Proof of Existence, Uniqueness, Causality, and Measurability of the Open-Loop Aggregate Solution

Throughout, $c=(\sigma,\gamma)$ denotes a generic transition label (there are $l(l-1)$ of them), and $\mathbb{Q}$ the rational numbers, a countable set which is dense in the real line (claim 1 there: every nonempty open interval contains a rational). For $x\in\mathbb{G}_N$ and $c=(\sigma,\gamma)$ put $\phi^{c}_x(s)=N\,x^{\sigma}\,\beta(\sigma,\gamma,x,a_s)$ for $s\in[0,T]$. We use the restricted Lebesgue measure $\lambda_{[0,T]}$ on $[0,T]$ and, for $t\in[0,T]$ and a bounded measurable $f:[0,T]\to\mathbb{R}$, we write $\int_{[0,t]}f\,ds$ for the Lebesgue integral of the restriction of $f$ to $[0,t]$; by claim 2 of the toolkit this equals the integral over $[0,T]$ of $\mathbf{1}_{[0,t]}f$, since both coincide with the integral over $\mathbb{R}$ of the common zero extension.

Step 0 (Preliminaries).

(P1) Counting paths. Let $q$ be a counting path and $j\ge1$ a natural number. For $u\ge0$ one has $\tau_j(q)\le u$ if and only if $q(u)\ge j$: if $q(u)\ge j$ then $u$ belongs to the set defining $\tau_j(q)$; conversely, if $\tau_j(q)\le u$ then for every $s>u$ there is $t$ with $\tau_j(q)\le t<s$ and $q(t)\ge j$ (by the definition of the greatest lower bound), whence $q(s)\ge j$ by monotonicity, and right-continuity gives $q(u)\ge j$. Consequently $q(u)\le j-1$ for $u<\tau_j(q)$, and if $\tau_j(q)<\infty$ then $q(\tau_j(q))=j$: indeed $q(\tau_j(q))\ge j$, while $q(\tau_j(q)-)\le j-1$ and the unit-jump condition give $q(\tau_j(q))\le j$. Moreover, if $u\ge0$ and $j=q(u)+1$, then $\tau_j(q)>u$, since $q(u)<j$. Finally, $\tau_j(q)$ is a jump time of $q$ when finite: $\tau_j(q)>0$, since $\tau_j(q)\le0$ would give $q(0)\ge j\ge1$, contradicting $q(0)=0$; and $q(\tau_j(q))=j>j-1\ge q(\tau_j(q)-)$.

(P2) Integrands. For $x\in\mathbb{G}_N$ and $c=(\sigma,\gamma)$ the map $\alpha\mapsto\beta(\sigma,\gamma,x,\alpha)$ is sequentially continuous on $\mathcal{A}$ (condition 2 of Transition-Rate Family with the constant sequence $\Sigma_n=x$), and the components of $a$ are measurable, so $s\mapsto\beta(\sigma,\gamma,x,a_s)$ is measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable; multiplying by the constant $Nx^{\sigma}$ (claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions) shows that $\phi^{c}_x$ is measurable, and $0\le\phi^{c}_x\le NB$ because $0\le x^{\sigma}\le1$ and $0\le\beta\le B$. Products of $\phi^{c}_x$ with indicators of measurable subsets of $[0,T]$ are measurable by claims 1 and 3 of the same lemma.

(P3) Indefinite integrals. Let $f:[0,T]\to[0,NB]$ be measurable and put $F(t)=\int_{[0,t]}f\,ds$. For $0\le s\le t\le T$, $F(t)-F(s)=\int_{[0,T]}\mathbf{1}_{(s,t]}f\,ds$ by linearity, and $0\le\mathbf{1}_{(s,t]}f\le NB\,\mathbf{1}_{(s,t]}$ gives, by monotonicity and the value $\lambda_{[0,T]}((s,t])=t-s$ of the restricted measure on an interval (claim 1 of the toolkit and claim 4 of Existence of Lebesgue Measure on the Real Line, the integral of an indicator being the measure of its set by Simple Function and Its Integral), that $0\le F(t)-F(s)\le NB(t-s)$. Hence $F$ is nondecreasing, and continuous on $[0,T]$. Moreover, if $g:[0,T]\to[0,NB]$ is measurable, $t\in[0,T]$, and $\{f\neq g\}\cap[0,t]$ is contained in a set $Z$ with $\lambda_{[0,T]}(Z)=0$, then $\int_{[0,u]}f=\int_{[0,u]}g$ for every $u\in[0,t]$: the functions $\mathbf{1}_{[0,u]}f$ and $\mathbf{1}_{[0,u]}g$ agree off $Z$, so claim 6 of the toolkit (applied on the co-null set $[0,T]\setminus Z$) gives equal integrals over $[0,T]$. In particular this holds when $\{f\neq g\}\cap[0,t]$ is finite, a finite set being a Borel set of measure $0$ by claim 4 of Existence of Lebesgue Measure on the Real Line (a single point is the interval $[s,s]$) and finite additivity of the measure.

