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 $\mathbb{E}[\Xi_t^{\,p}]<\infty$ for every natural $p\ge1$ by that part. For a real $\delta>0$ set

and likewise $\omega_G$. Each entry of $F$ is continuous on the compact interval $[0,T]$, hence uniformly continuous by the Heine-Cantor theorem, so $\omega_F(\delta)\to0$ as $\delta\to0$; 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 $0\le r\le t\le T$,

For the compensator identity: by condition 2 of the solution definition, at every $\omega$, $\tilde{A}^{i,\upsilon}_t$ 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,

and summing the integrals over $i$ (linearity of the Lebesgue integral) gives $\sum_i\tilde{A}^{i,\upsilon}_t=N\int_{[0,t]}\tilde{b}^\upsilon(\Sigma_s)\,ds$ on $\Omega_0$. By part (a) of the martingale decomposition theorem, $0\le\tilde{b}^\upsilon(\Sigma_s)\le\tilde{B}$; and $\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)=N^{-1}\sum_{i=1}^{N}\mathbf{1}_{\Omega_0}\tilde{\beta}(\sigma^i_s,\upsilon,\Sigma_s)$ pointwise in $(s,\omega)$ (both sides equal $N^{-1}$ times the displayed identity above when multiplied by $\mathbf{1}_{\Omega_0}$, since the identity holds on $\Omega_0$, and both sides vanish off $\Omega_0$); condition 2 of the solution definition already asserts product-measurability, for the product $\sigma$-algebra of the trace Borel $\sigma$-algebra on $[0,T]$ and $\mathcal{F}$, of each map $(s,\omega)\mapsto\mathbf{1}_{\Omega_0}\tilde{\beta}(\sigma^i_s,\upsilon,\Sigma_s)$ (this is exactly what makes $\tilde{A}^{i,\upsilon}_t$ a well-defined Lebesgue integral for every $t$ and a random variable, per that condition); summing over $i$ and dividing by $N$, $(s,\omega)\mapsto\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)$ is product-measurable, being a finite sum of product-measurable maps. Hence the compensator increment obeys, at every $\omega$ and for every $\gamma$,

and combining (0.1) and (0.2) gives the stated increment bound for $\tilde{J}^{F,\gamma}$ and the bound $|J^{F,\gamma}_t|\le\bar{F}N^{-1/2}\tilde{c}_t$ (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 $J^{F,\gamma}_{t'}$ 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 $\mathcal{F}^{\mathrm{sys}}_t$; hence on $\{\tau_j\le t\}$ the variables $\tau_j$ and $\upsilon_j$ are $\mathcal{F}^{\mathrm{sys}}_t$-measurable. Writing

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 $J^{F,\gamma}_t\,\mathbf{1}_{\{\tilde{c}_T=n\}}$ 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 $F^{\gamma\upsilon}_{\tau_j}$ (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 $\Omega$, $J^{F,\gamma}_t=\sum_{n\ge0}\mathbf{1}_{\{\tilde{c}_T=n\}}J^{F,\gamma}_t$ is $\mathcal{F}^{\mathrm{sys}}_t$-measurable, this sum having, at each $\omega$, exactly one nonzero summand.

For the compensator: the preceding paragraph shows $(s,\omega)\mapsto\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)$ is product-measurable and bounded by $\tilde{B}$; hence $(s,\omega)\mapsto\sum_{\upsilon}F^{\gamma\upsilon}_s\,\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)$ is product-measurable, being a finite sum of products of this map with the continuous (hence Borel-measurable) maps $s\mapsto F^{\gamma\upsilon}_s$, and bounded by $\bar{F}\tilde{l}\tilde{B}$; by Tonelli's theorem applied on $[0,T]\times[0,T]\times\Omega$ to its positive and negative parts, the map $(t,\omega)\mapsto N^{1/2}\int_{[0,t]}\sum_\upsilon F^{\gamma\upsilon}_s\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)\,ds=N^{1/2}\int_{[0,T]}\mathbf{1}_{\{s\le t\}}\sum_\upsilon F^{\gamma\upsilon}_s\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)\,ds$ (the integrand $(s,t,\omega)\mapsto\mathbf{1}_{\{s\le t\}}\sum_\upsilon F^{\gamma\upsilon}_s\mathbf{1}_{\Omega_0}\tilde{b}^\upsilon(\Sigma_s)$ being product-measurable, as a product of the Borel-measurable $\mathbf{1}_{\{s\le t\}}$ with the map just shown) is product-measurable in $(t,\omega)$, and in particular, for each fixed $t$, $\mathcal{F}^{\mathrm{sys}}_t$-measurable in $\omega$ (the states at times $\le t$ being adapted by part (iv) of the existence theorem, so the restriction of the integrand to $s\le t$ only involves $\mathcal{F}^{\mathrm{sys}}_t$-measurable data). Hence each $\tilde{J}^{F,\gamma}_t$ is a random variable, $\mathcal{F}^{\mathrm{sys}}_t$-measurable, and $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)\tilde{J}^{F,\gamma}_t(\omega)$ is product-measurable, being the difference of the product-measurable compensator just constructed and the map $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)J^{F,\gamma}_t(\omega)$, itself product-measurable: for each $j$, the set $\{(t,\omega):\tau_j(\omega)\le t\}$ is the preimage of the closed half-plane $\{(t,x)\in[0,T]\times\mathbb{R}:x\le t\}$ (closed, hence Borel, in the product) under the map $(t,\omega)\mapsto(t,\tau_j(\omega))$, product-measurable because $(t,\omega)\mapsto t$ and $(t,\omega)\mapsto\tau_j(\omega)$ are each measurable (composition with the coordinate projections); so $(t,\omega)\mapsto\mathbf{1}_{\{\tau_j\le t\}}$ is product-measurable, hence so is $(t,\omega)\mapsto\mathbf{1}_{\{\tau_j\le t\}}\mathbf{1}_{\{\upsilon_j=\upsilon\}}$ (the second factor not depending on $t$), and summing the finitely many nonzero terms on each $\{\tilde{c}_T=n\}\times[0,T]\times\Omega$ as above shows $(t,\omega)\mapsto J^{F,\gamma}_t(\omega)$, and a fortiori $\mathbf{1}_{\Omega_0}J^{F,\gamma}_t$, is product-measurable. Finally, at every $\omega$,

