Proof of Weighted Compensated Sums over the Observation Events of the Controlled N-Agent Dynamics

Fix a solution and adopt the notation of the statement. Throughout, $\Omega_0$ is the regular event; it has probability $1$, so expectations are unchanged when integrands are modified off $\Omega_0$, and we use this silently. Let $\Xi_t$ be the sum of all counters as in part (c) of the multiplier lemma, so that $\tilde{c}_t\le\Xi_t$ pointwise ($\tilde{c}_t$ omits the nonnegative transition counters) and $\mathbb{E}[\Xi_t^{\,k}]<\infty$ for every natural $k\ge1$ by that part. For a real $\delta>0$ set

$$\omega_F(\delta)=\sup\big\{|F^{p\upsilon}_s-F^{p\upsilon}_{s'}|:\ p\in\{1,\dots,d\},\ \upsilon\in\{1,\dots,\tilde{l}\},\ s,s'\in[0,T],\ |s-s'|\le\delta\big\},$$

and likewise $\omega_G$. Each entry of $F$ is continuous on $[0,T]$ (metric convention of the statement); the interval $[0,T]$ is compact in that metric by Closed Interval $[a,b]$ is Compact in $\mathbb{R}$, so each entry is uniformly continuous by the Heine-Cantor theorem. Given $\varepsilon>0$, choosing a uniform-continuity threshold for each of the $d\,\tilde{l}$ entries and taking the least gives a $\delta_0>0$ with $\omega_F(\delta)\le\varepsilon$ whenever $0<\delta\le\delta_0$; hence $\omega_F(\delta)\to0$ as $\delta\to0$, and likewise for $\omega_G$. Also $\omega_F\le2\bar{F}$ and $\omega_G\le2\bar{G}$ everywhere.

Step 0: part (a). Work at a fixed $\omega\in\Omega_0$ for the pathwise claims. By condition 3 of the solution definition, each observation counter $t\mapsto\tilde{N}^{i,\upsilon}_t$ and the observation total $t\mapsto\tilde{c}_t$ agree on $[0,T]$ with restrictions of counting paths; in particular all are nondecreasing, right-continuous, integer-valued, and vanish at $0$ (so $\tilde{J}^F_0=J^F_0=0$). As recorded in condition 5, at each jump time $\tau_j$ of the observation total exactly one observation counter jumps, its jump is exactly $1$, and its channel is $\upsilon_j$. Fix $\upsilon$ and consider $t\mapsto\Pi^\upsilon_t=\sum_i\tilde{N}^{i,\upsilon}_t$. Every point of increase of $\Pi^\upsilon$ is a point of increase of the observation total (the remaining summands of $\tilde{c}$ being nondecreasing), hence one of the $\tau_j$; and at $\tau_j$ the jump of $\Pi^\upsilon$ equals $1$ if $\upsilon_j=\upsilon$ and $0$ otherwise. This is the first claim. Consequently, for $0\le r\le t\le T$,

$$J^{F,p}_t-J^{F,p}_r=N^{-1/2}\sum_{j:\,r<\tau_j\le t}F^{p\upsilon_j}_{\tau_j}, \qquad\text{so}\qquad |J^{F,p}_t-J^{F,p}_r|\le\bar{F}N^{-1/2}(\tilde{c}_t-\tilde{c}_r). \tag{0.1}$$

For the compensator identity: by condition 2 of the solution definition, at every $\omega$, $\tilde{\mathcal{T}}^{i,\upsilon}_t$ is the Lebesgue integral over $[0,t]$ of $s\mapsto\mathbf{1}_{\Omega_0}\tilde{\beta}(\sigma^i_s,\upsilon,\Sigma_s)$. On $\Omega_0$, pointwise in $s$, using the occupation indicators $\eta^{i,\sigma}_s$ of the derived notation of the solution definition (each agent occupying exactly one state) and the formula of the aggregate observation drift,

$$\sum_{i=1}^{N}\tilde{\beta}(\sigma^i_s,\upsilon,\Sigma_s)=\sum_{\sigma=1}^{l}\Big(\sum_{i=1}^{N}\eta^{i,\sigma}_s\Big)\tilde{\beta}(\sigma,\upsilon,\Sigma_s)=N\sum_{\sigma=1}^{l}\Sigma^\sigma_s\,\tilde{\beta}(\sigma,\upsilon,\Sigma_s)=N\,\tilde{b}^\upsilon(\Sigma_s),$$

and summing the integrals over $i$ (linearity of the Lebesgue integral) gives $\sum_i\tilde{\mathcal{T}}^{i,\upsilon}_t=N\int_{[0,t]}\tilde{b}^\upsilon(\Sigma_s)\,ds$ on $\Omega_0$. At every $\omega$ and every $s$, $\Sigma_s\in\Delta^l$ (part (vii)(a) of the existence theorem), so its components are nonnegative and sum to $1$ (probability simplex); since every rate satisfies $0\le\tilde{\beta}(\sigma,\upsilon,\cdot)\le\tilde{B}$ (observation-rate family), the formula of the aggregate observation drift gives $0\le\tilde{b}^\upsilon(\Sigma_s)\le\tilde{B}$ everywhere. Hence the compensator increment obeys, at every $\omega$ and for every $p$,