Step 1 (Claim 1). Suppose the recursion has produced $\theta_k\in[0,T)$, $x^{(k)}\in\mathbb{G}_N$ and levels $\kappa^{c}_k\ge0$. The integrand defining $\mathsf{C}^{c,(k)}_t$ is $\mathbf{1}_{(\theta_k,T]}\phi^{c}_{x^{(k)}}$, measurable with values in $[0,NB]$ by (P2), so by (P3) the map $t\mapsto\mathsf{C}^{c,(k)}_t$ is nondecreasing, satisfies $|\mathsf{C}^{c,(k)}_t-\mathsf{C}^{c,(k)}_s|\le NB(t-s)$, and is continuous; and $\mathsf{C}^{c,(k)}_{\theta_k}=\kappa^{c}_k$ because the integrand vanishes on $[0,\theta_k]$ (applied at step $k+1$, this is the identity $\mathsf{C}^{c,(k+1)}_{\theta_{k+1}}=\kappa^{c}_{k+1}$ of claim 1). By (P1), $\lambda^{c}_k>\kappa^{c}_k$. For each label $c$ let $h^{c}_k$ be the greatest lower bound of $H^{c}_k=\{t\in[\theta_k,T]:\mathsf{C}^{c,(k)}_t\ge\lambda^{c}_k\}$ (equal to $+\infty$ if $H^{c}_k=\emptyset$), so that $\theta_{k+1}=\min(T,\min_c h^{c}_k)$. Since $\mathsf{C}^{c,(k)}$ is nondecreasing and continuous, $H^{c}_k$ is an interval of the form $[h^{c}_k,T]$ when nonempty: if $t\in H^{c}_k$ and $t\le t'\le T$ then $t'\in H^{c}_k$; and if $H^{c}_k\neq\emptyset$, choosing $t_n\in H^{c}_k$ with $t_n\to h^{c}_k$ gives $\mathsf{C}^{c,(k)}_{h^{c}_k}=\lim\mathsf{C}^{c,(k)}_{t_n}\ge\lambda^{c}_k$. Thus, for $u\in[\theta_k,T]$,

Since $\mathsf{C}^{c,(k)}_{\theta_k}=\kappa^{c}_k<\lambda^{c}_k$ and there are finitely many labels, continuity yields $\epsilon>0$ with $\mathsf{C}^{c,(k)}_t<\lambda^{c}_k$ for all $c$ and all $t\in[\theta_k,\min(T,\theta_k+\epsilon)]$, so $h^{c}_k\ge\min(T,\theta_k+\epsilon)$ for every $c$ by (1.1), and $\theta_{k+1}\ge\min(T,\theta_k+\epsilon)>\theta_k$. If $\theta_{k+1}<T$, then $\theta_{k+1}=h^{c}_k$ for some label $c$, and (1.1) with $u=h^{c}_k$ gives $c\in\mathcal{J}_k$. For any $c\in\mathcal{J}_k$ one has $h^{c}_k\le\theta_{k+1}$ by (1.1), while $\theta_{k+1}\le h^{c}_k$ by definition, so $h^{c}_k=\theta_{k+1}>\theta_k$; choosing $t_n\in[\theta_k,h^{c}_k)$ with $t_n\to h^{c}_k$ gives $\mathsf{C}^{c,(k)}_{t_n}<\lambda^{c}_k$, hence $\mathsf{C}^{c,(k)}_{\theta_{k+1}}\le\lambda^{c}_k$ by continuity, and together with $c\in\mathcal{J}_k$ this gives $\mathsf{C}^{c,(k)}_{\theta_{k+1}}=\lambda^{c}_k$. The bound $\kappa^{c}_{k+1}\le NB\theta_{k+1}$ follows by induction from $\kappa^{c}_0=0$ and $\kappa^{c}_{k+1}-\kappa^{c}_k=\mathsf{C}^{c,(k)}_{\theta_{k+1}}-\mathsf{C}^{c,(k)}_{\theta_k}\le NB(\theta_{k+1}-\theta_k)$.

