Proof of Ties in the Aggregate Recursion: Consumed Levels and First Hitting Times, Conflict-Freeness in the Absence of Ties, and the Point-Deletion Identity

Throughout, the theorem refers to Existence, Uniqueness, Causality, and Measurability of the Open-Loop Aggregate Solution, and we use freely its claim 1: at every step $k<K$ each $t\mapsto\mathsf{C}^{c,(k)}_t$ is nondecreasing and satisfies $\mathsf{C}^{c,(k)}_{\theta_k}=\kappa^{c}_k$ and $|\mathsf{C}^{c,(k)}_t-\mathsf{C}^{c,(k)}_s|\le NB(t-s)$ for $0\le s\le t\le T$ (hence is continuous); $\lambda^{c}_k>\kappa^{c}_k$; $\theta_{k+1}>\theta_k$; $\mathsf{C}^{c,(k)}_{\theta_{k+1}}=\lambda^{c}_k$ for $c\in\mathcal{J}_k$; and the recursion stops at a finite index $K$. Recall also that $\theta_{k+1}\le h^{c}_k$ for every label $c$ (by the definition $\theta_{k+1}=\min(T,\min_ch^{c}_k)$), and that $\kappa^{c}_{k+1}=\mathsf{C}^{c,(k)}_{\theta_{k+1}}$. Two elementary facts about a counting path $q$ (Counting Path and Its Jump Times) are used: (F1) for a natural number $j$, the set $\{t\ge0:q(t)\ge j\}$ is, if nonempty, the interval $[\tau_j(q),\infty)$ — it is an up-set by monotonicity, contains its greatest lower bound $\tau_j(q)$ by right-continuity ($q(\tau_j)$ is the greatest lower bound of the values $q(s)\ge j$, $s>\tau_j$), and so equals $[\tau_j(q),\infty)$; (F2) for a fixed step $k$ and label $c$, the hitting time $h^{c}_k$ is nondecreasing as a function of the next jump level it is computed from: if $\lambda\le\lambda'$ then $\{t\in[\theta_k,T]:\mathsf{C}^{c,(k)}_t\ge\lambda'\}\subseteq\{t\in[\theta_k,T]:\mathsf{C}^{c,(k)}_t\ge\lambda\}$, so the greatest lower bound of the former is at least that of the latter. Finally, since $\mathsf{C}^{c,(k)}$ is continuous and nondecreasing on $[\theta_k,T]$, the set $\{t\in[\theta_k,T]:\mathsf{C}^{c,(k)}_t\ge\lambda^{c}_k\}$ is, if nonempty, the interval $[h^{c}_k,T]$ (its greatest lower bound $h^{c}_k$ belongs to it by continuity, as the limit of a sequence in the set), so that (F3) $\mathsf{C}^{c,(k)}_t<\lambda^{c}_k$ for $t\in[\theta_k,T]$ with $t<h^{c}_k$, and $\mathsf{C}^{c,(k)}_{h^{c}_k}\ge\lambda^{c}_k$ when $h^{c}_k\le T$.