$$N^{1/2}\Big|\sum_{\upsilon=1}^{\tilde{l}}\int_{[r,t]}F^{p\upsilon}_s\,\tilde{b}^\upsilon(\Sigma_s)\,ds\Big|\ \le\ N^{1/2}\,\bar{F}\,\tilde{l}\,\tilde{B}\,(t-r),\tag{0.2}$$

and combining (0.1) and (0.2) gives the stated increment bound for $\tilde{J}^{F,p}$ and the bound $|J^{F,p}_t|\le\bar{F}N^{-1/2}\tilde{c}_t$ (case $r=0$). Right-continuity on $\Omega_0$: the finitely many jump times being isolated, for $t\in[0,T)$ and $t'\downarrow t$ the sum in $J^{F,p}_{t'}$ eventually acquires no new terms ($\tilde{c}_{t'}=\tilde{c}_t$ for $t'$ below the next jump time, by right-continuity and integrality of the counting path), while the compensator is Lipschitz in $t$ by (0.2).

Measurability and moments. Each $\tilde{c}_s$ is a finite sum of counters, hence $\mathcal{F}^{\mathrm{sys}}_s$-measurable. For $j\ge1$ define $\tau_j$ on all of $\Omega$ as the $j$-th jump time of the counting path agreeing with $\tilde{c}$ (equal to $+\infty$ off $\{\tilde{c}_T\ge j\}$); then $\{\tau_j\le s\}=\{\tilde{c}_s\ge j\}\in\mathcal{F}^{\mathrm{sys}}_s$ for every $s\in[0,T]$. By part (iv) of the existence theorem, the observation-event count, the event times, and the channels up to any time $t$ are measurable for the observation filtration, which is contained in $\mathcal{F}^{\mathrm{sys}}_t$ by part (vii)(e) of that theorem; hence on $\{\tau_j\le t\}$ the variables $\tau_j$ and $\upsilon_j$ are $\mathcal{F}^{\mathrm{sys}}_t$-measurable. Writing

$$J^{F,p}_t=N^{-1/2}\sum_{j\ge1}\mathbf{1}_{\{\tau_j\le t\}}\sum_{\upsilon=1}^{\tilde{l}}\mathbf{1}_{\{\upsilon_j=\upsilon\}}F^{p\upsilon}_{\tau_j}\qquad(\text{at most }\tilde{c}_T\text{ nonzero terms}),$$

fix $n\ge0$: on $\{\tilde{c}_T=n\}\in\mathcal{F}^{\mathrm{sys}}_t$ the sum has at most $n$ nonzero terms, indexed by $j\le n$, with $\{\tau_j\le t\}\cap\{\tilde{c}_T=n\}\in\mathcal{F}^{\mathrm{sys}}_t$ and $\{\upsilon_j=\upsilon\}\cap\{\tau_j\le t\}\cap\{\tilde{c}_T=n\}\in\mathcal{F}^{\mathrm{sys}}_t$ (both $\tau_j$ and $\upsilon_j$ being $\mathcal{F}^{\mathrm{sys}}_t$-measurable on $\{\tau_j\le t\}$, as just recorded), so $J^{F,p}_t\,\mathbf{1}_{\{\tilde{c}_T=n\}}$ is, on $\{\tilde{c}_T=n\}$, a finite sum of products of the $\mathcal{F}^{\mathrm{sys}}_t$-measurable indicators $\mathbf{1}_{\{\tau_j\le t\}}$, $\mathbf{1}_{\{\upsilon_j=\upsilon\}}$ ($j\le n$) with $F^{p\upsilon}_{\tau_j}$ (measurable by composition of the continuous entry with the measurable $\min(\tau_j,T)$), hence $\mathcal{F}^{\mathrm{sys}}_t$-measurable; since the events $\{\tilde{c}_T=n\}$ ($n\ge0$) partition $\Omega$, $J^{F,p}_t=\sum_{n\ge0}\mathbf{1}_{\{\tilde{c}_T=n\}}J^{F,p}_t$ is $\mathcal{F}^{\mathrm{sys}}_t$-measurable, this sum having, at each $\omega$, exactly one nonzero summand.

