,\end{equation*}

and consider the ESSF $X=(\Pi,\textbf{V})$ with characteristics $(d, \beta, \Lambda)$ and $c=0$. Then the process of marks

\begin{equation*}\tilde{\textbf{V}} = (\sum_k\delta_{\tilde{V}_k(t)},\,t\geq 0)\end{equation*}

has the same distribution as the growth-fragmentation $\textbf{X}$. Indeed, up to a time-change, we can assume that $\alpha=0$, and Proposition 5 implies that for all $t\geq 0$, $\mathbb{E} \sum_k \tilde{V}_k(t)^q = \textrm{e}^{t\kappa(q)}$, with

\begin{align*}\kappa(q) &= dq +\frac{\beta}{2}q^2 + \int_{\mathscr{Z}^\downarrow}\big(\sum_{i\geq 1} v_i^{q}\big) -1 -q \log v_1 \unicode{x1D7D9}_{\lvert {\log v_1} \rvert \leq 1} \,\Lambda(\textrm{d} \textbf{z})\\&= -k + dq +\frac{\beta}{2}q^2 + \int_{[-\log 2, 0)}\textrm{e}^{yq} + (1-\textrm{e}^{y})^{q} -1 -qy \,\lambda(\textrm{d} y)\\&= \varphi(q) + \int_{[-\log 2, 0)}(1-\textrm{e}^{y})^{q}\,\lambda(\textrm{d} y).\end{align*}

This shows that indeed $\tilde{\textbf{V}}\overset{(d)}{=} \textbf{X}$.

A.1 Proof of Proposition 1

Let us write $X=(\Pi,\textbf{V})$, as usual. First, note that $\Pi$ is an exchangeable partition with values in $\mathcal{M}^{\star}_{\infty}$, and therefore it has asymptotic frequencies—so $\lvert {X} \rvert^{\downarrow}$ exists a.s.—and the finite blocks of $\Pi$ (if any) are necessarily singletons. For the uniqueness part of the proposition, notice that any $\nu$ satisfying (2) must be equal to $\mathbb{P}(\lvert {X} \rvert^{\downarrow}\in\cdot)$.

For the existence, let $(U_k,k\geq 1)$ be an i.i.d. sequence of uniform random variables on [0,Reference Aldous and Pitman1], independent of X. For every $i\in \mathbb{N}$, let $Z_i=(U_{k(\Pi,i)},V_i)\in[0,1]\times S^1$, where $k(\Pi,i)$ is the label of the block of $\Pi$ containing i. (Recall that labels are defined as an enumeration of blocks in increasing order of their smallest element: $k(\pi,1)=1$ for any partition $\pi$, then $k(\pi,i)=2$ for the first i that is not in the same block as 1, etc.)

The key point of the proof is that the sequence $(Z_i,i\geq 1)$ is exchangeable, with values in a Polish space, which allows us to apply de Finetti’s theorem. To see that, consider a permutation $\sigma\,{:}\,\mathbb{N}\to\mathbb{N}$, and note that for each (deterministic) partition $\pi$, there exists a permutation $\varphi\,{:}\,\mathbb{N}\to\mathbb{N}$ such that

\begin{equation*} k(\pi,\sigma(i)) = \varphi(k(\pi^{\sigma}, i)). \end{equation*}

This permutation of labels may not be unique (in the case where $\pi$ has a finite number of blocks), but we can assume that such a map $\varphi$ is chosen deterministically as a function of $\pi$ and $\sigma$. Going back to our random sequence $(Z_i,i\geq 1)$, we see that permuting its elements according to $\sigma$ yields, for all $i\geq 1$,

\begin{align*} Z_{\sigma(i)} &= (U_{k(\Pi,\sigma(i))}, V_{\sigma(i)})\\ &= (U_{\varphi(k(\Pi^{\sigma},i))}, V_{\sigma(i)}), \end{align*}

where $\varphi$ is a random permutation whose randomness comes only from $\Pi$. Now since the sequence $(U_k,k\geq 1)$ is i.i.d. and is independent from $\Pi$, if we define $U'_k=U_{\varphi(k)}$ for each $k\geq 1$, then the sequence $(U'_k,k\geq 1)$ is still i.i.d. with the same distribution and independent from $\Pi$. Finally, using the exchangeability property of $X=(\Pi,\textbf{V})$, we have

\begin{equation*} (Z_{\sigma(i)},i\geq 1) = \big((U'_{k(\Pi^{\sigma},i)},V_{\sigma(i)}),\;i\geq 1\big) \overset{(d)}{=} \big((U'_{k(\Pi,i)},V_{i}),\;i\geq 1\big)\overset{(d)}{=} \big((U_{k(\Pi,i)},V_{i}),\;i\geq 1\big), \end{equation*}

which shows that the sequence $(Z_i,i\geq 1)$ is exchangeable.

Therefore we can use Finetti’s theorem: there exists a random probability measure $\theta \in \mathcal{M}_1([0,1]\times S^1)$ such that, conditional on $\theta$, the sequence $(Z_i,i\geq 1)$ is i.i.d. with distribution $\theta$. Let

\begin{equation*}(a_k,k\geq 1)=(u_k,v_k,\,k\geq 1)\end{equation*}

denote the collection of atoms of $\theta$. Note that $i\sim^{X}j$ iff $Z_i=Z_j$, and therefore the law of large numbers ensures that the blocks of X correspond to those atoms, i.e. for each $k\geq 1$, there is a block B of X with an asymptotic frequency $\lvert {B} \rvert=\theta(a_k)$ and a mark equal to $v_k$. Conversely, any block which is not reduced to a singleton must be formed in this way. Furthermore, note that singleton blocks have mark 0 because of the assumption that $X\in\mathcal{M}^{\star}_{\infty}$, so the knowledge of the atoms $(a_k,k\geq 1)$ and their mass is sufficient to reconstruct the sequence $(Z_i,i\geq 1)$, and therefore the marked partition X.

Now define, for all $k\in\mathbb{N}$, $z_k \,{:\!=}\, (\theta(a_k),v_k)$. Up to a reordering, we can assume that $\textbf{z}=(z_k,k\geq 1)$ is in $\mathscr{Z}^\downarrow$ (if the sequence of atoms is finite, we concatenate to $\textbf{z}$ infinitely many (0,0) terms). The previous discussion means that conditional on $\theta$, the asymptotic frequencies of X are exactly

\begin{equation*}\lvert {X} \rvert^{\downarrow} = \textbf{z} \in\mathscr{Z}^\downarrow,\end{equation*}

and conditional on $\textbf{z}$, the marked partition X is drawn according to $\varrho_{\textbf{z}}$. Note that the map $\theta\mapsto\textbf{z}$ is measurable. Indeed, by standard arguments on point processes (see e.g. [Reference Daley and Vere-Jones12, Lemma 9.1.XIII]), there exists a measurable enumeration $(a_k,k\geq 1)$ of the atoms of $\theta$, and it is elementary that the nonincreasing reordering of this sequence is measurable. Therefore, defining $\nu = \mathbb{P}(\lvert {X} \rvert^{\downarrow}\in\cdot)$, which is the push-forward of the distribution of $\theta$ by the map $\theta\mapsto\textbf{z}$, we see that it satisfies (2).

A.2 Proof of Proposition 3

In this section, it will be helpful to consider the restriction of an ESSF process X to a more general (and possibly random) subset $A\subset \mathbb{N}$, considered as a random variable living on the compact space $2^{\mathbb{N}}$. We first consider a fixed—non-random—$A\subset \mathbb{N}$ with cardinality $\#A\in\mathbb{N}\cup\{\infty\}$, and define a canonical enumeration of A by

\begin{equation*}\sigma_{A}:\begin{cases}[\#A] &\to \mathbb{N}\\i &\mapsto \min\{ n\in\mathbb{N}, \, \# (A\cap [n])=i \},\end{cases}\end{equation*}

such that $A = \{\sigma_{A}(1),\sigma_{A}(2), \ldots\}$, with $\sigma_A(1) < \sigma_{A}(2) < \ldots$. Now recall the definition of the action of injections on $\mathcal{M}_{\infty}$. For any $x\in\mathcal{M}_{\infty}$, one can see $x^{\sigma_A}\in\mathcal{M}_{\#A}$ as the restriction of x to the set A.