Step 1 (claim 1). Fix a label $c=(\sigma,\gamma)$. By the definition of the recursion consumed times, $\mathsf{C}^{\mathrm{rec},c}$ coincides on $[\theta_k,\theta_{k+1}]$ with $\mathsf{C}^{c,(k)}$ for $k<K$ and is constant equal to $\kappa^{c}_K$ on $[\theta_K,T]$; on each such interval it is nondecreasing and $NB$-Lipschitz by claim 1 of the theorem, and the pieces agree at the junction times $\theta_{k+1}$ (as noted in the theorem). Hence $\mathsf{C}^{\mathrm{rec},c}$ is nondecreasing on $[0,T]$, and for $s\le t$ lying in consecutive pieces $[\theta_{k},\theta_{k+1}]$, $\dots$, $[\theta_{k'},\theta_{k'+1}]$ (and possibly the terminal piece $[\theta_K,T]$, on which the map is constant) the Lipschitz bound follows by adding the bounds over the pieces (the intermediate junction times $\theta_{k+1},\dots,\theta_{k'}$ lying between $s$ and $t$); $\mathsf{C}^{\mathrm{rec},c}_0=\mathsf{C}^{c,(0)}_{\theta_0}=\kappa^{c}_0=0$. For $k<K$, $\kappa^{c}_{k+1}=\mathsf{C}^{c,(k)}_{\theta_{k+1}}\ge\mathsf{C}^{c,(k)}_{\theta_k}=\kappa^{c}_k$ by monotonicity, and by induction $\mathsf{C}^{\mathrm{rec},c}_t\le\kappa^{c}_{i+1}\le\kappa^{c}_{k+1}$ for $t\in[\theta_i,\theta_{i+1}]$ and $i\le k$, which gives $\mathsf{C}^{\mathrm{rec},c}_t\le\kappa^{c}_{k+1}$ on $[0,\theta_{k+1}]$. For $t\in[0,\theta_k]$ this yields $\mathsf{C}^{\mathrm{rec},c}_t\le\kappa^{c}_k<\lambda^{c}_k$, and for $t\in[\theta_k,\theta_{k+1})$ we have $t<\theta_{k+1}\le h^{c}_k$, so $\mathsf{C}^{\mathrm{rec},c}_t=\mathsf{C}^{c,(k)}_t<\lambda^{c}_k$ by (F3); together, $\mathsf{C}^{\mathrm{rec},c}_t<\lambda^{c}_k$ on $[0,\theta_{k+1})$.

Now let $k<K$ and $c\in\mathcal{J}_k$. Then $\kappa^{c}_{k+1}=\mathsf{C}^{c,(k)}_{\theta_{k+1}}=\lambda^{c}_k$ by claim 1 of the theorem. If $x^{(k),\sigma}$ were $0$, the integrand $\mathbf{1}_{(\theta_k,T]}(s)Nx^{(k),\sigma}\beta(\sigma,\gamma,x^{(k)},a_s)$ defining $\mathsf{C}^{c,(k)}$ would vanish identically, so $\mathsf{C}^{c,(k)}_t=\kappa^{c}_k<\lambda^{c}_k$ for every $t$ (the integral of the zero function being $0$ by the scalar rule of claim 1 of Linearity and Monotonicity of the Lebesgue Integral with scalar $0$), contradicting $\mathsf{C}^{c,(k)}_{\theta_{k+1}}\ge\lambda^{c}_k$; hence $x^{(k),\sigma}>0$. Since $\theta_{k+1}\in[\theta_k,T]$ and $\mathsf{C}^{c,(k)}_{\theta_{k+1}}\ge\lambda^{c}_k$, the time $\theta_{k+1}$ belongs to the set whose greatest lower bound is $h^{c}_k$, so $h^{c}_k\le\theta_{k+1}$; with $\theta_{k+1}\le h^{c}_k$ this gives $\theta_{k+1}=h^{c}_k$. Finally, the set $\{t\in[0,T]:\mathsf{C}^{\mathrm{rec},c}_t\ge\lambda^{c}_k\}$ contains $\theta_{k+1}$ (as $\mathsf{C}^{\mathrm{rec},c}_{\theta_{k+1}}=\mathsf{C}^{c,(k)}_{\theta_{k+1}}=\lambda^{c}_k$) and contains no $t<\theta_{k+1}$ (shown above), so its greatest lower bound is $\theta_{k+1}$.

Step 2 (claim 2). Assume no step $k<K$ is a tie. We show by induction on $k\le K$ that $x^{(k)}\in\mathbb{G}_N$. For $k=0$, $x^{(0)}=x_0\in\mathbb{G}_N$. Let $k<K$ with $x^{(k)}\in\mathbb{G}_N$. If $\mathcal{J}_k$ is empty, $x^{(k+1)}=x^{(k)}$. Otherwise $\mathcal{J}_k=\{c\}$ for a single label $c=(\sigma,\gamma)$, and $x^{(k+1)}=x^{(k)}+v_c/N$, which differs from $x^{(k)}$ only in the coordinates $\sigma$ and $\gamma$: $x^{(k+1),\sigma}=x^{(k),\sigma}-1/N$ and $x^{(k+1),\gamma}=x^{(k),\gamma}+1/N$. By Step 1, $x^{(k),\sigma}>0$, and $Nx^{(k),\sigma}\in\mathbb{N}_0$ by the definition of $\mathbb{G}_N$, so $Nx^{(k),\sigma}\ge1$ and $x^{(k+1),\sigma}\ge0$; thus all coordinates of $x^{(k+1)}$ are nonnegative, they sum to $1$ (the sum being unchanged), and $N$ times each is in $\mathbb{N}_0$; by the definition of the probability simplex and of $\mathbb{G}_N$, $x^{(k+1)}\in\mathbb{G}_N$. This completes the induction. The recursion stops at the index $K$ (claim 1 of the theorem) because $\theta_K=T$ or $x^{(K)}\notin\mathbb{G}_N$; the second alternative is excluded, so $\theta_K=T$ and $x^{(K)}\in\mathbb{G}_N$, i.e. the data are conflict-free.