Progressive measurability of the compensator and of $\tilde{J}^F$. We use the definition of progressive measurability and the progressive-measurability toolkit, both with respect to the filtration $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$, together with the measurable-arithmetic lemma. First, $\Omega_0\in\mathcal{F}^{\mathrm{sys}}_0$: the solution definition includes every $P$-null event of $\mathcal{F}$ in each $\mathcal{F}^{\mathrm{sys}}_t$, the complement of $\Omega_0$ is such an event, and a $\sigma$-algebra contains the complements of its members; so $\mathbf{1}_{\Omega_0}$ is $\mathcal{F}^{\mathrm{sys}}_0$-measurable (claim 1 of the arithmetic lemma). Fix $\upsilon$ and set $V^\upsilon_s=\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)$ for $s\in[0,T]$. The family $V^\upsilon$ is adapted: for each $s$ the components $\Sigma^{\sigma}_s$ are $\mathcal{F}^{\mathrm{sys}}_s$-measurable (part (iv) of the existence theorem) with $\Sigma_s\in\Delta^l$, each $\tilde{\beta}(\sigma,\upsilon,\Sigma_s)$ is $\mathcal{F}^{\mathrm{sys}}_s$-measurable by composition with the sequentially continuous map $\tilde{\beta}(\sigma,\upsilon,\cdot)$ on $E=\Delta^l$ (continuity clause of the observation-rate family), and $V^\upsilon_s=\mathbf{1}_{\Omega_0}\sum_{\sigma=1}^{l}\Sigma^\sigma_s\,\tilde{\beta}(\sigma,\upsilon,\Sigma_s)$ is $\mathcal{F}^{\mathrm{sys}}_s$-measurable by claims 2 and 3 of the arithmetic lemma. Every path of $V^\upsilon$ is right-continuous: at $\omega\in\Omega_0$ each state path is right-continuous with finitely many jump times (condition 1 of the solution definition), so for every $s\in[0,T)$ there is an $\varepsilon>0$ such that no agent has a jump time in $(s,s+\varepsilon]$; every state, hence $\Sigma$, hence $V^\upsilon$, is then constant on $[s,s+\varepsilon]\cap[0,T]$, and any sequence in $[s,T]$ converging to $s$ eventually enters this interval; for $s=T$ every sequence in $[T,T]$ is constant, so right-continuity there is trivial; and at $\omega\notin\Omega_0$ the path is identically $0$. By claim 2 of the toolkit, $V^\upsilon$ is progressively measurable, as claimed in (a). Each entry map $s\mapsto F^{p\upsilon}_s$ is continuous on $[0,T]$, so by claims 3 and 4 of the toolkit each weighted integrand $F^{p\upsilon}V^\upsilon$ is progressively measurable and bounded by $\bar{F}\tilde{B}$, and its indefinite time integral $t\mapsto\int_{[0,t]}F^{p\upsilon}_sV^\upsilon_s\,ds$ is adapted and progressively measurable with continuous paths; by claim 3 again, the compensator family $t\mapsto N^{1/2}\sum_{\upsilon=1}^{\tilde{l}}\int_{[0,t]}F^{p\upsilon}_s\,V^\upsilon_s\,ds$ is progressively measurable. Next, all counters vanish off $\Omega_0$ (part (vii)(c) of the existence theorem), so $\tilde{c}\equiv0$ and $J^{F,p}\equiv0$ off $\Omega_0$; at $\omega\in\Omega_0$ every path $t\mapsto J^{F,p}_t$ is right-continuous, as shown at the end of the pathwise part of this step (for $t'$ close enough to $t$ from the right the defining sum acquires no new terms); hence every path of $J^{F,p}$ is right-continuous, each $J^{F,p}_t$ is $\mathcal{F}^{\mathrm{sys}}_t$-measurable (previous paragraph), and $J^{F,p}$ is progressively measurable by claim 2 of the toolkit. Therefore $\tilde{J}^{F,p}$, the difference of $J^{F,p}$ and the compensator family, is progressively measurable by claim 3 of the toolkit; in particular each $\tilde{J}^{F,p}_t$ is an $\mathcal{F}^{\mathrm{sys}}_t$-measurable random variable, and $(t,\omega)\mapsto\tilde{J}^{F,p}_t(\omega)$ is measurable for the product $\sigma$-algebra of the trace Borel $\sigma$-algebra on $[0,T]$ and $\mathcal{F}^{\mathrm{sys}}_T$, hence for that of $[0,T]$ and $\mathcal{F}$ (claim 1 of the toolkit, the latter $\sigma$-algebra being the larger). Off $\Omega_0$ both $J^{F,p}$ and the compensator vanish identically, so $\tilde{J}^{F}\equiv0$ there; and $\tilde{J}^{F}_0=0$ everywhere ($\tilde{c}_0=0$ and the integral over $[0,0]$ is $0$). Finally, at every $\omega$,

$$|\tilde{J}^{F,p}_t|\le\bar{F}\big(N^{-1/2}\tilde{c}_T\,\mathbf{1}_{\Omega_0}+N^{1/2}\tilde{l}\tilde{B}T\big)\le\bar{F}\big(N^{-1/2}\Xi_T+N^{1/2}\tilde{l}\tilde{B}T\big)=:\Psi$$

(off $\Omega_0$ the left-hand side vanishes, and on $\Omega_0$ this is the case $r=0$ of the bounds (0.1)--(0.2) together with $\tilde{c}_T\le\Xi_T$). The majorant $\Psi$ does not depend on $t$ or $p$, and it has finite moments of every order: for each natural $k$, $(x+y)^k\le2^k(x^k+y^k)$ for $x,y\ge0$, and $\Xi_T$ has moments of every order. In particular $\mathbb{E}[|\tilde{J}^{F,p}_t|^k]<\infty$ for every natural number $k\ge1$. This proves (a).