Counting identity. We claim that for every step $k$ at which the recursion has not stopped and every label $c$,

Indeed, write $j=p^{c}(\kappa^{c}_k)+1$, so $\lambda^{c}_k=\tau_j(p^{c})$. If $c\in\mathcal{J}_k$ then $\kappa^{c}_{k+1}=\lambda^{c}_k$ is finite and $p^{c}(\kappa^{c}_{k+1})=j$ by (P1). If $c\notin\mathcal{J}_k$ then $\kappa^{c}_{k+1}=\mathsf{C}^{c,(k)}_{\theta_{k+1}}<\lambda^{c}_k$, so $p^{c}(\kappa^{c}_{k+1})\le j-1$ by (P1), while $p^{c}(\kappa^{c}_{k+1})\ge p^{c}(\kappa^{c}_k)=j-1$ by monotonicity and $\kappa^{c}_{k+1}\ge\kappa^{c}_k$. Summing (1.2) over the steps, $S_k=\sum_{c}p^{c}(\kappa^{c}_k)$ satisfies $S_{k+1}\ge S_k+1$ whenever $\mathcal{J}_k\neq\emptyset$, in particular whenever $\theta_{k+1}<T$. Since $\kappa^{c}_k\le NB\theta_k\le NBT$, monotonicity gives $S_k\le M:=\sum_{c}p^{c}(NBT)$. If the recursion has not stopped at step $k$, then $\theta_1,\dots,\theta_k<T$, so $S_k\ge S_0+k=k$, whence $k\le M$. Therefore the recursion stops at some $K\le M+1$.

Step 2 (Claim 2). Uniqueness and necessity of conflict-freeness. Let $\Sigma$ be an open-loop aggregate solution for $(p,a,x_0)$, with consumed clock times $\mathsf{C}^{c}$ and counters $\mathsf{N}^{c}$. We show by induction on $k$ that, as long as $\theta_k$ is defined (the recursion not having stopped before step $k$): (i) if $\theta_k<T$ then $\Sigma_t=x^{(k)}$ for $t\in[\theta_k,\theta_{k+1})$ and $\mathsf{C}^{c}_t=\mathsf{C}^{c,(k)}_t$ for $t\in[\theta_k,\theta_{k+1}]$ and all $c$, while if $\theta_k=T$ then $\Sigma_T=x^{(k)}$; and (ii) $x^{(k+1)}\in\mathbb{G}_N$ whenever $\theta_{k+1}$ is defined. The starting point is $\mathsf{C}^{c}_0=0=\kappa^{c}_0$ and $\Sigma_0=x_0=x^{(0)}$, the latter from the state identity and $\mathsf{N}^{c}_0=p^{c}(0)=0$.

Assume the recursion has reached step $k$ with $\theta_k<T$ and $x^{(k)}\in\mathbb{G}_N$, and that $\mathsf{C}^{c}_{\theta_k}=\kappa^{c}_k$ for all $c$ and $\Sigma_{\theta_k}=x^{(k)}$ (for $k=0$ this was just checked; for $k\ge1$ it is part of the inductive conclusion below). Let $t^{*}$ be the least upper bound of the set of $t\in[\theta_k,T]$ such that $\Sigma_s=x^{(k)}$ for all $s\in[\theta_k,t]$; this set contains $\theta_k$, and by condition 1 of the definition (piecewise constancy on left-closed, right-open pieces $[t_i,t_{i+1})$ and on $[t_n,T]$, so that each piece contains a right neighbourhood, relative to $[0,T]$, of each of its points) it contains $[\theta_k,\theta_k+\epsilon]$ for some $\epsilon>0$ with $\theta_k+\epsilon\le T$, so $t^{*}>\theta_k$ and $\Sigma_s=x^{(k)}$ for $s\in[\theta_k,t^{*})$. For $t\in[\theta_k,t^{*})$ the integrand $s\mapsto N\Sigma^{\sigma}_s\beta(\sigma,\gamma,\Sigma_s,a_s)$ defining $\mathsf{C}^{c}$ in the definition (measurable with values in $[0,NB]$, as noted there) agrees on $(\theta_k,t]$ with $\phi^{c}_{x^{(k)}}$, so, splitting the integral over $[0,t]$ as in (P3),