Step 3 (the deleted path). Let $c_0$, $u$ and $p^{-}$ be as in claim 3, and write $q=p^{c_0}$, $q^{-}=p^{-,c_0}$. Since $u$ is a jump time of $q$, $q(u)>q(u-)$, and by the unit-jump property $q(u)=q(u-)+1$, both values being integers with difference in $(0,1]$ (the value $q(u-)$ is the least upper bound of a nonempty set of elements of $\mathbb{N}_0$ bounded above by $q(u)$, hence its maximum and in particular an integer, with $q(u-)\ge q(0)=0$). We check that $q^{-}$ is a counting path. $q^{-}(0)=q(0)=0$ (as $u>0$), and $q^{-}(t)$ is an integer; it is nonnegative since for $t<u$ it equals $q(t)\ge0$, and for $t\ge u$, $q(t)\ge q(u)=q(u-)+1\ge1$. Monotonicity: for $s\le t$, if both are $<u$ or both $\ge u$ then $q^{-}(t)-q^{-}(s)=q(t)-q(s)\ge0$; if $s<u\le t$ then $q^{-}(t)-q^{-}(s)=q(t)-1-q(s)\ge q(u)-1-q(u-)=0$, using $q(s)\le q(u-)$ (the least upper bound over $[0,u)$) and $q(t)\ge q(u)$. Right-continuity at $t$: the greatest lower bound of $\{q^{-}(s):s>t\}$ equals that of $\{q(s):s>t\}$ minus the constant $\mathbf{1}\{t\ge u\}$ when $t\ge u$ (all $s>t$ then satisfy $s\ge u$) and when $t<u$ (the values for $s\in(t,u)$ already realise the greatest lower bound $q(t)$ of $\{q(s):s>t\}$, $q$ being nondecreasing, and every $q^{-}(s)$ with $s>t$ is at least $q^{-}(t)=q(t)$ by the monotonicity just proved), so it equals $q^{-}(t)$. Unit jumps: for $t\ne u$ the jump $q^{-}(t)-q^{-}(t-)$ equals $q(t)-q(t-)\le1$ (for $t>u$ the subtracted constant is the same on a left neighbourhood; for $t<u$ nothing is subtracted on $[0,t]$); at $t=u$, $q^{-}(u-)=q(u-)$ and $q^{-}(u)=q(u)-1=q(u-)$, so the jump is $0$. Hence $q^{-}$ is a counting path, and $q^{-}=q$ on $[0,u)$, $q^{-}=q-1$ on $[u,\infty)$.

Consequently, for a consumed level $\kappa<u$ and $j=q(\kappa)+1=q^{-}(\kappa)+1$: (a) $q(u)\ge q(u-)+1\ge q(\kappa)+1=j$, so $\tau_j(q)\le u$ by (F1); (b) if $\tau_j(q)<u$ then $\tau_j(q^{-})=\tau_j(q)$: indeed $q^{-}\le q$ gives $\{q^{-}\ge j\}\subseteq\{q\ge j\}$ and so $\tau_j(q^{-})\ge\tau_j(q)$, while $q^{-}(\tau_j(q))=q(\tau_j(q))\ge j$ (by (F1), as $\tau_j(q)<u$) gives $\tau_j(q^{-})\le\tau_j(q)$; (c) if $\tau_j(q)=u$ then $\tau_j(q^{-})>u$: for $t\in[\kappa,u)$, $q^{-}(t)=q(t)<j$ (as $t<\tau_j(q)$, by (F1)), and $q^{-}(u)=q(u)-1=q(u-)$, where $q(u-)$ is the least upper bound of the values $q(t)$, $t<u$, all of which are $\le j-1$ (for $t<\kappa$ by monotonicity, since $q(\kappa)=j-1$; for $t\in[\kappa,u)$ as just shown), so $q^{-}(u)\le j-1<j$; thus no $t\le u$ lies in $\{q^{-}\ge j\}$, and by (F1) $\tau_j(q^{-})>u$ (the set being $[\tau_j(q^{-}),\infty)$ or empty, with $\tau_j(q^{-})=+\infty$ in the latter case). In either case $\tau_j(q^{-})\ge\tau_j(q)$.