As an inverse operation, for any $x'=(\pi',\textbf{v}')\in\mathcal{M}_{\#A},x''=(\pi'',\textbf{v}'')\in\mathcal{M}_{\#A^{\textrm{c}}}$, where $A^{\textrm{c}}\,{:\!=}\,\mathbb{N}\setminus A$, we can define

\begin{equation*}x'\overset{A}{\oplus}x''\in\mathcal{M}_{\infty}\end{equation*}

as the pair $(\pi,\textbf{v})$ such that

\begin{gather*}\forall i,j \in \mathbb{N}, \qquad i\sim^{\pi}j \iff \begin{cases}&i,j\in A\text{ and }\sigma_{A}^{-1}(i) \sim ^{\pi'}\sigma_{A}^{-1}(\,j)\\[3pt] \text{or} & i,j\in A^{\textrm{c}}\text{ and }\sigma_{A^{\textrm{c}}}^{-1}(i) \sim ^{\pi''}\sigma_{A^{\textrm{c}}}^{-1}(\,j)\end{cases} \\[3pt] \text{and }{v}_{i} = \begin{cases}{v}'_{\sigma_{A}^{-1}(i)} & \text{ if } i\in A,\\{v}''_{\sigma_{A^{\textrm{c}}}^{-1}(i)} & \text{ if } i\in A^{\textrm{c}}.\end{cases}\end{gather*}

Similarly, for processes $X'=(X'(t),\, t\geq 0)$ and $X^{\prime\prime}$, we write for conciseness

\begin{equation*}X' \overset{A}{\oplus} X'' \,{:\!=}\, \big(X'(t) \overset{A}{\oplus} X''(t), \, t\geq 0\big ).\end{equation*}

For $x=(\pi,\textbf{v})\in \mathcal{M}_{\infty}$ and $A\subset \mathbb{N}$, we will say that A is x-compatible if it is a union of a family of blocks of $\pi$—i.e. if A is such that $i\in A,j\notin A \implies i\nsim^{\pi} j$. These definitions enable us to reformulate the branching property as follows.

Lemma 2. Let X be an ESSF process, $x=(\pi,\textbf{v})\in\mathcal{M}_{\infty}$, and $A\subset \mathbb{N}$ an x-compatible set. If we define $X'\,{:\!=}\,X^{\sigma_A}$ and $X''\,{:\!=}\,X^{\sigma_{A^{\textrm{c}}}}$, then under $\mathbb{P}_x$, $X^{\prime}$ and $X^{\prime\prime}$ are two independent copies of the process X, respectively started at $x^{\sigma_A}$ and $x^{\sigma_{A^{\textrm{c}}}}$, and

\begin{equation*} X = X' \overset{A}{\oplus} X''. \end{equation*}

Proof. This is an immediate consequence of the branching property (ii) of Definition 2 and of the definition of the ssFrag operator.

Let us now tackle the proof of the Markov property for stopping lines (3). We write as usual $X=(\Pi,\textbf{V})$. We first assume that there exist $0\leq t_1 < t_2 < \ldots < t_k \leq \infty$ such that for all $i\in\mathbb{N}$, $L_i$ takes values in the finite set $\{t_1, \ldots t_k\}$. We prove the Markov property for such stopping lines by induction on k. For $k=1$ and $t_1<\infty$, this amounts to the simple Markov property, so (3) holds by definition. If $t_1=\infty$, then for all $t\geq 0$, for all $i\geq 1$, $L_i+t\equiv \infty$. By convention $\textbf{V}(\infty)$ is the null vector, and by definition $\textrm{ssFrag}\!\big(X(L),X^{(\cdot)}\big)$ is the process which is a.s. constant and equal to $X(\infty)=(\Pi(\infty),\textbf{V}(\infty))$, so (3) holds again.

Now assume that $k>1$, and that the stopping line Markov property has been proven for all stopping lines taking at most $k-1$ distinct values. By Remark 3, $L\wedge t_{k-1}$ is a stopping line taking at most $k-1$ distinct values. Therefore, one can apply the induction hypothesis, which says that conditional on $\mathcal{G}_{L\wedge t_{k-1}}$, the process $X(L\wedge t_{k-1}+\cdot)$ has the distribution of a copy of X started from $X(L\wedge t_{k-1})$. Now we define the random set

\begin{equation*}A \,{:\!=}\, \{i\in \mathbb{N}, \, L_i = t_{k}\},\end{equation*}

which is $\mathcal{G}_{L\wedge t_{k-1}}$-measurable. Indeed let us show that $\{i\in A\}\in \mathcal{G}_{L\wedge t_{k-1}}$. Since $\{L_i=t_{k}\}=\{L_i>t_{k-1}\}\in \mathcal{G}_i(t_{k-1})$, one can write the indicator of this event as

\begin{equation*}\unicode{x1D7D9}_{\{L_i>t_{k-1}\}}=F(B_i(s),V_i(s),\,s\in[0,t_{k-1}])\end{equation*}

for a measurable functional F—recall that $\mathcal{G}_i(t_{k-1})=\sigma(B_i(s),V_i(s),\,s\in[0,t_{k-1}]),$ where $B_i(s)$ is the block of $\Pi(s)$ containing i. Now on the event $\{L_i \geq t_{k-1}\}$, we have

\begin{equation*}F(B_i(s),V_i(s),\,s\in[0,t_{k-1}])=F\big(B_i(s\wedge L_i \wedge t_{k-1}),V_i(s\wedge L_i \wedge t_{k-1}),\,s\in[0,t_{k-1}]\big);\end{equation*}

therefore we can write

\begin{equation*}\{i\in A\} = \{L_i \geq t_{k-1}\}\cap\big\{ F\big(B_i(s\wedge L_i \wedge t_{k-1}),V_i(s\wedge L_i \wedge t_{k-1}),\,s\in[0,t_{k-1}]\big) = 1\big\} \in \mathcal{G}_{L\wedge t_{k-1}},\end{equation*}

so finally A is $\mathcal{G}_{L\wedge t_{k-1}}$-measurable. Now notice that because L is a stopping line, A is compatible with $\Pi(L\wedge t_{k-1})$ in the sense that A is necessarily a union of blocks of $\Pi(L\wedge t_{k-1})$. Therefore, it is immediate by definition that

\begin{equation*}X(L\wedge t_{k-1}+\cdot) = X' \overset{A}{\oplus}X'',\end{equation*}

with

\begin{align*}&X' = X(L\wedge t_{k-1}+\cdot)^{\sigma_{A}} \\[3pt] \text{ and } &X'' = X(L\wedge t_{k-1}+\cdot)^{\sigma_{A^{\textrm{c}}}}.\end{align*}

Now by Lemma 2, conditional on $\mathcal{G}_{L\wedge t_{k-1}}$, $X^{\prime}$ and $X^{\prime\prime}$ are two independent copies of X started respectively from $X(L\wedge t_{k-1})^{\sigma_{A}}$ and $X(L\wedge t_{k-1})^{\sigma_{A^{\textrm{c}}}}$.

Also, notice that by definition of the random set A, we have the equality

(7) \begin{equation} X(L+\cdot) = X'(t_{k}-t_{k-1}+\cdot) \overset{A}{\oplus} X'',\end{equation}

and for the same reason, the following equality between $\sigma$-algebras holds:

\begin{equation*}\mathcal{G}_{L} = \mathcal{G}_{L\wedge t_{k-1}} \vee \sigma\big(X'(s),\, s\in [0,t_k-t_{k-1}] \big).\end{equation*}

Clearly $X^{\prime}$ and $X^{\prime\prime}$ are still independent conditional on $\mathcal{G}_{L}$, and the distribution of $X'(t_{k}-t_{k-1}{+}\cdot)$ conditional on $\mathcal{G}_{L}$ is by the Markov property at time $t_k-t_{k-1}$ the law of X started from $X'(t_{k}-t_{k-1})=X(L)^{\sigma_{A}}$. Finally, again using Lemma 2, conditional on $\mathcal{G}_{L}$, the process

\begin{equation*}X'(t_{k}-t_{k-1}+\cdot) \overset{A}{\oplus} X''\end{equation*}

has simply the distribution of a copy of X started at X(L). So by (7) the Markov property for stopping lines holds for L, and so by induction it holds for all stopping lines taking at most a finite number of values.

Now fix a general stopping line L and a time $t\geq 0$, and let us assume that our probability space contains an independent sequence $X^{(\cdot)}$ of i.i.d. copies of the process started from $(\textbf{1},1)$. To conclude, it is enough to prove that conditional on $\mathcal{G}_L$,

(8) \begin{equation} X(L+t) \overset{(d)}{=} \textrm{ssFrag}_\alpha\!\big(X(L),X^{(\cdot)}\big)(t),\end{equation}

because then for any $0 \leq t_1 < t_2 < \ldots t_k$ one can apply (8) successively to the stopping lines $L+t_i$, $1\leq i \leq k$, which implies that (3) holds for any finite-dimensional distributions. Therefore it remains only to prove

\begin{equation*}\mathbb{E}\left[F(X(L+t)) Z\right] = \mathbb{E}\left[F\left (\textrm{ssFrag}_\alpha\!\big(X(L),X^{(\cdot)}\big)(t)\right ) Z\right]\end{equation*}

for any fixed continuous bounded map $F\,{:}\,\mathcal{M}^{\star}_{\infty}\to\mathbb{R}$ and $\mathcal{G}_{L}$-measurable bounded random variable Z. To show this, consider the sequence of stopping lines $(L^{n},n\geq 1)$ defined by

\begin{equation*}L^{n}_i = 2^{-n}\lceil2^{n}L_i\rceil \unicode{x1D7D9}_{L_i \leq n} + \infty \unicode{x1D7D9}_{L_i > n}, \quad i\geq 1,\end{equation*}

where $\lceil\cdot\rceil$ denotes the usual ceiling function. This is a classical transformation for stopping times, and it is easily checked that $L^n$ is a stopping line for all $n\geq 1$. Furthermore, for all $i\geq 1$, $L^n_i$ is nonincreasing and tends to $L_i$ as $n\to\infty$, so since X has right-continuous sample paths, for any $t'\geq 0$, we have $V_i(L^n_i+t') \to V_i(L_i+t')$ and $B_i(L^n_i+t') \to B_i(L_i+t')$ a.s. (recall that $B_i(u)$ denotes the block of $\Pi(u)$ containing i). In other words, for any $t'\geq 0$, the convergence $X(L^n+t')\to X(L+t')$ holds a.s. Therefore

\begin{equation*}\mathbb{E}\left[F(X(L^{n}+t)) Z\right] \underset{n\to\infty}{\longrightarrow} \mathbb{E}\left[F(X(L+t)) Z\right].\end{equation*}

