Proof of Mean-Square Linearization Residual of the Observation Fluctuation Process

Throughout, fix the data of the statement. All Lebesgue integrals over subintervals of $[0,T]$ are those of the restricted Lebesgue measure and integral toolkit, and we use the linearity and monotonicity of the integral without further comment; every bounded measurable real function on a compact interval is Lebesgue integrable there, by the Cauchy-Schwarz claim of the toolkit applied with the constant function $1$; all integrals over $[0,t]$ below are $0$ for $t=0$; and product-measurability always refers to the product $\sigma$-algebra of the trace Borel $\sigma$-algebra on $[0,T]$ and $\mathcal{F}$. The trace Lebesgue measure of the toolkit and $P$ are finite, hence $\sigma$-finite, measures, so the Tonelli theorem applies to nonnegative product-measurable maps below. Write $d$ for the Euclidean distance.

Step 1 (Simplex geometry and elementary norm estimates). By the definition of a solution of the controlled $N$-agent dynamics, the empirical state measure satisfies $\Sigma_s(\omega)\in\Delta^l$ at every point of $[0,T]\times\Omega$; and $S_s\in\Delta^l$ for every $s$ because $(S,A)$ is a mean-field trajectory pair. For $x\in\mathbb{R}^l$,

the second estimate using $2uv\le u^2+v^2$; so $|x|\le\sum_\gamma|x^\gamma|\le\sqrt{l}\,|x|$. Applying this to $x=\Sigma_s-S_s$, whose components satisfy $|\Sigma^\gamma_s-S^\gamma_s|\le\Sigma^\gamma_s+S^\gamma_s$ with $\sum_\gamma\Sigma^\gamma_s=\sum_\gamma S^\gamma_s=1$ (both points lying in the probability simplex),

so $|\mathfrak{s}_s|=\sqrt{N}\,|\Sigma_s-S_s|\le2\sqrt{N}$ at every point of $[0,T]\times\Omega$. Moreover, for $\tau\in[0,1]$ the point $S_s+\tau(\Sigma_s-S_s)=(1-\tau)S_s+\tau\Sigma_s$ has nonnegative components with sum $(1-\tau)+\tau=1$, so the segment from $S_s$ to $\Sigma_s$ lies in $\Delta^l$, and $\Delta^l\subseteq\tilde{U}$ by the extension definition. Finally, for any $y\in\mathbb{R}^{\tilde{l}}$ we use $|y|^2=\sum_\upsilon(y^\upsilon)^2\le\tilde{l}\,\max_\upsilon(y^\upsilon)^2$.

Step 2 (Proof of (a)). Fix $(s,\omega)\in[0,T]\times\Omega$ and $\upsilon\in\{1,\dots,\tilde{l}\}$, and write $h=\Sigma_s-S_s$, so that $\sqrt{N}\,h=\mathfrak{s}_s$. By part (i) of the regularity of the extended aggregate observation drift, $\bar{\tilde{b}}^\upsilon$ is a $C^1$ map on the open set $\tilde{U}$, each $\partial_\gamma\bar{\tilde{b}}^\upsilon$ is again a $C^1$ map there, and $\bar{\tilde{b}}^\upsilon$ agrees with $\tilde{b}^\upsilon$ on $\Delta^l$; by part (iii), $|\partial_\delta\partial_\gamma\bar{\tilde{b}}^\upsilon|\le3\tilde{K}$ at every point of $\Delta^l$, hence at every point of the segment from $S_s$ to $\Sigma_s$ by Step 1. Part (ii) of the multivariate Taylor expansion lemma, applied with $n=l$, $W=\tilde{U}$, $f=\bar{\tilde{b}}^\upsilon$, $x=S_s$, $y=\Sigma_s$, and $M_2=3\tilde{K}$, gives

Multiplying by $\sqrt{N}$ and using $\sqrt{N}\,h^\gamma=\mathfrak{s}^\gamma_s$ and $\sqrt{N}\,|h|^2=N^{-1/2}\,|\mathfrak{s}_s|^2$,

and the last estimate of Step 1 yields $|\tilde{e}_s|\le\sqrt{\tilde{l}}\,\max_\upsilon|\tilde{e}^\upsilon_s|\le\frac{3\,l\,\sqrt{\tilde{l}}\,\tilde{K}}{2\,\sqrt{N}}\,|\mathfrak{s}_s|^2$, the first bound of (a). For the linear bound, part (ii) of the regularity lemma gives $|\tilde{b}^\upsilon(\Sigma_s)-\tilde{b}^\upsilon(S_s)|\le\sqrt{l}\,(\tilde{B}+\tilde{K})\,d(\Sigma_s,S_s)$, so

and it also gives $|(\tilde{\mathcal{E}}_s)_{\upsilon\gamma}|=|\partial_\gamma\bar{\tilde{b}}^\upsilon(S_s)|\le\tilde{B}+\tilde{K}$ for all indices, so by Step 1