and by continuity of both sides in $t$ ((P3) applied to the integrand of the definition, and Step 1) the identity $\mathsf{C}^{c}_t=\mathsf{C}^{c,(k)}_t$ extends to $t=t^{*}$. Hence $\mathsf{N}^{c}_t=p^{c}(\mathsf{C}^{c,(k)}_t)$ for $t\in[\theta_k,t^{*}]$. Suppose, for contradiction, that $t^{*}<\theta_{k+1}$. For $t\in[\theta_k,t^{*}]$ and every $c$ we have $t<\theta_{k+1}\le h^{c}_k$, so $\mathsf{C}^{c,(k)}_t<\lambda^{c}_k$ by (1.1), and (P1) with monotonicity gives $p^{c}(\mathsf{C}^{c,(k)}_t)=p^{c}(\kappa^{c}_k)$, i.e. $\mathsf{N}^{c}_t=\mathsf{N}^{c}_{\theta_k}$; the state identity then gives $\Sigma_t=\Sigma_{\theta_k}=x^{(k)}$ for all $t\in[\theta_k,t^{*}]$, in particular at $t^{*}<T$, and condition 1 of the definition (the piece containing $t^{*}$ contains a right neighbourhood of it) yields $\epsilon>0$ with $\Sigma=x^{(k)}$ on $[t^{*},t^{*}+\epsilon]$, contradicting the definition of $t^{*}$. Hence $t^{*}\ge\theta_{k+1}$: $\Sigma_t=x^{(k)}$ on $[\theta_k,\theta_{k+1})$ and $\mathsf{C}^{c}_t=\mathsf{C}^{c,(k)}_t$ on $[\theta_k,\theta_{k+1}]$, so $\mathsf{C}^{c}_{\theta_{k+1}}=\kappa^{c}_{k+1}$. By the counting identity (1.2) applied at all steps $i\le k$, $p^{c}(\kappa^{c}_{k+1})=\#\{i\le k:c\in\mathcal{J}_i\}$, so the state identity at $\theta_{k+1}$ reads

and $\Sigma_{\theta_{k+1}}\in\mathbb{G}_N$ by condition 1, so $x^{(k+1)}\in\mathbb{G}_N$. This establishes the inductive step, including the hypotheses $\mathsf{C}^{c}_{\theta_{k+1}}=\kappa^{c}_{k+1}$ and $\Sigma_{\theta_{k+1}}=x^{(k+1)}$ for the next step. The induction shows that the recursion never stops through $x^{(k)}\notin\mathbb{G}_N$, hence stops with $\theta_K=T$: the data are conflict-free, $\Sigma=\Sigma^{\mathrm{rec}}$ on $[0,T)$, and $\Sigma_T=x^{(K)}=\Sigma^{\mathrm{rec}}_T$. Any two solutions therefore coincide, and a solution exists only if the data are conflict-free; the identities $\mathsf{C}^{c}_t=\mathsf{C}^{c,(k)}_t$ on $[\theta_k,\theta_{k+1}]$ were obtained along the way.