Step 1: partitions and the increment decomposition. Fix $0\le r<t\le T$, a natural $n\ge1$, and set $\delta=(t-r)/n$, $t_k=r+k\delta$ ($0\le k\le n$), $I_k=(t_k,t_{k+1}]$. Write $\Delta_kX=X_{t_{k+1}}-X_{t_k}$ for any process $X$, $\Delta_k\tilde{c}=\tilde{c}_{t_{k+1}}-\tilde{c}_{t_k}$, and define

$$S^{F,p}_k=N^{-1/2}\sum_{\upsilon=1}^{\tilde{l}}F^{p\upsilon}_{t_k}\sum_{i=1}^{N}\Delta_kM^{(i,\upsilon)},\qquad \rho^{F,p}_k=\Delta_k\tilde{J}^{F,p}-S^{F,p}_k .$$

On $\Omega_0$: by the jump correspondence of Step 0, $\Delta_kJ^{F,p}=N^{-1/2}\sum_{j:\tau_j\in I_k}F^{p\upsilon_j}_{\tau_j}$, while $N^{-1/2}\sum_\upsilon F^{p\upsilon}_{t_k}\sum_i\Delta_k\tilde{N}^{i,\upsilon}=N^{-1/2}\sum_{j:\tau_j\in I_k}F^{p\upsilon_j}_{t_k}$, and each $|\tau_j-t_k|\le\delta$ for $\tau_j\in I_k$; by the compensator identity of Step 0, $N^{-1/2}\sum_\upsilon F^{p\upsilon}_{t_k}\sum_i\Delta_k\tilde{\mathcal{T}}^{i,\upsilon}=N^{1/2}\sum_\upsilon F^{p\upsilon}_{t_k}\int_{I_k}\tilde{b}^\upsilon(\Sigma_s)ds$. Hence, with $\lambda_k=N^{-1/2}\Delta_k\tilde{c}+N^{1/2}\tilde{l}\tilde{B}\delta$,

$$|\rho^{F,p}_k|\le\omega_F(\delta)\,\lambda_k,\qquad |S^{F,p}_k|\le\bar{F}\,\lambda_k,\qquad |\Delta_k\tilde{J}^{F,p}|\le\bar{F}\,\lambda_k\qquad\text{on }\Omega_0,\tag{1.1}$$

the last by (a). Note $\sum_{k=0}^{n-1}\lambda_k=N^{-1/2}(\tilde{c}_t-\tilde{c}_r)+N^{1/2}\tilde{l}\tilde{B}(t-r)$ and, since $(x+y)^w\le2^w(x^w+y^w)$ for $x,y\ge0$ (the exponent written $w$, the letter $m$ being the control dimension of the statement),

$$\sum_{k=0}^{n-1}\lambda_k^w\ \le\ 2^w\Big(N^{-w/2}\tilde{c}_T^{\,w}+n\,\big(N^{1/2}\tilde{l}\tilde{B}\delta\big)^w\Big)\qquad(w\in\{2,3,4\})\tag{1.2}$$

on $\Omega_0$, using $\sum_k(\Delta_k\tilde{c})^w\le(\sum_k\Delta_k\tilde{c})^{w-1}\max_k\Delta_k\tilde{c}\le\tilde{c}_T^{\,w}$ for the first part and $n\delta^w=(t-r)\delta^{w-1}$ for the second. All variables $\lambda_k$, $\tilde{c}_T$, and $\Xi_T$ have moments of every order (Step 0).

Step 2: part (b). Let $Z$ be $\mathcal{F}^{\mathrm{sys}}_r$-measurable and square-integrable. Integrability of $Z\,(\tilde{J}^{F,p}_t-\tilde{J}^{F,p}_r)$ holds by the Cauchy-Schwarz inequality for the mean-square norm and (a). Telescoping and using the decomposition of Step 1,

$$\mathbb{E}\big[Z(\tilde{J}^{F,p}_t-\tilde{J}^{F,p}_r)\big]=\sum_{k=0}^{n-1}\mathbb{E}[Z\,S^{F,p}_k]+\sum_{k=0}^{n-1}\mathbb{E}[Z\,\rho^{F,p}_k].$$

For each $k$, $Z$ is $\mathcal{F}^{\mathrm{sys}}_{t_k}$-measurable (the filtration is increasing) and square-integrable, so $\mathbb{E}[Z\,\Delta_kM^{(i,\upsilon)}]=0$ for every observation clock label by part (a) of the multiplier lemma; by linearity $\mathbb{E}[Z\,S^{F,p}_k]=0$. By (1.1) and the Cauchy-Schwarz inequality,

$$\Big|\sum_k\mathbb{E}[Z\rho^{F,p}_k]\Big|\le\omega_F(\delta)\,\mathbb{E}\Big[|Z|\sum_k\lambda_k\Big]\le\omega_F(\delta)\Big(N^{-1/2}\,\mathbb{E}[|Z|\,\tilde{c}_T]+N^{1/2}\tilde{l}\tilde{B}\,T\,\mathbb{E}[|Z|]\Big),$$

which is finite and tends to $0$ as $n\to\infty$ since $\omega_F(\delta)\to0$. The left-hand side does not depend on $n$, so it vanishes. Taking $Z=\mathbf{1}_D$ with $D\in\mathcal{F}^{\mathrm{sys}}_r$ yields the averaged martingale property of the definition of a square-integrable martingale; adaptedness and square-integrability hold by (a), and $\tilde{J}^F_0=0$ everywhere (Step 0). This proves (b).