Step 4 (claim 3). Keep the notation of claim 3. We prove by induction on $i\in\{0,\dots,k\}$ that the recursion for $p^{-}$ performs step $i$ with $\theta^{-}_i=\theta_i$, $x^{-(i)}=x^{(i)}$ and $\kappa^{-,c}_i=\kappa^{c}_i$ for all $c$, and that, for $i<k$, also $\theta^{-}_{i+1}=\theta_{i+1}$, $\mathcal{J}^{-}_i=\mathcal{J}_i$, $x^{-(i+1)}=x^{(i+1)}$ and $\kappa^{-,c}_{i+1}=\kappa^{c}_{i+1}$. The initial data agree. Let $i<k$ and assume agreement at step $i$; since $i<k<K$, the recursion for $p$ has not stopped at any index $\le i$, i.e. $\theta_i<T$ and $x^{(i)}\in\mathbb{G}_N$, hence neither has the recursion for $p^{-}$, and both perform step $i$. The functions $\mathsf{C}^{c,(i)}$ and $\mathsf{C}^{-,c,(i)}$ coincide for every $c$ (their definition involves only $\kappa^{c}_i$, $\theta_i$, $x^{(i)}$ and $a$). For $c\ne c_0$, $p^{-,c}=p^{c}$ gives $\lambda^{-,c}_i=\lambda^{c}_i$ and $h^{-,c}_i=h^{c}_i$. For $c_0$: $\kappa^{c_0}_i\le\kappa^{c_0}_k<\lambda^{c_0}_k=u$ (Step 1 and claim 1 of the theorem), so Step 3 applies with $\kappa=\kappa^{c_0}_i$ and $j=q(\kappa^{c_0}_i)+1$, for which $\lambda^{c_0}_i=\tau_j(q)$ and $\lambda^{-,c_0}_i=\tau_j(q^{-})$: either $\lambda^{c_0}_i<u$ and $\lambda^{-,c_0}_i=\lambda^{c_0}_i$, in which case all quantities of step $i$ coincide; or $\lambda^{c_0}_i=u$ and $\lambda^{-,c_0}_i>u$. In the latter case $c_0\notin\mathcal{J}_i$, since otherwise $\kappa^{c_0}_{i+1}=\lambda^{c_0}_i=u$ (Step 1), whereas $\kappa^{c_0}_{i+1}\le\kappa^{c_0}_k<u$; hence $\mathsf{C}^{c_0,(i)}_{\theta_{i+1}}<\lambda^{c_0}_i$, so $\theta_{i+1}<h^{c_0}_i$ (if $\theta_{i+1}=h^{c_0}_i\le T$, then $\mathsf{C}^{c_0,(i)}_{\theta_{i+1}}\ge\lambda^{c_0}_i$ by (F3)), and $h^{-,c_0}_i\ge h^{c_0}_i>\theta_{i+1}$ by (F2). Since $h^{c_0}_i>\theta_{i+1}$ and $h^{-,c_0}_i>\theta_{i+1}$, the label $c_0$ may be dropped from both minima defining $\theta_{i+1}$ and $\theta^{-}_{i+1}$, and $h^{-,c}_i=h^{c}_i$ for $c\ne c_0$; hence $\theta^{-}_{i+1}=\theta_{i+1}$. Then $\mathcal{J}^{-}_i=\mathcal{J}_i$: for $c\ne c_0$ the membership conditions coincide, and $c_0$ belongs to neither ($\mathsf{C}^{-,c_0,(i)}_{\theta_{i+1}}=\mathsf{C}^{c_0,(i)}_{\theta_{i+1}}<\lambda^{c_0}_i<\lambda^{-,c_0}_i$). Hence $x^{-(i+1)}=x^{(i+1)}$ and $\kappa^{-,c}_{i+1}=\mathsf{C}^{-,c,(i)}_{\theta_{i+1}}=\kappa^{c}_{i+1}$, completing the induction.