Existence for conflict-free data. Assume the data are conflict-free, so that $\theta_0<\theta_1<\dots<\theta_K=T$ and every $x^{(k)}\in\mathbb{G}_N$; put $\Sigma=\Sigma^{\mathrm{rec}}$. Here $K\ge1$, since $\theta_0=0<T=\theta_K$, and $0<\theta_1<\dots<\theta_K=T$ by Step 1. Condition 1 of the definition holds with $n=K$ and $t_i=\theta_i$: $\Sigma$ takes values in $\mathbb{G}_N$ and is constant on $[0,\theta_1)$, on each $[\theta_i,\theta_{i+1})$, and on $[\theta_K,T]=\{T\}$. The integrand $s\mapsto N\Sigma^{\sigma}_s\beta(\sigma,\gamma,\Sigma_s,a_s)$ defining the consumed clock times equals $\sum_{k<K}\mathbf{1}_{[\theta_k,\theta_{k+1})}\phi^{c}_{x^{(k)}}+\mathbf{1}_{\{T\}}\phi^{c}_{x^{(K)}}$, which is measurable by (P2) and claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; let $\mathsf{C}^{c}_t$ be its integral over $[0,t]$. We show $\mathsf{C}^{c}_t=\mathsf{C}^{c,(k)}_t$ for $t\in[\theta_k,\theta_{k+1}]$ by induction on $k$. For $t\in[\theta_k,\theta_{k+1}]$, by linearity, $\mathsf{C}^{c}_t=\mathsf{C}^{c}_{\theta_k}+\int_{[0,T]}\mathbf{1}_{(\theta_k,t]}\,N\Sigma^{\sigma}_s\beta(\sigma,\gamma,\Sigma_s,a_s)\,ds$, and on $(\theta_k,t]$ the integrand differs from $\phi^{c}_{x^{(k)}}$ at most at the single point $t=\theta_{k+1}$, so by (P3) the integral equals $\int_{[0,t]}\mathbf{1}_{(\theta_k,T]}\phi^{c}_{x^{(k)}}\,ds$; with the inductive hypothesis $\mathsf{C}^{c}_{\theta_k}=\mathsf{C}^{c,(k-1)}_{\theta_k}=\kappa^{c}_k$ (for $k=0$, $\mathsf{C}^{c}_0=0=\kappa^{c}_0$) this gives $\mathsf{C}^{c}_t=\mathsf{C}^{c,(k)}_t$. Condition 2 (state identity): with the counters $\mathsf{N}^{c}_t=p^{c}(\mathsf{C}^{c}_t)$ of the definition, for $t\in[\theta_k,\theta_{k+1})$, (1.1) and $t<h^{c}_k$ give $\mathsf{C}^{c,(k)}_t<\lambda^{c}_k$, so $\mathsf{N}^{c}_t=p^{c}(\kappa^{c}_k)=\#\{i<k:c\in\mathcal{J}_i\}$ by (P1) and (1.2), and the state identity $x_0+\frac1N\sum_c v_c\mathsf{N}^{c}_t=x^{(k)}=\Sigma_t$ follows as displayed above; at $t=T=\theta_K$ the same computation with $\mathsf{N}^{c}_T=p^{c}(\kappa^{c}_K)=\#\{i<K:c\in\mathcal{J}_i\}$ gives $\Sigma_T=x^{(K)}$. Thus $\Sigma^{\mathrm{rec}}$ is a solution.