Step 3: part (c). Let $Z$ be $\mathcal{F}^{\mathrm{sys}}_r$-measurable with $Z^2$ square-integrable; then $Z$ is square-integrable ($\mathbb{E}[Z^2]\le1+\mathbb{E}[Z^4]$). Set $X_k=\tilde{J}^{F,p}_{t_k}-\tilde{J}^{F,p}_r$ and $Y_k=\tilde{J}^{G,q}_{t_k}-\tilde{J}^{G,q}_r$, so $X_0=Y_0=0$ and $X_n$, $Y_n$ are the full increments. All products handled below are integrable: for instance $\mathbb{E}[|Z\,X\,Y|]\le\Vert ZX\Vert_2\Vert Y\Vert_2$ and $\mathbb{E}[(ZX)^2]=\mathbb{E}[Z^2X^2]\le\Vert Z^2\Vert_2\Vert X^2\Vert_2$, finite by hypothesis and (a), with the Cauchy-Schwarz inequality. Telescoping,

$$X_nY_n=\sum_{k=0}^{n-1}\big(X_k\,\Delta_kY+Y_k\,\Delta_kX+\Delta_kX\,\Delta_kY\big),\qquad \Delta_kX=\Delta_k\tilde{J}^{F,p},\ \Delta_kY=\Delta_k\tilde{J}^{G,q}.$$

Multiply by $Z$ and take expectations termwise. The multipliers $ZX_k$ and $ZY_k$ are $\mathcal{F}^{\mathrm{sys}}_{t_k}$-measurable (by (a)) and square-integrable (as just computed), so $\mathbb{E}[Z\,X_k\,\Delta_kY]=0$ and $\mathbb{E}[Z\,Y_k\,\Delta_kX]=0$ by part (b) applied on $[t_k,t_{k+1}]$. For the quadratic term, write $\Delta_kX=S^{F,p}_k+\rho^{F,p}_k$ and $\Delta_kY=S^{G,q}_k+\rho^{G,q}_k$. First,

$$\mathbb{E}\big[Z\,S^{F,p}_kS^{G,q}_k\big]=N^{-1}\sum_{(i,\upsilon)}\sum_{(i',\upsilon')}F^{p\upsilon}_{t_k}G^{q\upsilon'}_{t_k}\,\mathbb{E}\big[Z\,\Delta_kM^{(i,\upsilon)}\Delta_kM^{(i',\upsilon')}\big]=N^{-1}\sum_{(i,\upsilon)}F^{p\upsilon}_{t_k}G^{q\upsilon}_{t_k}\,\mathbb{E}\big[Z\,\Delta_k\tilde{\mathcal{T}}^{i,\upsilon}\big]$$

by part (b) of the multiplier lemma, whose hypotheses hold: $Z$, $Z\,M^{(i,\upsilon)}_{t_k}$, and $Z\,M^{(i',\upsilon')}_{t_k}$ are square-integrable, since $\mathbb{E}[(ZM^a_{t_k})^2]\le\Vert Z^2\Vert_2\Vert(M^a_{t_k})^2\Vert_2<\infty$ by part (c) of that lemma. By the compensator identity of Step 0 (on $\Omega_0$),

$$N^{-1}\sum_{(i,\upsilon)}F^{p\upsilon}_{t_k}G^{q\upsilon}_{t_k}\,\Delta_k\tilde{\mathcal{T}}^{i,\upsilon}=\int_{I_k}\sum_{\upsilon}F^{p\upsilon}_{t_k}G^{q\upsilon}_{t_k}\,\tilde{b}^\upsilon(\Sigma_s)\,ds,$$

and replacing the frozen weights by the running ones costs, per $k$, at most $\mathbb{E}[|Z|]\,(\omega_F(\delta)\bar{G}+\bar{F}\omega_G(\delta))\,\tilde{l}\tilde{B}\,\delta$ in absolute value. Second, by (1.1),

$$\big|\mathbb{E}\big[Z\big(S^{F,p}_k\rho^{G,q}_k+\rho^{F,p}_kS^{G,q}_k+\rho^{F,p}_k\rho^{G,q}_k\big)\big]\big|\ \le\ \big(\omega_F(\delta)+\omega_G(\delta)\big)\,\big(\bar{F}+\bar{G}\big)\,\mathbb{E}\big[|Z|\,\lambda_k^2\big].$$