Now, because $L^{n}$ only takes values in a finite set for all n, we can apply (3), so

(9) \begin{equation}\mathbb{E}\left[F(X(L^{n}+t)) Z\right] = \mathbb{E}\left[F\left (\textrm{ssFrag}_\alpha\!\big(X(L^{n}),X^{(\cdot)}\big)(t)\right ) Z\right].\end{equation}

This holds because Z is $\mathcal{G}_{L^{n}}$-measurable, since $\mathcal{G}_L \subset\mathcal{G}_{L^{n}}$ for all $n\geq 1$. For the convergence of the right-hand side, recall that $X(L^{n}+t)\to X(L+t)$, and note also that if $V_i(L_i)=0$ for $i\geq 1$, then since $L_i$ is a stopping time, by the strong Markov property $V_i(L_i+t)$ is also zero for all $t\geq 0$, and in particular $V_i(L^{n}_i)=0$ for all $n\geq 1$. Now by definition an ESSF process is stochastically continuous, so in particular for any $i\geq 1$, on the event $\{V_i(L_i) >0\}$, we have

\begin{equation*}X^{(\,j)}(V_i(L_i)^{\alpha}t) \underset{n\to\infty}{\longrightarrow} X^{(\,j)}(V_i(L_i)^{\alpha}t) \qquad \forall j\geq 1 \; \text{ a.s.}\end{equation*}

We can now invoke the continuity property of the operator $\textrm{ssFrag}_\alpha$ pointed out in Remark 1 and deduce

\begin{equation*}\textrm{ssFrag}_\alpha\!\big(X(L^{n}),X^{(\cdot)}\big)(t) \underset{n\to\infty}{\longrightarrow}\textrm{ssFrag}_\alpha\!\big(X(L),X^{(\cdot)}\big)(t) \qquad \text{ a.s.}\end{equation*}

Taking limits in (9) yields the equality needed to end the proof.

A.3. Proof of Proposition 4

Let $X=(\Pi,\textbf{V})$ be an $\alpha$-ESSF process, let $t\geq 0$, and recall the definition of $\tau^{\beta}(t)$ as in the proposition, i.e.

\begin{equation*}\tau^{\beta}_i(t) \;=\; \left (\int_{0}^{\cdot}V_i(s)^{\beta} \,\textrm{d} s\right )^{-1}\!\!\!(t) \;=\; \inf \bigg\{u\geq 0,\,\int_{0}^{u}V_i(s)^{\beta} \,\textrm{d} s=t\bigg\},\qquad i\geq 1,\end{equation*}

with the convention $\inf \varnothing = \infty$. For conciseness and because $\beta$ is fixed, let us write simply $\tau$ instead of $\tau^{\beta}$ throughout the proof. First, let us see that $\tau(t)$ is a stopping line. Fix $i\in\mathbb{N}$, and note that for $T\geq 0$,

\begin{equation*}\{\tau_i(t) \leq T\} = \Big\{ \int_{0}^{T}V_{i}(s)^{\beta}\,\textrm{d} s \geq t \,\Big\} \in \mathcal{G}_i(T),\end{equation*}

so $\tau_i(t)$ is a $\mathcal{G}_i$-stopping time. Furthermore, conditional on $\tau_i(t)$, for any $i,j\in\mathbb{N}$, if $i\sim j$ in $\Pi(\tau_i(t))$, then a.s. for all $0\leq s\leq \tau_i(t)$ we have $i\sim j$ in $\Pi(s)$, so $V_i(s)=V_j(s)$, and necessarily $\tau_j(t) = \tau_i(t)$.

Therefore $\tau(t)$ is indeed a stopping line, so the process $X\circ\tau = (X(\tau(t),t\geq 0)$ is well-defined. We claim that its sample paths are càdlàg in $\mathcal{M}_{\infty}$. Indeed, by definition, for each $i\in\mathbb{N}$, $\tau_i$ is a nondecreasing right-continuous map. Now, because X also has càdlàg sample paths by definition, the following holds a.s.: for each $i\in\mathbb{N}$, and $t\in[0,\infty)$,

\begin{equation*}X(\tau_i(s)) \underset{s\downarrow t}{\longrightarrow} X(\tau_i(t)) \quad \text{ and }\quad X(\tau_i(s)) \underset{s\uparrow t}{\longrightarrow} X(\tau_i(t)-) \quad \text{ if } t>0.\end{equation*}

Now note that for each stopping line L and integer $n\in\mathbb{N}$, by definition $X(L)_{|[n]}$ is a (deterministic) continuous functional of the variables $(X(L_1), \ldots, X(L_n))$. Applying this to $L=\tau(s)$ and letting $s\to t$, it follows that a.s.,

\begin{align*}&X(\tau(s))_{|[n]} \underset{s\downarrow t}{\longrightarrow} X(\tau(t))_{|[n]}\\\text{ and }&X(\tau(s))_{|[n]} \text{ converges in }\mathcal{M}_{n} \text{ when }s\uparrow t, \text{ in the case } t>0.\end{align*}

The integer n being generic, this shows that $X\circ\tau$ is an a.s. càdlàg process.

Let $t\geq 0$ be fixed. Since $\tau(t)$ is a stopping line, we can apply Proposition 3, and assume that the process $X(\tau(t)+\cdot)$ is given by

\begin{equation*}\textrm{ssFrag}_\alpha\!\big(X(\tau(t)), X^{(\cdot)}\big),\end{equation*}

where $X^{(\cdot)}$ is an independent sequence of i.i.d. copies of the process started from $(\textbf{1},1)$. For each $k\in\mathbb{N}$, let $(\tau^{(k)}(s),s\geq 0)$ denote the stopping lines corresponding to $X^{(k)}$, i.e.

\begin{equation*}\tau_i^{(k)}(s) = \bigg(\int_{0}^{\cdot}V_i^{(k)}(u)^{\beta} \,\textrm{d} u\bigg)^{-1}(s), \qquad i\geq 1, \, s\geq 0.\end{equation*}

Our aim is to show that

(10) \begin{equation} X(\tau(t+\cdot)) = \textrm{ssFrag}_{\alpha-\beta}\!\big(X(\tau(t)),X^{(\cdot)}\circ\tau^{(\cdot)}\big).\end{equation}

Now let us fix $i\in\mathbb{N}$, and work conditional on $\mathcal{G}_{\tau(t)}$. On the event $\{V_i(\tau_i(t))=0\}$, by definition of the operators $\textrm{ssFrag}_\alpha$, the block containing i is constant in time and has mark 0 in both processes in (10), so there is equality for the index i. Now we condition on $V_i(\tau_i(t))=v$ with $v>0$. Note that $\{V_i(\tau_i(t))>0\}\subset\{\tau_i(t)<\infty\}$, so in that case we have $\tau_i(t)<\infty$ a.s., and there is the equality

\begin{equation*}\int_{0}^{\tau_i(t)}V_i(s)^{\beta} \,\textrm{d} s = t.\end{equation*}

Therefore we can write, for $s\geq 0$,

\begin{align*}\tau_i(t+s) &= \left (\int_{0}^{\cdot}V_i(u)^{\beta} \,\textrm{d} u\right )^{-1}(t+s)\\&= \tau_i(t)+ \left (\int_{0}^{\cdot}V_i(\tau_i(t)+u)^{\beta} \,\textrm{d} u\right )^{-1}(s).\end{align*}