Properties (a)--(c). (a) follows from Step 1 and $\mathsf{C}^{c}_t=\mathsf{C}^{c,(k)}_t$ on the pieces, together with $\kappa^{c}_k\le NB\theta_k$: $\mathsf{C}^{c}_t\le\kappa^{c}_k+NB(t-\theta_k)\le NBt$, and the Lipschitz bound across pieces is obtained by adding the bounds on the sub-pieces. (b): $\mathsf{N}^{c}_t=p^{c}(\mathsf{C}^{c}_t)$ is nondecreasing as a composition of nondecreasing maps, takes values in $\mathbb{N}_0$, vanishes at $0$, and $\mathsf{N}^{c}_t\le p^{c}(NBt)$ by (a); for the right-continuity at $t\in[0,T)$, let $\ell$ be the greatest lower bound of $\{\mathsf{N}^{c}_s:t<s\le T\}$, so $\ell\ge\mathsf{N}^{c}_t$ by monotonicity; if $\mathsf{C}^{c}_s=\mathsf{C}^{c}_t$ for some $s>t$ then $\mathsf{N}^{c}_s=\mathsf{N}^{c}_t$ and $\ell=\mathsf{N}^{c}_t$; otherwise $\mathsf{C}^{c}_s>\mathsf{C}^{c}_t$ for all $s\in(t,T]$, and by (a) the values $\mathsf{C}^{c}_s$, $s\in(t,T]$, come arbitrarily close to $\mathsf{C}^{c}_t$ from above, so that condition 3 of Counting Path and Its Jump Times (right-continuity of $p^{c}$ at the level $\mathsf{C}^{c}_t$) together with monotonicity of $p^{c}$ gives $\ell=p^{c}(\mathsf{C}^{c}_t)=\mathsf{N}^{c}_t$. (c): by construction $\Sigma=x^{(k)}$ on $[\theta_k,\theta_{k+1})$ and $\Sigma_{\theta_{k+1}}=x^{(k+1)}$, so $\Sigma_{\theta_{k+1}}-\Sigma_s=x^{(k+1)}-x^{(k)}=\frac1N\sum_{c\in\mathcal{J}_k}v_c$ for $s\in[\theta_k,\theta_{k+1})$; for $c\in\mathcal{J}_k$, $\mathsf{C}^{c}_{\theta_{k+1}}=\mathsf{C}^{c,(k)}_{\theta_{k+1}}=\lambda^{c}_k=\tau_j(p^{c})$ with $j=p^{c}(\kappa^{c}_k)+1$ by Step 1, which is a jump time of $p^{c}$ by (P1). A time $t\in(0,T]$ with $\Sigma_t\neq\Sigma_s$ for all $s<t$ close to $t$ cannot lie in the interior of a constancy interval $[\theta_k,\theta_{k+1})$, hence is one of $\theta_1,\dots,\theta_K$; and since $x^{(k+1)}\neq x^{(k)}$ forces $\mathcal{J}_k\neq\emptyset$, which by (1.2) increases $S_k$ by at least one, there are at most $M=\sum_c p^{c}(NBT)$ such times.

Step 3 (Claim 3). We show by induction on $k$ that, as long as $\theta_k\le t$ and neither recursion has stopped before step $k$, the two recursions have the same $\theta_k$, $x^{(k)}$ and $\kappa^{c}_k$, and moreover $\theta_{k+1}\wedge t=\theta'_{k+1}\wedge t$ and $\mathsf{C}^{c,(k)}_s=\mathsf{C}'^{c,(k)}_s$ for $s\in[0,t]$. Suppose the data of step $k$ agree and $\theta_k\le t$. The integrands $\mathbf{1}_{(\theta_k,T]}\phi^{c}_{x^{(k)}}$ and $\mathbf{1}_{(\theta_k,T]}\phi'^{c}_{x^{(k)}}$ (formed with $a'$) differ on $[0,t]$ only where $a'_s\neq a_s$, a subset of a $\lambda_{[0,T]}$-null set, so (P3) gives $\mathsf{C}^{c,(k)}_s=\mathsf{C}'^{c,(k)}_s$ for $s\le t$. Since $\kappa^{c}_k\le NB\theta_k\le NBt$, $p^{c}(\kappa^{c}_k)=p'^{c}(\kappa^{c}_k)$, so the index $j$ is common; if $\lambda^{c}_k=\tau_j(p^{c})\le NBt$ then $p'^{c}(\lambda^{c}_k)=p^{c}(\lambda^{c}_k)\ge j$ and $p'^{c}(u)=p^{c}(u)\le j-1$ for $u<\lambda^{c}_k$, so $\lambda'^{c}_k=\lambda^{c}_k$ by (P1); if $\lambda^{c}_k>NBt$ then $p^{c}(u)\le j-1$ for $u\le NBt$, so also $p'^{c}(u)\le j-1$ there and $\lambda'^{c}_k>NBt$. In either case, for $s\le t$ we have $\mathsf{C}^{c,(k)}_s\le NBs\le NBt$, and (1.1) shows that $h^{c}_k\le s$ if and only if $h'^{c}_k\le s$ for every $s\in[\theta_k,t]$; hence $\theta_{k+1}\wedge t=\theta'_{k+1}\wedge t$, and if $\theta_{k+1}\le t$ then $\theta_{k+1}=\theta'_{k+1}$, $\mathcal{J}_k=\mathcal{J}'_k$ (by (1.1) at $u=\theta_{k+1}$), $x^{(k+1)}=x'^{(k+1)}$ and $\kappa^{c}_{k+1}=\mathsf{C}^{c,(k)}_{\theta_{k+1}}=\kappa'^{c}_{k+1}$. This proves the inductive assertion; in particular, whether the recursion stops at a step $k$ with $\theta_k\le t$ (through $\theta_k=T$ or $x^{(k)}\notin\mathbb{G}_N$) is decided by data common to both recursions, so if one recursion stops at an index $K$ with $\theta_K\le t$ then so does the other, with the same $x^{(K)}$ and $\kappa^{c}_K$. The recursion paths agree on $[0,t]$ because, on $[\theta_k,\theta_{k+1}\wedge t]$ for the common steps and on $[\theta_K,t]$ after a common stop, each is determined by the common data; the recursion consumed times agree on $[0,t]$ for the same reason, being $\mathsf{C}^{c,(k)}_s=\mathsf{C}'^{c,(k)}_s$ on the common pieces and the common $\kappa^{c}_K$ after a stop; and the recursion counters agree on $[0,t]$ because they are $p^{c}$, respectively $p'^{c}$, read at the common levels $\mathsf{C}^{\mathrm{rec},c}_s\le NBs\le NBt$ (Step 1), where the two clock families agree. The final assertion follows from claim 2, the solutions being the recursion paths with the recursion consumed times and counters.