Summing over $k$ and using (1.2) with $w=2$ together with the Cauchy-Schwarz inequality ($\mathbb{E}[|Z|\tilde{c}_T^2]\le\Vert Z\Vert_2\Vert\tilde{c}_T^2\Vert_2<\infty$), the total error from both sources is at most

$$\big(\omega_F(\delta)+\omega_G(\delta)\big)\Big[(\bar{F}+\bar{G})\Big(4N^{-1}\mathbb{E}[|Z|\tilde{c}_T^2]+4N(\tilde{l}\tilde{B})^2T\delta\,\mathbb{E}|Z|\Big)+\big(\bar{G}+\bar{F}\big)\tilde{l}\tilde{B}T\,\mathbb{E}[|Z|]\Big]\ \longrightarrow\ 0\qquad(n\to\infty),$$

since $\omega_F(\delta)+\omega_G(\delta)\to0$ and the bracket stays bounded. The main terms add up, by additivity of the Lebesgue integral over adjacent intervals and linearity of the expectation, to $\mathbb{E}[Z\int_{[r,t]}\sum_\upsilon F^{p\upsilon}_sG^{q\upsilon}_s\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)ds]$ (the indicator changes nothing under the expectation, $\Omega_0$ having probability $1$, and makes the inner integral the one displayed in the statement, defined at every $\omega$). As the left-hand side $\mathbb{E}[ZX_nY_n]$ does not depend on $n$, the identity of (c) follows. The recorded special case is the case $Z=1$, $r=0$, rewritten with the diagonal matrix $D(\Sigma_s)$ (the $(p,q)$ entry of $F_sD(\Sigma_s)G_s^{\top}$ being exactly $\sum_\upsilon F^{p\upsilon}_sG^{q\upsilon}_s\tilde{b}^\upsilon(\Sigma_s)$ by the formulas for matrix products and the transpose), and the second-moment bound follows from $0\le\tilde{b}^\upsilon\le\tilde{B}$ and monotonicity of the integral.

Step 4: part (d). Let $Z$ be as in (c). Recall from part (b) of the martingale decomposition that $M^\gamma_t=\Sigma^\gamma_t-\Sigma^\gamma_0-\int_{[0,t]}\mathbf{1}_{\Omega_0}\,b^\gamma(\Sigma_s,\alpha_s)ds$ (the indicator equal to $1$ on $\Omega_0$, where we work), so that $|M^\gamma_t-M^\gamma_s|\le\frac{1}{N}\,\Delta\Xi_{(s,t]}+2(l-1)B(t-s)$ at every $\omega\in\Omega_0$ and all $s\le t$, where $\Delta\Xi_{(s,t]}=\Xi_t-\Xi_s$: indeed each state-transition event changes each $\Sigma^\gamma$ by at most $1/N$ (condition 6 of the solution definition), and $|b^\gamma|\le2(l-1)B$ by part (a) of the decomposition theorem. In particular $M^\gamma$ is bounded by $2+2(l-1)BT$ on $\Omega_0$. By part (c) of the counter moment lemma, almost surely $N\,M^\gamma_t=\sum_{\sigma:\sigma\neq\gamma}(\mathfrak{M}^{\sigma\gamma}_t-\mathfrak{M}^{\gamma\sigma}_t)$ for all $t$, where $\mathfrak{M}^{\sigma\gamma}_t=\sum_iM^{i,\sigma\gamma}_t$ is the aggregate compensated counter over the transition clock labels; hence, for any square-integrable $\mathcal{F}^{\mathrm{sys}}_{s}$-measurable $W$ and $t\ge s$,

$$\mathbb{E}\big[W\,(M^\gamma_t-M^\gamma_s)\big]=\frac1N\sum_{\sigma\neq\gamma}\sum_{i}\Big(\mathbb{E}\big[W\Delta M^{i,\sigma\gamma}\big]-\mathbb{E}\big[W\Delta M^{i,\gamma\sigma}\big]\Big)=0\tag{4.1}$$

by part (a) of the multiplier lemma. Now telescope with $X_k=M^\gamma_{t_k}-M^\gamma_r$ and $Y_k=\tilde{J}^{F,p}_{t_k}-\tilde{J}^{F,p}_r$ as in Step 3. The terms $\mathbb{E}[Z\,X_k\,\Delta_kY]$ vanish by part (b) with the square-integrable multiplier $ZX_k$ ($X_k$ bounded, and $\mathcal{F}^{\mathrm{sys}}_{t_k}$-measurable by adaptedness of the martingale $M^\gamma$); the terms $\mathbb{E}[Z\,Y_k\,\Delta_kX]$ vanish by (4.1) with $W=ZY_k$ (square-integrable as in Step 3, $\mathcal{F}^{\mathrm{sys}}_{t_k}$-measurable by (a)). For the quadratic terms, write $\Delta_kY=S^{F,p}_k+\rho^{F,p}_k$ and expand $\Delta_kX$ by the almost sure representation above:

$$\mathbb{E}\big[Z\,\Delta_kX\,S^{F,p}_k\big]=N^{-3/2}\sum_{\sigma\neq\gamma}\sum_{i}\sum_{(i',\upsilon)}F^{p\upsilon}_{t_k}\Big(\mathbb{E}\big[Z\,\Delta M^{i,\sigma\gamma}\Delta M^{(i',\upsilon)}\big]-\mathbb{E}\big[Z\,\Delta M^{i,\gamma\sigma}\Delta M^{(i',\upsilon)}\big]\Big)=0,$$

since every pair consists of a transition clock label and an observation clock label, which are distinct, so each expectation vanishes by part (b) of the multiplier lemma (hypotheses verified as in Step 3). Finally, by the pathwise bounds and (1.1),