Now for all $u\geq 0$, because $X(\tau(t)+\cdot)=\textrm{ssFrag}_\alpha(X(\tau(t)), X^{(\cdot)})$, we have $V_i(\tau_i(t)+u)=vV^{(k)}_i(v^{\alpha}u)$, for k such that i is in the kth block of $\Pi(\tau(t))$. This implies

\begin{align*}\tau_i(t+s)-\tau_i(t) &=\left (\int_{0}^{\cdot}V_i(\tau_i(t)+u)^{\beta} \,\textrm{d} u\right )^{-1}(s) \\&= \left (\int_{0}^{\cdot}v^{\beta}V^{(k)}_i(v^{\alpha}u)^{\beta} \,\textrm{d} u\right )^{-1}(s) \\&= \left (v^{\beta-\alpha}\!\!\int_{0}^{v^{\alpha}\cdot}V^{(k)}_i(u)^{\beta} \,\textrm{d} u\right )^{-1}(s) \\&= v^{-\alpha}\left (\int_{0}^{\cdot}V^{(k)}_i(u)^{\beta} \,\textrm{d} u\right )^{-1}(v^{\alpha-\beta}s) \\&= v^{-\alpha}\tau^{(k)}_i(v^{\alpha-\beta}s).\end{align*}

Note that $L_i(s) \,{:\!=}\, \tau_i(t+s)-\tau_i(t)$ is a stopping time for the natural filtration of the process $V_i(\tau_i(t)+\cdot) = vV_i^{(k)}(v^\alpha \cdot)$, because for all $i\geq 1$ and $u\geq 0$, we have

\begin{equation*} \{L_i(s) \leq y\} \;=\; \Big\{\int_0^y \big(vV_i^{(k)}(v^\alpha u)\big)^{\beta}\,\textrm{d} u \geq s \Big\} \;\in\; \sigma\big(vV_i^{(k)}(v^\alpha u), \, u\in [0,y]\big). \end{equation*}

This implies that $L(s) = (L_i(s),i\geq 1)$ is a stopping line for the process $X(\tau(t)+\cdot)$, and the equality $X(\tau(t)+L(s))=X(\tau(t+s))$ holds. We can then write

\begin{equation*} X(\tau(t+s)) = \textrm{ssFrag}_\alpha\!\big(X(\tau(t)), X^{(\cdot)}\big)(L(s)) \end{equation*}

and compute the mark attached to the block containing i in this marked partition: by definition of the operator $\textrm{ssFrag}_\alpha$ and using the fact that $L_i(s) = v^{-\alpha}\tau^{(k)}_i(v^{\alpha-\beta}s)$, we have

\begin{align*} V_i(\tau_i(t+s)) &= v V^{(k)}_i(v^\alpha L_i(s))\\ &= v V^{(k)}_i(\tau^{(k)}_i(v^{\alpha-\beta}s)). \end{align*}

We now recognize the mark attached to the block containing i in $\textrm{ssFrag}_{\alpha-\beta}\!\big(X(\tau(t)), X^{(\cdot)}\circ\tau^{(\cdot)})\big)(s)$. The same argument allows us to check that $B_i(\tau_i(t+s))$ is also the block of $\textrm{ssFrag}_{\alpha-\beta}\!\big(X(\tau(t)), X^{(\cdot)}\circ\tau^{(\cdot)})\big)(s)$ containing i, so in the end we have exactly

\begin{equation*} X\circ\tau(t+\cdot) = \textrm{ssFrag}_{\alpha-\beta}\ \big(X\circ\tau(t), X^{(\cdot)}\circ\tau^{(\cdot)})\big), \end{equation*}

which shows that $X\circ \tau$ is an $(\alpha-\beta)$-ESSF.

Finally, note that if X is non-degenerate, then for all $x\in\mathcal{M}^{\star}_{\infty}$, $\mathbb{P}_x$-a.s. for any time $t\geq 0$, any block of X(t) either is infinite or has mark 0. So $\mathbb{P}_x$-a.s., for all possible stopping lines L, the blocks of X(L) satisfy the same condition, i.e. $X(L)\in\mathcal{M}^{\star}_{\infty}$. Therefore, $\mathbb{P}_x$-a.s. $X\circ \tau$ has sample paths with values in $\mathcal{M}^{\star}_{\infty}$.

A.4. Proof of Theorem 1

Note that the whole process $(X(t),t\geq 0)$ defines a coupling of all $\xi_n$ for $n\geq 1$. By definition, one has

\begin{equation*}\xi_{n+1}(t) = \xi_n(t), \quad \forall t\leq T_{n+1},\end{equation*}

and at time $T_{n+1}$, either $\xi_n$ is killed on the event $\{T_n = T_{n+1}\}$, or, conditional on $\{T_n > T_{n+1}\}$, the process $\xi_n$ jumps, independently of the past, according to the probability

\begin{equation*}\eta_{n+1}(\cdot) \,{:\!=}\, \mathcal{D}_{n+1}\big(\log v_1 \in \cdot \mid \pi_{|[n]} = \textbf{1}_n \text{ and } v_1\neq 0\big),\end{equation*}

and goes on independently of the past, its remaining part

\begin{equation*}(\xi_n(T_{n+1}+t)-\xi_n(T_{n+1}),0\leq t\leq T_n - T_{n+1})\end{equation*}

being independent from $\xi_{n+1}$ and equal in distribution to $\xi_n$ by the strong Markov property. Let us first compute the probability $p_n$ that $T_{n+1} = T_n$, which is by construction

(11) \begin{equation} p_n = \mathcal{D}_{n+1}(\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0).\end{equation}

From the previous description, one can write

\begin{equation*}T_n = T_{n+1} + ZT'_n,\end{equation*}

where $Z = \unicode{x1D7D9}_{\{T_{n}\neq T_{n+1}\}}\sim Be(1-p_n)$ is a Bernoulli random variable with parameter $(1-p_n)$, and $T'_n$ is a random variable equal in distribution to $T_n$, and independent from $T_{n+1}$ and Z. Then $T_{n+1},Z$ and $T'_n$ are independent because Z is simply a function of the marked partition $D_{n+1}=(\Pi(T_{n+1}),\textbf{V}(T_{n+1})/V_1(T_{n+1}-))_{|[n+1]}$ which is independent from $\xi_{n+1}$, and $T'_n$ is independent of $(\xi_{n+1}, D_{n+1})$ because of the strong Markov property and the fact that $\alpha=0$. Taking expectations yields

\begin{equation*}\frac{1}{J_n} = \frac{1}{J_{n+1}} + \frac{1-p_n}{J_n},\end{equation*}

which implies that $p_n = J_n / J_{n+1}$.

Now let us rebuild the coupling between $\xi_n$ and $\xi_{n+1}$ to show that their respective Lévy measures $\lambda_n$ and $\lambda_{n+1}$ satisfy

(12) \begin{equation} \lambda_n = \lambda_{n+1} + (J_{n+1}-J_n)\widetilde{\eta}_{n+1},\end{equation}

where

(13) \begin{equation} \widetilde{\eta}_{n+1} \,{:\!=}\, \eta_{n+1}\big(\cdot\cap\;\mathbb{R}\!\setminus\!\{0\}\big) = \mathcal{D}_{n+1}\big(\{\log v_1 \in \cdot\}\cap\{v_1\neq 1\} \mid \pi_{|[n]} = \textbf{1}_n \text{ and } v_1\neq 0\big).\end{equation}

Consider the process $\xi_{n+1}$ a Lévy process with characteristic exponent $\psi_{n+1}$, and let $T_n\sim Exp(J_n)$ be independent. Now independently define $T'\sim Exp(J_{n+1}-J_n)$, and let $T_{n+1} \,{:\!=}\, T_n \wedge T'$. We see that $T_{n+1}\sim Exp(J_{n+1})$ and that it is independent of $\xi_{n+1}$ and of the event $\{T_n = T_{n+1}\}$, which has probability $J_n/J_{n+1}=p_n$. Now conditional on $(T_n,T_{n+1})$, let us define

(14) \begin{equation} D_{n+1} = \begin{cases}D' &\text{if }T_n=T_{n+1}, \quad \text{where } D' \sim \mathcal{D}_{n+1}(\cdot \mid \pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0),\\[3pt] D'' &\text{if }T_n\neq T_{n+1}, \quad \text{where } D'' \sim \mathcal{D}_{n+1}(\cdot \mid \pi_{|[n]}= \textbf{1}_n \text{ and }v_1 \neq 0),\end{cases}\end{equation}

where $D^{\prime}$ and $D^{\prime\prime}$ are mutually independent and independent of everything else. Note that $D_{n+1}$ is independent of $T_{n+1}$ and of $\xi_{n+1}$, and because of (11), $D_{n+1}$ indeed has distribution $\mathcal{D}_{n+1}$. Let us define $J=\log v_1$, where $v_1$ is the mark associated with the integer 1 in the marked partition $D_{n+1}$, and let $\xi_n$ be a Lévy process with characteristic exponent $\psi_n$. Now, putting everything together, define $\tilde{\xi}_{n+1}$ as the killed Lévy process $(\xi_{n+1}(t), 0\leq t < T_{n+1})$, and define $\tilde{\xi}_n$ as

\begin{equation*}\tilde{\xi}_n(t) = \begin{cases}\xi_{n+1}(t) &\text{ if }t<T_{n+1},\\[3pt] \xi_{n+1}(T_{n+1})+J + \xi_n(t-T_{n+1}) & \text{ if }T_{n+1}\leq t < T_n.\end{cases}\end{equation*}