At step $k$ (performed by both recursions, by the same argument): $\mathsf{C}^{-,c,(k)}=\mathsf{C}^{c,(k)}$ for every $c$; $\lambda^{-,c}_k=\lambda^{c}_k$ and $h^{-,c}_k=h^{c}_k$ for $c\ne c_0$; and for $c_0$, $\kappa^{c_0}_k<u=\lambda^{c_0}_k=\tau_j(q)$ with $j=q(\kappa^{c_0}_k)+1$; since $\kappa^{-,c_0}_k=\kappa^{c_0}_k<u$ we have $q^{-}(\kappa^{c_0}_k)=q(\kappa^{c_0}_k)$, so $\lambda^{-,c_0}_k=\tau_j(q^{-})$, and by Step 3(c) $\lambda^{-,c_0}_k>u$, whence $h^{-,c_0}_k\ge h^{c_0}_k$ by (F2). Since $c'\in\mathcal{J}_k$, Step 1 gives $h^{c'}_k=\theta_{k+1}$, so $h^{-,c'}_k=\theta_{k+1}$; every other $h^{-,c}_k$ is at least $\theta_{k+1}$ (equal to $h^{c}_k\ge\theta_{k+1}$ for $c\ne c_0$, and $h^{-,c_0}_k\ge h^{c_0}_k\ge\theta_{k+1}$), and $T\ge\theta_{k+1}$; therefore $\theta^{-}_{k+1}=\min(T,\min_ch^{-,c}_k)=\theta_{k+1}$. The recursion consumed times of $p^{-}$ on $[0,\theta_{k+1}]$ are determined by the steps $0,\dots,k$ (on $[\theta^{-}_i,\theta^{-}_{i+1}]$ they equal $\mathsf{C}^{-,c,(i)}$), which coincide with those of $p$; hence $\mathsf{C}^{-,\mathrm{rec},c}_t=\mathsf{C}^{\mathrm{rec},c}_t$ for all $c$ and $t\in[0,\theta_{k+1}]$. In particular $\mathsf{C}^{-,\mathrm{rec},c_0}_{\theta_{k+1}}=\mathsf{C}^{\mathrm{rec},c_0}_{\theta_{k+1}}=\mathsf{C}^{c_0,(k)}_{\theta_{k+1}}=\lambda^{c_0}_k=u$ (claim 1 of the theorem, as $c_0\in\mathcal{J}_k$). Finally, by Step 1 applied to $c'\in\mathcal{J}_k$ with $v=\lambda^{c'}_k$ (a jump time of $p^{c'}$: it is $\tau_{j'}(p^{c'})$ with $j'=p^{c'}(\kappa^{c'}_k)+1\ge1$, is finite since $c'\in\mathcal{J}_k$ gives $\mathsf{C}^{c',(k)}_{\theta_{k+1}}\ge\lambda^{c'}_k$, is positive since $p^{c'}(\tau_{j'})\ge j'\ge1>0=p^{c'}(0)$, and satisfies $p^{c'}(\tau_{j'})\ge j'>p^{c'}(\tau_{j'}-)$, since by (F1) and the definition of $\tau_{j'}$ every value $p^{c'}(s)$ with $s<\tau_{j'}$ satisfies $p^{c'}(s)<j'$, hence $p^{c'}(s)\le j'-1$ by integer-valuedness, so that the least upper bound $p^{c'}(\tau_{j'}-)$ is at most $j'-1$), $\theta_{k+1}$ is the greatest lower bound of $\{t\in[0,T]:\mathsf{C}^{\mathrm{rec},c'}_t\ge v\}$: no $t<\theta_{k+1}$ belongs to this set and $\theta_{k+1}$ does. As $\mathsf{C}^{-,\mathrm{rec},c'}$ agrees with $\mathsf{C}^{\mathrm{rec},c'}$ on $[0,\theta_{k+1}]$, the same two facts hold for $\{t\in[0,T]:\mathsf{C}^{-,\mathrm{rec},c'}_t\ge v\}$, whose greatest lower bound is therefore $\theta_{k+1}$ as well. $\blacksquare$