and for each natural $p$, $(x+y)^p\le2^p(x^p+y^p)$ for $x,y\ge0$, so $\mathbb{E}[|\tilde{J}^{F,\gamma}_t|^p]<\infty$ by the moments of $\Xi_T$. 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

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

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 $(x+y)^q\le2^q(x^q+y^q)$,

on $\Omega_0$, using $\sum_k(\Delta_k\tilde{c})^q\le(\sum_k\Delta_k\tilde{c})^{q-1}\max_k\Delta_k\tilde{c}\le\tilde{c}_T^{\,q}$ for the first part and $n\delta^q=(t-r)\delta^{q-1}$ 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 $Z\,(\tilde{J}^{F,\gamma}_t-\tilde{J}^{F,\gamma}_r)$ holds by the Cauchy-Schwarz inequality for the mean-square norm and (a). Telescoping and using the decomposition of Step 1,

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 $\mathbb{E}[Z\,S^{F,\gamma}_k]=0$. By (1.1) and the Cauchy-Schwarz inequality,

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 $\tilde{J}^F_0=0$ on $\Omega_0$, an event of probability $1$. 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 $X_k=\tilde{J}^{F,\gamma}_{t_k}-\tilde{J}^{F,\gamma}_r$ and $Y_k=\tilde{J}^{G,\delta}_{t_k}-\tilde{J}^{G,\delta}_r$, 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,

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 $\Delta_kX=S^{F,\gamma}_k+\rho^{F,\gamma}_k$ and $\Delta_kY=S^{G,\delta}_k+\rho^{G,\delta}_k$. First,

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 $\Omega_0$),

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),

Summing over $k$ and using (1.2) with $q=2$ 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

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 $\mathbb{E}[Z\int_{[r,t]}\sum_\upsilon F^{\gamma\upsilon}_sG^{\delta\upsilon}_s\tilde{b}^\upsilon(\Sigma_s)ds]$. 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 $(\gamma,\delta)$ entry of $F_sD(\Sigma_s)G_s^{\top}$ being exactly $\sum_\upsilon F^{\gamma\upsilon}_sG^{\delta\upsilon}_s\tilde{b}^\upsilon(\Sigma_s)$ 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 $M^\gamma_t=\Sigma^\gamma_t-\Sigma^\gamma_0-\int_{[0,t]}b^\gamma(\Sigma_s,\alpha_s)ds$, 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 $t\ge s$,

by part (a) of the multiplier lemma. Now telescope with $X_k=M^\gamma_{t_k}-M^\gamma_r$ and $Y_k=\tilde{J}^{F,\delta}_{t_k}-\tilde{J}^{F,\delta}_r$ 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 $\Delta_kY=S^{F,\delta}_k+\rho^{F,\delta}_k$ and expand $\Delta_kX$ by the almost sure representation above:

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),

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 $\gamma$ and abbreviate $X_s=\tilde{J}^{F,\gamma}_s$. 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,

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 $f_a=F^{\gamma\upsilon}_{t_k}$ 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 $0\le\tilde{b}^\upsilon\le\tilde{B}$:

Linear multiplier. Write $\Delta_kX=S_k+\rho_k$ with $S_k=S^{F,\gamma}_k$, $\rho_k=\rho^{F,\gamma}_k$. 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 $\omega_F\le2\bar{F}$, $|3S_k^2\rho_k+3S_k\rho_k^2+\rho_k^3|\le7(3\bar{F})^2\,\omega_F(\delta)\,\lambda_k^3$. Using (1.2) with $q=3$ and the Cauchy-Schwarz inequality, the total contribution of these remainder terms over all $k$ is at most $\omega_F(\delta)$ times a finite constant independent of $n$, hence tends to $0$. 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 $\mathbb{E}[X_{t_k}(\Delta_kM^a)^3]=\mathbb{E}[X_{t_k}((\Delta_kM^a)^3-\Delta_k A^a)]+\mathbb{E}[X_{t_k}\Delta_kA^a]$ (writing $A^a=\tilde{A}^{i,\upsilon}$ 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, $\sum_{a\in\mathcal{O}}\Delta_kA^a\le N\tilde{l}\tilde{B}\delta$ pathwise on $\Omega_0$ (compensator identity and $\tilde{b}^\upsilon\le\tilde{B}$), so

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); by (1.2) with $q=4$ the total of these remainders tends to $0$ as before. 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 $k=4$, $\mathbb{E}[(\Delta_kM^a)^4]\le\mathbb{E}[\Delta_kA^a]+C_\star\delta^2$, 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 $\bar{F}^4\tilde{l}\tilde{B}\,t+\bar{F}^4\tilde{l}C_\star t\,\delta$.