By construction, the joint distribution of $(\tilde{\xi}_{n}, \tilde{\xi}_{n+1})$ is equal to the one we get from the original process X, and it should now be clear that the point process of jumps of $\xi_n$ is equal in distribution to the point process of jumps of $\xi_{n+1}$ with additional jumps distributed as $J=\log v_1$, arising at rate $(J_{n+1}-J_n)$. Note that by construction, J has distribution $\eta_{n+1}$, so finally we have proven (12). The fact that $(\lambda_n,n\geq 1)$ is a nonincreasing sequence of $\sigma$-finite measures ensures the existence of a limiting measure $\lambda_\infty$ on $\mathbb{R}\setminus\{0\}$ such that for all $n\in\mathbb{N}$,

(15) \begin{equation} \lambda_n = \lambda_\infty + \sum_{k>n}(J_{k}-J_{k-1})\widetilde{\eta}_k.\end{equation}

Recall that we wrote the characteristic exponent of $\xi_n$ in the following way:

\begin{equation*}\psi_n(\theta) = id_n\theta -\frac{\beta_n}{2}\theta^2 + \int_{\mathbb{R}}\Big(\textrm{e}^{i\theta y} -1-i\theta y\unicode{x1D7D9}_{\lvert {y} \rvert\leq 1} \Big) \,\lambda_n(\textrm{d} y).\end{equation*}

From the previous discussion, one can construct a coupling between the two Lévy processes such that $(\xi_n(t)-\xi_{n+1}(t),t\geq 0)$ is simply a compound Poisson process with jump measure $(J_{n+1}-J_n)\widetilde{\eta}_{n+1}$ which is independent from $\xi_{n+1}$. It is then clear that necessarily $\beta_{n}=\beta_{n+1}$, and

\begin{equation*}d_n = d_{n+1} + (J_{n+1}-J_n)\int_{\lvert {y} \rvert\leq 1}y\,\widetilde{\eta}_{n+1}(\textrm{d} y) = d_{n+1} + \int_{\lvert {y} \rvert\leq 1}y\,(\lambda_n-\lambda_{n+1})(\textrm{d} y).\end{equation*}

To summarize, letting $\beta \,{:\!=}\, \beta_1$, the following holds for all $n\in\mathbb{N}$:

(16) \begin{equation} \beta_n=\beta \quad \text{ and } \quad d_n = d_1 - \int_{\lvert {y} \rvert\leq 1}y\,(\lambda_1-\lambda_n)(\textrm{d} y),\end{equation}

where $(\lambda_1 - \lambda_n)$ denotes the positive measure given by

\begin{equation*}(\lambda_1 - \lambda_n) = \sum_{k=2}^{n}(J_{k}-J_{k-1})\widetilde{\eta}_k.\end{equation*}

Let us now examine the consistency properties of the measures $\mathcal{D}_n$. From this point on, for the sake of clarity, we decompose the proof into a series of steps.

Step 1. We prove the existence and uniqueness of a measure $\mathcal{D}$ on $\mathcal{M}_{\infty}$ satisfying

(17) \begin{equation} \mathcal{D}(\pi=\textbf{1}\text{ and }v_1\neq 0) = 0\end{equation}

and such that for all $n\in\mathbb{N}$,

(18) \begin{equation} \mathcal{D}\big(\{x_{|[n]}\in\cdot\}\cap\{\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\}\big) = J_n \mathcal{D}_n;\end{equation}

then we show that this measure is exchangeable.

First, note that for the construction above to be consistent—with $D_{n+1}$ defined in (14) as $D_{n+1} = D'$ on the event $\{T_n=T_{n+1}\}$, where $D'\sim \mathcal{D}_{n+1}(\cdot \mid \pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0)$—the random variable $D'_{|[n]}$ must have distribution $\mathcal{D}_n$. Indeed, on the event $\{T_{n+1}<T_n\}$, the strong Markov property at time $T_{n+1}$ implies that the process $X_{|[n]}$ jumps according to $\mathcal{D}_n$, independently of the past, so on the complement this must hold as well; thus

\begin{equation*}\mathcal{D}_{n+1}\big(x_{|[n]}\in \cdot \mid \pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\big) = \mathcal{D}_n,\end{equation*}

which can be rewritten as

(19) \begin{equation} J_{n+1}\mathcal{D}_{n+1}\big(\{x_{|[n]}\in\cdot\}\cap\{\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\}\big) = J_n \mathcal{D}_n.\end{equation}

Now, for all integers $n\leq m$, let us define a measure on $\mathcal{M}_{m}$ by

\begin{equation*}\mu_{n}^{m} \,{:\!=}\, J_m \mathcal{D}_m\big( \cdot\cap \{\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\}\big), \quad m\geq n.\end{equation*}

Let us prove that for all integers $n\leq k \leq m$, we have

(20) \begin{equation} \mu_{n}^{m}(x_{|[k]} \in \cdot) = \mu_{n}^{k}.\end{equation}

Note that there is nothing to prove in the case $k=m$. Now suppose this is proven for fixed $n\leq k\leq m$. Then

\begin{align*}\mu_{n}^{m+1}(x_{|[k]}\in\cdot) &= J_{m+1}\mathcal{D}_{m+1}\big(\{x_{|[k]}\in\cdot\}\cap\{\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\}\big)\\&= J_{m+1}\mathcal{D}_{m+1}\big(\{(x_{|[m]})_{|[k]}\in\cdot\} \cap\{(\pi_{|[m]})_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\} \\& \phantom{=J_{m+1}\mathcal{D}_{m+1}\big(\{(x_{|[m]})_{|[k]}\in\cdot\}} \;\cap\{\pi_{|[m]}\neq \textbf{1}_m \text{ or }v_1 = 0\}\big)\\&= J_m \mathcal{D}_m\big( \{x_{|[k]}\in\cdot\} \cap\{\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\}\big)\\&= \mu_n^{m}(x_{|[k]}\in\cdot) = \mu_n^k,\end{align*}

where we have used (19) and the fact that

\begin{equation*}\{\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\}\subset \{\pi_{|[m]}\neq \textbf{1}_m \text{ or }v_1 = 0\}.\end{equation*}

By induction on m this proves (20) for any integers $n\leq k\leq m$. Note that in particular, taking $k=n$, we see that the total mass of $\mu_n^{m}$ is equal to that of $\mu_n^{n}$, which is $J_n$. In summary, for any $n\in\mathbb{N}$, the sequence $(\mathcal{M}_{m},\mu_n^{m}/J_n,m\geq n)$ defines a inverse system of compact probability spaces, and by the Kolmogorov extension theorem, there exists a unique measure $\mu_n$ (with total mass $J_n$) on the inverse limit $\varprojlim_{m}\mathcal{M}_{m} = \mathcal{M}_{\infty}$ such that for each $m\geq n$, $\mu_n(x_{|[m]}\in\cdot)=\mu_{n}^{m}$. Now notice that by definition, for any integers $n_1 \leq n_2 \leq m$, we have

\begin{equation*}\mu_{n_2}^{m}(\cdot \cap \{\pi_{|[n_1]}\neq \textbf{1}_{n_1} \text{ or }v_1 = 0\}) = \mu_{n_1}^{m},\end{equation*}

which implies by construction

\begin{equation*}\mu_{n_2}(\cdot \cap \{\pi_{|[n_1]}\neq \textbf{1}_{n_1} \text{ or }v_1 = 0\}) = \mu_{n_1}.\end{equation*}

This means that the sequence of measures $(\mu_n, n\geq 1)$ on $\mathcal{M}_{\infty}$ is increasing, and one can define the limit as $\mathcal{D}$. This measure then satisfies by construction

\begin{align*}\mathcal{D}\big(\{x_{|[n]}\in\cdot\}\cap\{\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0\}\big) &= \mu_{n}(x_{|[n]}\in\cdot) \\&= \mu_n^{n} \\&= J_n \mathcal{D}_n,\end{align*}

which is indeed (18).

Secondly, note that since for any $n\in\mathbb{N}$, clearly $\mu_n(\pi=\textbf{1}\text{ and }v_1\neq 0) = 0$, where $\mu_n$ are the measures defined above, so in the limit (17) holds. Let us now show uniqueness. If a measure $\mathcal{D}'$ on $\mathcal{M}_{\infty}$ satisfies (18) and (17) then

\begin{equation*}\mathcal{D}'(\cdot\cap\{\pi_{|[n]}\neq \textbf{1}_{n} \text{ or }v_1 = 0\}) = \mu_{n},\end{equation*}

and letting $n\to\infty$, $\mathcal{D}'=\mathcal{D}$, which proves uniqueness.