Step 4 (Claim 4). Sections of $\mathsf{a}$. Fix $r\in\mathsf{R}$ and a component $\mathsf{a}^{i}$ of $\mathsf{a}$. Let $\mathrm{pt}_{r}$ be the probability measure on $(\mathsf{R},\mathcal{R})$ with $\mathrm{pt}_{r}(A)=1$ if $r\in A$ and $\mathrm{pt}_{r}(A)=0$ otherwise (a measure, countable additivity holding because a point lies in at most one member of a disjoint family). For a Borel set $E\subseteq\mathbb{R}$ the set $A=\{(s,r'):\mathsf{a}^{i}(s,r')\in E\}$ belongs to $\mathcal{B}_{[0,T]}\otimes\mathcal{R}$, so by the section clause of Tonelli and Fubini Theorems on $([0,T]\times\mathsf{R},\mathcal{B}_{[0,T]}\otimes\mathcal{R},\lambda_{[0,T]}\otimes\mathrm{pt}_{r})$ (both factors finite measures), applied to the indicator $\mathbf{1}_{A}$, the section $s\mapsto\mathbf{1}_{A}(s,r)$ is $\mathcal{B}_{[0,T]}$-measurable, i.e. $\{s:\mathsf{a}^{i}(s,r)\in E\}\in\mathcal{B}_{[0,T]}$. Hence $s\mapsto\mathsf{a}(s,r)$ is a control path.