$$\big|\mathbb{E}\big[Z\,\Delta_kX\,\rho^{F,p}_k\big]\big|\ \le\ \omega_F(\delta)\,\mathbb{E}\Big[|Z|\Big(\tfrac1N\Delta_k\Xi+2(l-1)B\delta\Big)\lambda_k\Big],$$

and summing over $k$, using $\Delta_k\tilde{c}\le\Delta_k\Xi$, $\sum_k(\Delta_k\Xi)^2\le\Xi_T^2$, $\sum_k\Delta_k\Xi\le\Xi_T$, $n\delta^2=(t-r)\delta$, and the Cauchy-Schwarz inequality, the total is at most $\omega_F(\delta)$ times a finite constant (depending on $N$, $T$, $B$, $\tilde{B}$, $\tilde{l}$, $l$, $\mathbb{E}[Z^2]$, and the moments of $\Xi_T$, but not on $n$), which tends to $0$. Since $\mathbb{E}[Z\,X_nY_n]$ does not depend on $n$, part (d) follows.

Step 5: part (e). Fix $p\in\{1,\dots,d\}$ and abbreviate $X_s=\tilde{J}^{F,p}_s$. By (c), $m_2=\sup\{\mathbb{E}[X_s^2]:s\in[0,T]\}\le\bar{F}^2\tilde{l}\tilde{B}T$. Fix $t\in(0,T]$ (the case $t=0$ is trivial) and partition $[0,t]$ as in Step 1 with $r=0$. Telescoping fourth powers and expanding by the binomial theorem,

$$\mathbb{E}[X_t^4]=\sum_{k=0}^{n-1}\Big(4\,\mathbb{E}[X_{t_k}^3\Delta_kX]+6\,\mathbb{E}[X_{t_k}^2(\Delta_kX)^2]+4\,\mathbb{E}[X_{t_k}(\Delta_kX)^3]+\mathbb{E}[(\Delta_kX)^4]\Big),$$

all terms integrable by (a). We bound the four groups; write $\mathcal{O}$ for the set of the $N\tilde{l}$ observation clock labels and $f_a=F^{p\upsilon}_{t_k}$ for $a=(i,\upsilon)\in\mathcal{O}$, and let $C_\star$ be the constant of the multiplier lemma.

Cubic multiplier. $\mathbb{E}[X_{t_k}^3\Delta_kX]=0$ by part (b) with the square-integrable $\mathcal{F}^{\mathrm{sys}}_{t_k}$-measurable multiplier $X_{t_k}^3$.

Quadratic multiplier. By part (c) with $Z=X_{t_k}^2$ (its square $X_{t_k}^4$ is square-integrable by (a)) on $[t_k,t_{k+1}]$, and $0\le\tilde{b}^\upsilon\le\tilde{B}$:

$$\mathbb{E}\big[X_{t_k}^2(\Delta_kX)^2\big]=\mathbb{E}\Big[X_{t_k}^2\int_{I_k}\sum_\upsilon(F^{p\upsilon}_s)^2\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)ds\Big]\le\bar{F}^2\tilde{l}\tilde{B}\,\delta\;\mathbb{E}[X_{t_k}^2]\le\bar{F}^2\tilde{l}\tilde{B}\,m_2\,\delta .$$

Linear multiplier. Write $\Delta_kX=S_k+\rho_k$ with $S_k=S^{F,p}_k$, $\rho_k=\rho^{F,p}_k$. Then $(\Delta_kX)^3=S_k^3+(3S_k^2\rho_k+3S_k\rho_k^2+\rho_k^3)$, and by (1.1) with $\omega_F\le2\bar{F}$, $|3S_k^2\rho_k+3S_k\rho_k^2+\rho_k^3|\le7(3\bar{F})^2\,\omega_F(\delta)\,\lambda_k^3$. By (a), $|X_{t_k}|\le\Psi$ at every $\omega$, with $\Psi$ the majorant of Step 0; pulling this $k$-independent majorant out before summing over $k$, the total contribution of these remainder terms is at most $7(3\bar{F})^{2}\,\omega_F(\delta)\,\mathbb{E}\big[\Psi\sum_{k=0}^{n-1}\lambda_k^3\big]$, and by (1.2) with $w=3$,

$$\mathbb{E}\Big[\Psi\sum_{k=0}^{n-1}\lambda_k^3\Big]\ \le\ 8\,N^{-3/2}\,\mathbb{E}\big[\Psi\,\Xi_T^{3}\big]\ +\ 8\,N^{3/2}(\tilde{l}\tilde{B})^{3}\,T\,\delta^{2}\,\mathbb{E}[\Psi]$$