Finally, $\mathcal{D}$ is exchangeable. Indeed if $\sigma\,{:}\,\mathbb{N}\to\mathbb{N}$ is a permutation, let $m\in\mathbb{N}$ be such that $\sigma(k)=k$ for all $k\geq m$. Now, for all $n\geq m$, using the exchangeability of the probability measures $(\mathcal{D}_k,k\geq 1)$, we get

\begin{align*}\mathcal{D}((x^{\sigma})_{|[n]}\in\cdot) &= \lim_{k\to\infty}\mathcal{D}\big(\{(x^{\sigma})_{|[n]}\in\cdot\}\cap\{\pi_{|[k]}\neq \textbf{1}_{k} \text{ or }v_1 = 0\}\big)\\[3pt]&= \lim_{k\to\infty}J_k\mathcal{D}_k((x^{\sigma})_{|[n]}\in\cdot)\\[3pt]&= \lim_{k\to\infty}J_k\mathcal{D}_k(x_{|[n]}\in\cdot)\\[3pt]&= \lim_{k\to\infty}\mathcal{D}(\{x_{|[n]}\in\cdot\}\cap\{\pi_{|[k]}\neq \textbf{1}_{k} \text{ or }v_1 = 0\})\\[3pt]&= \mathcal{D}(x_{|[n]}\in\cdot).\end{align*}

As this is true for all $n\geq m$, necessarily $\mathcal{D}(x^{\sigma}\in\cdot)=\mathcal{D}$, i.e. $\mathcal{D}$ is exchangeable.

Step 2. We prove that $\mathcal{D}(\mathcal{M}_{\infty}\setminus\mathcal{M}^{\star}_{\infty})=0$ by using that X is non-degenerate. For this, we need to show first that the process $(B_1(t),t\geq 0)$ of the block of X containing 1 is equal in distribution to a process $(B(t),t\geq 0)$ constructed from a Poisson point process of intensity $\textrm{d} t \otimes \mathcal{D}$.

More precisely, define $\mathcal{N}$ to be a Poisson point process on $[0,\infty)\times\mathcal{M}_{\infty}$ with intensity $\textrm{d} t\otimes (\mathcal{D}+\widetilde{\lambda}_\infty)$, where $\widetilde{\lambda}_\infty$ is the push-forward of $\lambda_\infty$ by the map $y \in\mathbb{R} \mapsto (\textbf{1},\textrm{e}^{y}) \in\mathcal{M}^{\star}_{\infty}$. Let us define

\begin{equation*}\mathcal{N}' \,{:\!=}\, \big\{(t,\log v_1), \; (t,x)\in\mathcal{N} \text{ with }x =(\pi,\textbf{v}) \text{ and } v_1\notin \{0,1\}\big\},\end{equation*}

which is then a Poisson point process on $[0,\infty)\times \mathbb{R}$ with intensity

\begin{equation*}\textrm{d} t \otimes \Big(\mathcal{D} \big(\{\log v_1\in\cdot\}\cap\{v_1\notin \{0,1\}\}\big) + \lambda_\infty \Big).\end{equation*}

However, note that using (17), (18), and finally (13), we get

\begin{align*}\mathcal{D}&\big(\{\log v_1\in\cdot\}\cap\{v_1\notin \{0,1\}\}\big)\\[3pt]&= \sum_{n\in\mathbb{N}}\mathcal{D}\big(\{\log v_1\in\cdot\}\cap\{\pi_{|[n]} = \textbf{1}_n\text{ and }v_1\notin \{0,1\}\}\cap\{\pi_{|[n+1]} \neq \textbf{1}_{n+1}\}\big)\\[3pt]&= \sum_{n\in\mathbb{N}}J_{n+1}\mathcal{D}_{n+1}\big(\{\log v_1\in\cdot\}\cap\{\pi_{|[n]} = \textbf{1}_n\text{ and }v_1\notin \{0,1\}\}\big)\\[3pt]&= \sum_{n\in\mathbb{N}}(J_{n+1}-J_n) \widetilde{\eta}_{n+1},\end{align*}

and by (15), we find $\mathcal{D} \big(\{\log v_1\in\cdot\}\cap\{v_1\notin \{0,1\}\}\big) + \lambda_\infty = \lambda_1$. Therefore, it is possible to define a Lévy process $\xi$ with characteristic exponent $\psi_1$, such that the point process of its jumps is precisely $\mathcal{N}'$. Let us also define a process $B=(B(t),t\geq 0)$ with càdlàg sample paths with values in $2^{\mathbb{N}}$ (the subsets of $\mathbb{N}$), such that B has the distribution of $(B_1(t),t\geq 0)$, the block containing 1 in X. First define T as the first time $t\in[0,\infty)$ such that there is an atom $(t,(\pi,\textbf{v}))\in \mathcal{N}$ with $v_1=0$. If there is none, then let $T=\infty$. Then, for each $n\in\mathbb{N}$, let $(t_1,x_1),(t_2,x_2),\ldots$ be the whole sequence (finite or infinite) of atoms of $\mathcal{N}$ with $t_1 < t_2 < \ldots \leq T$ such that for each i, $x_i=(\pi_i,\textbf{v}_i)$ with $(\pi_i)_{|[n]}\neq \textbf{1}_n$ and $(v_i)_1>0$ (or such that $(v_i)_1=0$ for the possible last atom, at time T). We can define $\widetilde{B}^{n}_0 = [n]$ and, inductively, for each $i\geq 1$,

\begin{equation*}\widetilde{B}^{n}_{i} \,{:\!=}\, \widetilde{B}^{n}_{i-1} \cap (A_i \cap [n]),\end{equation*}

where $A_i$ is the block of $\pi_i$ containing 1. Now define, for $t\in[0,\infty)$,

\begin{equation*}B^{n}(t) = \widetilde{B}^{n}_i \text{ if }t\in[t_i,t_{i+1}),\end{equation*}

where we let $t_0\,{:\!=}\,0$, and in the case $T < \infty$, i.e. if the sequence of atoms is finite, say with length $k\in\mathbb{N}$, we let $t_{k+1} \,{:\!=}\, \infty$. It is readily checked that this construction is consistent in the sense that for each $t\geq 0$ there is a single $B(t)\in 2^{\mathbb{N}}$ such that $B^{n}(t)=B(t)\cap[n]$. Let us show that this process $(B(t),\xi(t),t\geq 0)$ has the same distribution as the marked block containing 1 in X, i.e. $(B_1(t), \log V_1(t),t\geq 0)$. For fixed $n\in\mathbb{N}$ and $x\in\mathcal{M}_{n}$, recall that

\begin{equation*}X_{|[n]} \text{ under } \mathbb{P}_{x} \overset{(d)}{=} \textrm{ssFrag}_0(x,X^{(\cdot)}),\end{equation*}

where $X^{(\cdot)}$ is an independent i.i.d. sequence of copies of X started from $(\textbf{1},1)$. Using the same notation, for any $A\subset [n]$ with $1\in A$ and $v>0$, the law of the process $(B_1(t)\cap[n],\log V_1(t),t\geq 0)$ started from $(A,\log v)$ can be deduced from that of $X^{(1)}=(\Pi^{(1)},\textbf{V}^{(1)})$. More precisely, $\log V_1(t)$ behaves as a Lévy process with characteristic exponent $\psi_n$ started from $\log v$, until an independent time $T_n\sim Exp(J_n)$ when $\Pi^{(1)}_{|[n]}$ first jumps. At that time, $D_n^{(1)}=(\Pi^{(1)}(T_n),\textbf{V}^{(1)}(T_n)/V_1(T_n-))_{|[n]}$ is independently drawn according to $\mathcal{D}_n$, and then, writing $D_n^{(1)}=(\pi,\textbf{v})$,

\begin{equation*}\log V_1(T_n) = \log V_1(T_n-)+\log v_1 \quad\text{ and }\quad B_1(T_n) = B_1(T_n-)\cap A_1=A\cap A_1,\end{equation*}

where $A_1$ is the block of $\pi$ containing 1. Note that there is a non-zero probability that $B_1(T_n) = B_1(T_n-)$ (even with $v_1>0$) when $A\neq [n]$. Now in our construction, if $(t_1,x)$ is the first atom of $\mathcal{N}$ such that $\pi\neq \textbf{1}_n$ or $v_1 = 0$, where $x=(\pi,\textbf{v})$, then $t_1$ is exponentially distributed with parameter $\mathcal{D}(\pi_{|[n]}\neq \textbf{1}_n \text{ or }v_1 = 0) = J_n$, and x is independent of $t_1$, distributed as

\begin{equation*}\frac{1}{J_n}\mathcal{D}(\cdot \cap \{\pi\neq \textbf{1}_n \text{ or }v_1 = 0\} ),\end{equation*}

so that $x_{|[n]}$ has distribution $\mathcal{D}_n$. It remains only to show that $(\xi(s),0\leq s< t_1)$ is distributed as a Lévy process with characteristic exponent $\psi_n$ (killed at $t_1$). The point process of its jumps is

\begin{equation*}\mathcal{N}'\cap [0,t_1)\times \mathbb{R},\end{equation*}

which conditional on $t_1$ has intensity

\begin{equation*}\textrm{d} t\otimes\Big(\mathcal{D}\big(\{\log v_1\in\cdot\}\cap\{\pi_{|[n]} = \textbf{1}_n\text{ and }v_1\notin \{0,1\}\}\big)+\lambda_\infty\Big).\end{equation*}