Reduction to stopped clocks. Fix $t_0\in[0,T]$ and put $u_0=NBt_0$. Let $\tilde{\mathsf{P}}^{c}_u=\mathsf{P}^{c}_{\min(u,u_0)}$; every path of $\tilde{\mathsf{P}}^{c}$ is a counting path (it is $\mathsf{P}^{c}(\omega)$ frozen after level $u_0$), $\tilde{\mathsf{P}}^{c}_u$ is $\mathcal{H}_{t_0}$-measurable for every $u\ge0$, and $\tilde{\mathsf{P}}^{c}$ agrees with $\mathsf{P}^{c}$ on $[0,u_0]$. By Step 3, the recursion for $(\tilde{\mathsf{P}}(\omega),\mathsf{a}(\cdot,r),x_0)$ produces the same $\theta_k$, $x^{(k)}$, $\kappa^{c}_k$ as the recursion for $(\mathsf{P}(\omega),\mathsf{a}(\cdot,r),x_0)$ at all steps with $\theta_k\le t_0$, the same recursion path on $[0,t_0]$, and the same recursion consumed times on $[0,t_0]$; since these consumed times are at most $u_0$, the recursion counters on $[0,t_0]$ also agree. It therefore suffices to prove: if $\mathcal{E}\subseteq\mathcal{F}$ is a $\sigma$-algebra such that $\mathsf{Q}^{c}_u$ is $\mathcal{E}$-measurable for every $u\ge0$ and every label, where $\mathsf{Q}=(\mathsf{Q}^{c})$ is any family of processes with counting paths, then all quantities of the recursion for $(\mathsf{Q}(\omega),\mathsf{a}(\cdot,r),x_0)$, including the set of conflict-free pairs and, for every $t$, the recursion path, consumed times and counters at time $t$, are $\mathcal{R}\otimes\mathcal{E}$-measurable functions of $(r,\omega)$, and the recursion path, consumed times and counters are $\mathcal{B}_{[0,T]}\otimes(\mathcal{R}\otimes\mathcal{E})$-measurable functions of $(t,r,\omega)$, and to apply this with $\mathsf{Q}=\tilde{\mathsf{P}}$ and $\mathcal{E}=\mathcal{H}_{t_0}$ (for the statements at time $t=t_0$), with $\mathsf{Q}=\mathsf{P}$ stopped at level $NBT$ and $\mathcal{E}=\mathcal{H}_T$ (for $\mathsf{G}$), and with $\mathsf{Q}=\mathsf{P}$ and $\mathcal{E}=\mathcal{F}$ (for the joint measurability in $(t,r,\omega)$). Here measurability of a map with values in $[0,+\infty]$ or in a finite subset of $\mathbb{R}^l$ means measurability of the sets $\{\cdot\le u\}$ for real $u$, respectively of the level sets; for a real-valued map this is measurability with respect to $\mathcal{B}(\mathbb{R})$ by Generator Criterion for Measurability, the intervals $(-\infty,u]$ generating $\mathcal{B}(\mathbb{R})$; a map which is measurable on each member of a finite or countable measurable partition of its domain is measurable; and all partial integrals produced by Tonelli and Fubini Theorems below are real-valued, being bounded by $NBT$. We write $\mathcal{M}=\mathcal{R}\otimes\mathcal{E}$, fix a point $r_0\in\mathsf{R}$ (nonempty by hypothesis), and let $\mathrm{pt}_{r_0}\otimes P|_{\mathcal{E}}$ be the product measure on $\mathcal{M}$ of the probability measures $\mathrm{pt}_{r_0}$ (defined above) and the restriction of $P$ to $\mathcal{E}$, so that Tonelli and Fubini Theorems is available on $([0,T]\times(\mathsf{R}\times\Omega),\mathcal{B}_{[0,T]}\otimes\mathcal{M},\lambda_{[0,T]}\otimes(\mathrm{pt}_{r_0}\otimes P|_{\mathcal{E}}))$, all measures involved being finite.

Jump times of the clocks. For a label $c$ and $j\ge1$, $\{\tau_j(\mathsf{Q}^{c})\le u\}=\{\mathsf{Q}^{c}_u\ge j\}\in\mathcal{E}$ for $u\ge0$ by (P1), so $\omega\mapsto\tau_j(\mathsf{Q}^{c}(\omega))$ is $\mathcal{E}$-measurable; and for a $\mathcal{M}$-measurable $w:\mathsf{R}\times\Omega\to[0,\infty)$, $\{\mathsf{Q}^{c}_{w}\ge j\}=\{\tau_j(\mathsf{Q}^{c})\le w\}=\bigcap_{n\ge1}\bigcup_{q\in\mathbb{Q},\,q\ge0}\bigl(\{\tau_j(\mathsf{Q}^{c})\le q\}\cap\{q<w+1/n\}\bigr)$ is in $\mathcal{M}$ as a countable intersection of countable unions (if $\tau_j\le w$, every $n$ admits, by density, a rational $q\in[\tau_j,w+1/n)$, which is $\ge0$; conversely the right side forces $\tau_j<w+1/n$ for all $n$); hence $(r,\omega)\mapsto\mathsf{Q}^{c}_{w(r,\omega)}(\omega)$ is $\mathcal{M}$-measurable. More generally, for $\mathcal{M}$-measurable $f,g:\mathsf{R}\times\Omega\to[0,+\infty]$ the set $\{f\le g\}=\bigcap_{n\ge1}\Bigl(\bigcup_{q\in\mathbb{Q}}\bigl(\{f\le q\}\cap\{q<g+1/n\}\bigr)\cup\{g=+\infty\}\Bigr)$ is in $\mathcal{M}$ by the same argument (density of $\mathbb{Q}$ in the interval $[f,g+1/n)$ when $f\le g<+\infty$); we use this comparison principle below without further comment.