(using $\tilde{c}_T\le\Xi_T$ and $n\delta^3=t\delta^2\le T\delta^2$), which is bounded uniformly in $n$ because $\Psi$ and $\Xi_T$ have moments of every order; hence the total tends to $0$ with $\omega_F(\delta)$. Expand $S_k^3=N^{-3/2}\sum_{a,b,c\in\mathcal{O}}f_af_bf_c\,\Delta_kM^a\Delta_kM^b\Delta_kM^c$. For the triples not all equal, part (f) of the multiplier lemma with the integrable multiplier $X_{t_k}$ bounds each term by $C_\star\mathbb{E}[|X_{t_k}|]\delta^2\le C_\star m_2^{1/2}\delta^2$; there are fewer than $(N\tilde{l})^3$ of them, so with the prefactor $N^{-3/2}\bar{F}^3$ their total over all $k$ is at most $N^{3/2}\tilde{l}^3\bar{F}^3C_\star m_2^{1/2}\,t\,\delta\to0$. For the $N\tilde{l}$ diagonal triples, split $\mathbb{E}[X_{t_k}(\Delta_kM^a)^3]=\mathbb{E}[X_{t_k}((\Delta_kM^a)^3-\Delta_k \mathcal{T}^a)]+\mathbb{E}[X_{t_k}\Delta_k\mathcal{T}^a]$ (writing $\mathcal{T}^a=\tilde{\mathcal{T}}^{i,\upsilon}$ for $a=(i,\upsilon)$): by part (e) of the multiplier lemma with the square-integrable multiplier $X_{t_k}$, the first expectation is at most $C_\star(1+m_2)\delta^{3/2}$ in absolute value, so these parts total at most $N^{-1/2}\tilde{l}\bar{F}^3C_\star(1+m_2)\,t\,\delta^{1/2}\to0$; for the second parts, $\sum_{a\in\mathcal{O}}\Delta_k\mathcal{T}^a\le N\tilde{l}\tilde{B}\delta$ pathwise on $\Omega_0$ (compensator identity and $\tilde{b}^\upsilon\le\tilde{B}$), so

$$N^{-3/2}\Big|\sum_{a\in\mathcal{O}}f_a^3\,\mathbb{E}[X_{t_k}\Delta_k\mathcal{T}^a]\Big|\le N^{-1/2}\bar{F}^3\tilde{l}\tilde{B}\,\delta\;\mathbb{E}[|X_{t_k}|]\le\bar{F}^3\tilde{l}\tilde{B}\,m_2^{1/2}\,\delta,$$

using $N\ge1$ and $\mathbb{E}[|X_{t_k}|]\le m_2^{1/2}$ (Cauchy-Schwarz against the constant $1$).

Constant multiplier. $(\Delta_kX)^4=S_k^4+\big((\Delta_kX)^4-S_k^4\big)$, and $|(\Delta_kX)^4-S_k^4|\le15(3\bar{F})^3\omega_F(\delta)\lambda_k^4$ by (1.1); summing over $k$ and using (1.2) with $w=4$ together with the moments of $\Xi_T$ (and $n\delta^4=t\delta^3\le T\delta^3$), the total of these remainders is $\omega_F(\delta)$ times a quantity bounded uniformly in $n$, hence tends to $0$. Expand $S_k^4=N^{-2}\sum_{a,b,c,d\in\mathcal{O}}f_af_bf_cf_d\,\Delta_kM^a\Delta_kM^b\Delta_kM^c\Delta_kM^d$. Quadruples not all equal: part (f) with $Z=1$ bounds each by $C_\star\delta^2$; fewer than $(N\tilde{l})^4$ of them, prefactor $N^{-2}\bar{F}^4$, total over $k$ at most $N^2\tilde{l}^4\bar{F}^4C_\star t\,\delta\to0$. Diagonal quadruples: by part (d) of the multiplier lemma with $Z=1$ (bounded by $\zeta=1$) and its power parameter equal to $4$, $\mathbb{E}[(\Delta_kM^a)^4]\le\mathbb{E}[\Delta_k\mathcal{T}^a]+C_\star\delta^2$, so their total per interval is at most $N^{-2}\bar{F}^4(N\tilde{l}\tilde{B}\delta+N\tilde{l}C_\star\delta^2)\le\bar{F}^4\tilde{l}\tilde{B}\delta+\bar{F}^4\tilde{l}C_\star\delta^2$ ($N\ge1$), and over all $k$ at most $\bar{F}^4\tilde{l}\tilde{B}\,t+\bar{F}^4\tilde{l}C_\star t\,\delta$.

Summation. Summing the surviving bounds over $k$ (each carries a factor $\delta$ and there are $n=t/\delta$ intervals) and letting $n\to\infty$, the vanishing groups disappear and

$$\mathbb{E}[X_t^4]\ \le\ 6\,\bar{F}^2\tilde{l}\tilde{B}\,t\,m_2+4\,\bar{F}^3\tilde{l}\tilde{B}\,t\,m_2^{1/2}+\bar{F}^4\tilde{l}\tilde{B}\,t .$$

With $y=\tilde{l}\tilde{B}T\ge\tilde{l}\tilde{B}t$ and $m_2\le\bar{F}^2y$, $m_2^{1/2}\le\bar{F}y^{1/2}$, the right-hand side is at most $\bar{F}^4(6y^2+4y^{3/2}+y)$. If $y\ge1$ this is at most $11\bar{F}^4y^2$; if $y<1$ it is at most $11\bar{F}^4y$; in either case at most $11\bar{F}^4(1+y)^2\le11(1+\bar{F})^4(1+\tilde{l}\tilde{B}T)^2$. The constants involved depend only on $\bar{F}$, $\tilde{l}$, $\tilde{B}$, and $T$, proving (e). $\blacksquare$