Let us show that this intensity is equal to $\textrm{d} t\otimes\lambda_n$. Note that

\begin{align*}&\mathcal{D}\big(\{\log v_1\in\cdot\}\cap\{\pi_{|[n]} = \textbf{1}_n\text{ and }v_1\notin \{0,1\}\}\big)\\[3pt]&= \sum_{m\geq n}\mathcal{D}\big(\{\log v_1\in\cdot\}\cap\{\pi_{|[m]} = \textbf{1}_m\text{ and }v_1\notin \{0,1\}\}\cap\{\pi_{|[m+1]} \neq \textbf{1}_{m+1}\}\big)\\[3pt]&= \sum_{m\geq n}J_{m+1}\mathcal{D}_{m+1}\big(\{\log v_1\in\cdot\}\cap\{\pi_{|[m]} = \textbf{1}_m\text{ and }v_1\notin \{0,1\}\}\big)\\[3pt]&= \sum_{m\geq n}(J_{m+1}-J_m) \widetilde{\eta}_{m+1},\end{align*}

so (15) shows that

(21) \begin{equation} \lambda_n=\lambda_\infty+\mathcal{D}\big(\{\log v_1\in\cdot\}\cap\{\pi_{|[n]} = \textbf{1}_n\text{ and }v_1\notin \{0,1\}\}\big);\end{equation}

therefore the Lévy measure of $(\xi(s),0\leq s< t_1)$ is indeed $\lambda_n$. In the end, we have shown that

\begin{equation*}(B(t),\xi(t),t\geq 0)\overset{(d)}{=} (B_1(t), \log V_1(t),t\geq 0).\end{equation*}

From this construction, we see that for each atom $(t,x)\in \mathcal{N}$, the process B jumps, with

\begin{equation*}B(t) = B(t-)\cap A,\end{equation*}

where A is the block of x containing 1. Let us show that this implies $\mathcal{D}(\mathcal{M}_{\infty}\setminus\mathcal{M}^{\star}_{\infty})=0$. Assuming the opposite, there is a non-zero probability that there is an atom $(t,x)\in \mathcal{N}$ with $t< T$ such that x contains a finite block with mark not equal to zero. By exchangeability of $\mathcal{D}$, and from the description of the jumps of B, there is a non-zero probability that there is a jump $B(t) = B(t-) \cap A$ where A is finite and $t < T$. This contradicts the assumption of non-degeneracy of X, as then we would have $X(t)\in \mathcal{M}_{\infty}\setminus\mathcal{M}^{\star}_{\infty}$.

From now on, we view $\mathcal{D}$ as an exchangeable measure on $\mathcal{M}^{\star}_{\infty}$, satisfying (17) and the $\sigma$-finiteness assumption

\begin{equation*}\forall n\in\mathbb{N}, \quad \mathcal{D}\big(\pi_{|[n]}\neq \textbf{1}_n\text{ or } v_1=0\big) = J_n < \infty.\end{equation*}

It remains essentially to study $\mathcal{D}$ in order to express it as a mixture of paintbox processes.

Step 3. Let us decompose

\begin{equation*}\mathcal{D} = \mathcal{D}\big(\cdot\cap\{\pi=\textbf{1}\}\big)+\mathcal{D}\big(\cdot\cap\{\pi\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}=\textbf{1}\}\big)+\mathcal{D}\big(\cdot\cap\{\pi\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big)\\\end{equation*}

and show that there exist a constant $c\geq 0$ and a $\sigma$-finite measure $\Lambda'$ on $\mathscr{Z}^\downarrow$ such that

1. (a) $\mathcal{D}\big(\cdot\cap\{\pi=\textbf{1}\}\big)=\mathcal{D}(\pi=\textbf{1})\delta_{(\textbf{1},0)};$

2. (b) $\displaystyle\mathcal{D}\big(\cdot\cap\{\pi\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}=\textbf{1}\}\big)=c \sum_{n\in\mathbb{N}}\delta_{\mathfrak{e}_n},$ where

   \begin{equation*} \mathfrak{e}_n \,{:\!=}\, \Big(\big\{\{n\},\mathbb{N}\setminus\{n\}\big\},\, (1,\ldots,1,\underbrace{0}_{n-\text{th index}},1,\ldots) \Big) \in \mathcal{M}^{\star}_{\infty}; \end{equation*}

3. (c) $\displaystyle\mathcal{D}\big(\cdot\cap\{\pi\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big)= \int_{\mathscr{Z}^\downarrow}\varrho_{\textbf{z}}(\cdot)\,\Lambda'(\textrm{d} \textbf{z}).$

We use similar arguments as in [5, Theorem 3.1], as we have already done in the context of nested fragmentations [Reference Duchamps13, Proposition 19]. First note that by (17), $\mathcal{D}$-a.e. on the event $\{\pi=\textbf{1}\}$ we have $x=(\textbf{1},0)$; in other words 1 holds. Note also that $\mathcal{D}(\pi=\textbf{1})\leq \mathcal{D}(v_1=0)=J_1<\infty$.

Let us now study the measure $\mathcal{D}(\cdot\cap\{\pi\neq \textbf{1}\})$. Note that $\mathcal{D}(\{\pi\in\cdot\}\cap\{\pi\neq \textbf{1}\})$ is an exchangeable measure on $\mathcal{P}_{\infty}$ satisfying, for all $n\geq 1$,

\begin{equation*}\mathcal{D}(\pi_{|[n]}\neq \textbf{1}_n) < \infty.\end{equation*}

A consequence of [5, Theorem 3.1] is that $\pi$ has asymptotic frequencies $\mathcal{D}$-a.e.–recall that $\lvert {\pi} \rvert^{\downarrow}\in\mathscr{S}^\downarrow\subset[0,1]^{\mathbb{N}}$ denotes the nonincreasing reordering of asymptotic frequencies of blocks of $\pi$ – and one can write

\begin{equation*}\mathcal{D}(\{\pi\in\cdot\}\cap\{\pi\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow} = (1,0,0,\ldots)\}) = c \sum_{n\in\mathbb{N}}\delta_{\{\{n\},\mathbb{N}\setminus\{n\}\}},\end{equation*}

where $c\geq 0$. For conciseness—and again with some abuse of notation—we will from now on let $\textbf{1}\,{:\!=}\,(1,0,0\ldots)\in\mathscr{S}^\downarrow$. Now let us examine the distribution of $x=(\pi,\textbf{v})$ on the event $\big\{\pi=\{\{n\},\mathbb{N}\setminus\{n\}\}\big\}$. Since $x\in\mathcal{M}^{\star}_{\infty}$ $\mathcal{D}$-a.e., the singleton block must have mark 0, while the other block may have a positive mark. Let $\eta$ be the distribution of this mark on $S^1$, that is,

\begin{equation*}\eta \,{:\!=}\, \mathcal{D}\big(\{v_1\in\cdot\}\cap\big\{\pi=\{\{n\},\mathbb{N}\setminus\{n\}\}\big\}\big),\end{equation*}

which is a measure of total mass c, for any fixed $n>1$ (by exchangeability, $\eta$ does not depend on the value of n). First, note that $\eta(\{0\}) = 0$. Indeed, since the events

\begin{equation*}\big\{\pi=\{\{n\},\mathbb{N}\setminus\{n\}\}\big\}, \qquad n > 1,\end{equation*}

are disjoint, the following holds:

\begin{align*}\sum_{n>1} \eta(\{0\}) &= \sum_{n>1}\mathcal{D}\big(\big\{v_1=0\text{ and }\pi=\{\{n\},\mathbb{N}\setminus\{n\}\}\big\}\big) \\&= \mathcal{D}\Big(\bigcup_{n>1}\big\{v_1=0\text{ and }\pi=\{\{n\},\mathbb{N}\setminus\{n\}\}\big\}\Big)\\&\leq \mathcal{D}(v_1=0)\\&=J_1 < \infty.\end{align*}

This implies necessarily $\eta(\{0\}) = 0$. Now recall that $\mathcal{D}\big(\{\log v_1\in\cdot\}\cap\{v_1 \notin \{0,1\}\}\big)$ is a Lévy measure, so for all $\varepsilon > 0$,

\begin{align*}\sum_{n>1} \eta(\lvert {\log v_1} \rvert > \varepsilon) &= \sum_{n>1}\mathcal{D}\big(\big\{\lvert {\log v_1} \rvert > \varepsilon\text{ and }\pi=\{\{n\},\mathbb{N}\setminus\{n\}\}\big\}\big) \\&= \mathcal{D}\Big(\bigcup_{n>1}\big\{\lvert {\log v_1} \rvert > \varepsilon\text{ and }\pi=\{\{n\},\mathbb{N}\setminus\{n\}\}\big\}\Big)\\&\leq \mathcal{D}\big(\lvert {\log v_1} \rvert>\varepsilon\text{ and }v_1 \neq 0\big) < \infty,\end{align*}

which implies necessarily $\eta(\lvert {\log v_1} \rvert > \varepsilon) = 0$. Letting $\varepsilon\to 0$, we have $\eta(\lvert {\log v_1} \rvert > 0) = 0$, so in the end $\eta = c\delta_1$, and 2 follows.

Let us now decompose the measure $\mathcal{D}(\cdot\cap\{\lvert {\pi} \rvert^{\downarrow} \neq \textbf{1}\})$. Recall that by construction,

\begin{equation*}\mathcal{D}(\cdot\cap \{\pi_{|[n]}\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\})\leq \mathcal{D}(\cdot\cap \{\pi_{|[n]}\neq \textbf{1}\text{ or }v_1=0\})=\mu_n,\end{equation*}

which is a finite measure with total mass $J_n$. Now let us introduce the injection $\theta_n\,{:}\,\mathbb{N}\to\mathbb{N}, \, k\mapsto n+k$, and consider

\begin{equation*}\overset{\leftarrow}{\mu}_n \,{:\!=}\, \mathcal{D}\big(\{x^{\theta_n}\in\cdot\}\cap\{\pi_{|[n]}\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big),\end{equation*}

which can be seen as the distribution of the marked partition restricted to $\{n+1,n+2,\ldots\}$, on the event $\{\pi_{|[n]}\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}$. It is readily checked that this measure is exchangeable on $\mathcal{M}^{\star}_{\infty}$, and since it is finite, by Proposition 1,

\begin{equation*}\overset{\leftarrow}{\mu}_n = \int_{\mathscr{Z}^\downarrow}\varrho_{\textbf{z}}(\cdot)\,\Lambda_n(\textrm{d} \textbf{z}),\end{equation*}

with $\Lambda_n = \overset{\leftarrow}{\mu}_n(\lvert {x} \rvert^{\downarrow}\in\cdot)$ a finite measure on $\mathscr{Z}^\downarrow$. Asymptotic frequencies are such that $\lvert {x} \rvert^{\downarrow}=\lvert {x^{\theta_n}} \rvert^{\downarrow}$ for all $x\in\mathcal{M}^{\star}_{\infty}$; therefore $\Lambda_n$ can also be written

\begin{equation*}\Lambda_n = \mathcal{D}\big(\{\lvert {x} \rvert^{\downarrow}\in\cdot\}\cap\{\pi_{|[n]}\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big),\end{equation*}

and taking nondecreasing limits one can define

\begin{equation*}\Lambda' \,{:\!=}\, \mathcal{D}\big(\{\lvert {x} \rvert^{\downarrow}\in\cdot\}\cap\{\pi\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big).\end{equation*}

To show C, fix $k,n\in\mathbb{N}$ and consider the permutation $\tau\,{:}\,\mathbb{N}\to\mathbb{N}$ given by

\begin{equation*}\tau(i) = \begin{cases}i+k&\text{ if }i\leq n,\\i-n&\text{ if }n<i\leq n+k,\\i&\text{ otherwise.}\end{cases}\end{equation*}

Now notice that

\begin{equation*}\mathcal{D}\big(\{x_{|[k]}\in\cdot\}\cap\{\pi\neq \textbf{1}\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big) = \lim_{n\to\infty}\mathcal{D}\big(\{x_{|[k]}\in\cdot\}\cap\{(\pi^{\theta_k})_{|[n]}\neq \textbf{1}_n\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big),\end{equation*}

which can be written, using the exchangeability of $\mathcal{D}$, as

\begin{align*}&\mathcal{D}\big(\{x_{|[k]}\in\cdot\}\cap\{(\pi^{\theta_k})_{|[n]}\neq \textbf{1}_n\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big)\\[3pt]&=\mathcal{D}\big(\{(x^{\tau\circ\theta_n})_{|[k]}\in\cdot\}\cap\{(\pi^{\tau})_{|[n]}\neq \textbf{1}_n\text{ and }\lvert {\pi^{\tau}} \rvert^{\downarrow}\neq \textbf{1}\}\big)\\[3pt]&=\mathcal{D}\big(\{(x^{\theta_n})_{|[k]}\in\cdot\}\cap\{\pi_{|[n]}\neq \textbf{1}_n\text{ and }\lvert {\pi} \rvert^{\downarrow}\neq \textbf{1}\}\big)\\[3pt]&=\int_{\mathscr{Z}^\downarrow}\varrho^k_{\textbf{z}}(\cdot)\,\Lambda_n(\textrm{d} \textbf{z}),\end{align*}

where $\varrho_{\textbf{z}}^{k}$ is the paintbox process restricted to k elements defined in Section 1.2. Taking limits and because k is generic, we indeed have C.

Step 4. It remains to define the measure $\Lambda$ correctly and we will be able to complete the proof of Theorem 1. Recall the definition of $\widetilde{\lambda}_\infty$ as the push-forward of $\lambda_\infty$ by the map $y \in\mathbb{R} \mapsto (\textbf{1},\textrm{e}^{y}) \in\mathcal{M}^{\star}_{\infty}$, and note that

\begin{equation*}\widetilde{\lambda}_\infty = \int_{\mathscr{Z}^\downarrow}\varrho_{\textbf{z}}(\cdot)\,\widehat{\lambda}_\infty(\textrm{d} \textbf{z}),\end{equation*}

where $\widehat{\lambda}_\infty$ is the push-forward of $\lambda_\infty$ by the map $y \in\mathbb{R} \mapsto (\textbf{1},\textrm{e}^{y}) \in\mathscr{Z}^\downarrow$. Finally, let us define

\begin{equation*}\Lambda = \Lambda'+\mathcal{D}(\pi=\textbf{1})\delta_{(\textbf{1},0)}+\widehat{\lambda}_\infty.\end{equation*}

Putting everything together, we have

\begin{equation*}\mathcal{D}+\widetilde{\lambda}_\infty = c \sum_{n\in\mathbb{N}}\delta_{\mathfrak{e}_n}+\int_{\mathscr{Z}^\downarrow}\varrho_{\textbf{z}}(\cdot)\,\Lambda(\textrm{d} \textbf{z}).\end{equation*}

We can now almost complete the proof. First, fix $n\in\mathbb{N}$ and recall that by definition $J_n = \mathcal{D}(\pi_{|[n]}\neq\textbf{1}_n\text{ or }v_1=0)$. Since by definition $\widetilde{\lambda}_\infty(\pi_{|[n]}\neq\textbf{1}_n\text{ or }v_1=0)=0$ for all n, Part (ii) of Theorem 1 is proven:

\begin{equation*}J_n = nc + \int_{\mathscr{Z}^\downarrow}\Big (1-\sum_{\substack{i\geq 1\\ v_i>0}}s_i^n\Big )\,\Lambda(\textrm{d} \textbf{z}).\end{equation*}

This implies that

\begin{equation*}\big(1-\sum_{\substack{i\geq 1\\ v_i>0}}s_i^n\big)\end{equation*}

is $\Lambda$-integrable for any $n\geq 1$. Furthermore, note that for any $\textbf{z}=(\textbf{s},\textbf{v})\in\mathscr{Z}^\downarrow$,

\begin{equation*}1-s_1\leq 1-s_1\Big(\sum_{\substack{i\geq 1\\v_i>0}}s_i\Big)\leq 1-\sum_{\substack{i\geq 1\\v_i>0}}s_i^2, \quad \text{and} \quad s_1\unicode{x1D7D9}_{v_1=0} \leq 1-\sum_{\substack{i\geq 1\\v_i>0}}s_i;\end{equation*}

therefore summing the two expressions yields

(22) \begin{equation}\int_{\mathscr{Z}^\downarrow}(1-s_1\unicode{x1D7D9}_{v_1>0})\,\Lambda(\textrm{d} \textbf{z}) < \infty.\end{equation}

We keep this in mind for later use and go back to the expression for the measures $\mathcal{D}_n$. If $J_n>0$, then $\mathcal{D}_n=\mathcal{D}\big(\{x_{|[n]}\in\cdot\}\cap\{\pi_{|[n]}\neq\textbf{1}_n\text{ or }v_1=0\}\big)/J_n$, so Part (iii) of Theorem 1 is proven:

\begin{equation*}\mathcal{D}_n=\frac{1}{J_n}\Big(\sum_{i=1}^{n}c\delta_{\mathfrak{e}^{n}_i} + \int_{\mathscr{Z}^\downarrow}\varrho^{n}_{\textbf{z}}\big(\cdot\cap\{\pi\neq\textbf{1}_n\text{ or }v_1=0\}\big)\,\Lambda(\textrm{d}\textbf{z})\Big),\end{equation*}

where $\mathfrak{e}^{n}_i\in\mathcal{M}_{n}$ is defined as

\begin{equation*}\mathfrak{e}^{n}_i \,{:\!=}\, \Big(\big\{[n]\setminus\{i\},\{i\}\big\},\, (1,\ldots,1,\underbrace{0}_{i-\text{th index}},1,\ldots,1) \Big).\end{equation*}

It remains for the first part of the theorem to express $\psi_n$ correctly and to show the integrability condition (4). Recall from (21) that