Hence $|\tilde{e}^\upsilon_s|\le|\tilde{g}^\upsilon_s|+|(\tilde{\mathcal{E}}_s\mathfrak{s}_s)^\upsilon|\le2\,\sqrt{l}\,(\tilde{B}+\tilde{K})\,|\mathfrak{s}_s|$ and $|\tilde{e}_s|\le\sqrt{\tilde{l}}\cdot2\,\sqrt{l}\,(\tilde{B}+\tilde{K})\,|\mathfrak{s}_s|=2\,\tilde{\Lambda}\,|\mathfrak{s}_s|$. Combining the linear bound with $|\mathfrak{s}_s|\le2\sqrt{N}$ from Step 1 gives $|\tilde{e}_s|\le4\,\tilde{\Lambda}\,\sqrt{N}$ everywhere. This proves (a).

Step 3 (Proof of (b)). The map $(s,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)$ is product-measurable, being the indicator of the measurable rectangle $[0,T]\times\Omega_0$ (with $\Omega_0\in\mathcal{F}$ by the definition of a solution); we write $\mathbf{1}_{\Omega_0}$ also for this map. Define, componentwise,

Each $(s,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)\,\Sigma^\gamma_s(\omega)$ is product-measurable by the joint measurability of the state and control; each $(s,\omega)\mapsto(1-\mathbf{1}_{\Omega_0}(\omega))\,S^\gamma_s$ is product-measurable by the continuous-composition lemma applied to the sequentially continuous map $(u,v)\mapsto(1-u)\,v$ of the product-measurable pair $(\mathbf{1}_{\Omega_0},S^\gamma_s)$ (the map $(s,\omega)\mapsto S^\gamma_s$ being continuous in $s$ by the trajectory-pair definition, hence product-measurable); and sums of product-measurable maps are again product-measurable (the continuous-composition lemma with $(u,v)\mapsto u+v$). By Step 1, $\Sigma'_s(\omega)\in\Delta^l$ at every point. Define $G^\upsilon:[0,T]\times\Delta^l\to\mathbb{R}$ by

$G^\upsilon$ is sequentially continuous: if $(s_k,x_k)\to(s,x)$ in $[0,T]\times\Delta^l$, then $S_{s_k}\to S_s$ because $S$ is continuous, and $\bar{\tilde{b}}^\upsilon$ and every $\partial_\gamma\bar{\tilde{b}}^\upsilon$ are continuous on $\tilde{U}$ by part (i) of the regularity lemma, so each term converges. The pair map $(s,\omega)\mapsto(s,\Sigma'_s(\omega))\in[0,T]\times\Delta^l\subseteq\mathbb{R}^{1+l}$ is product-measurable in each coordinate, so $(s,\omega)\mapsto G^\upsilon(s,\Sigma'_s(\omega))$ is product-measurable by the continuous-composition lemma. At every point of $[0,T]\times\Omega$,

for $\omega\in\Omega_0$ we have $\Sigma'_s=\Sigma_s$ and, since $\bar{\tilde{b}}^\upsilon=\tilde{b}^\upsilon$ on $\Delta^l$, $G^\upsilon(s,\Sigma_s)=\tilde{g}^\upsilon_s-(\tilde{\mathcal{E}}_s\mathfrak{s}_s)^\upsilon=\tilde{e}^\upsilon_s$; and both sides vanish off $\Omega_0$. The right-hand side is product-measurable (the continuous-composition lemma with $(u,v)\mapsto uv$), which proves the measurability claim of (b).

Let $p^\upsilon$ and $q^\upsilon$ be the positive and negative parts of $\mathbf{1}_{\Omega_0}\tilde{e}^\upsilon$; they are nonnegative, product-measurable (the continuous-composition lemma with $u\mapsto\max(u,0)$ and $u\mapsto\max(-u,0)$), and bounded by $4\tilde{\Lambda}\sqrt{N}$ by (a). By the sections part of the Tonelli theorem for $[0,\infty]$-valued product-measurable maps, at each $\omega$ the sections $s\mapsto p^\upsilon(s,\omega)$ and $s\mapsto q^\upsilon(s,\omega)$ are measurable on $[0,T]$, hence so is their difference $s\mapsto\mathbf{1}_{\Omega_0}(\omega)\,\tilde{e}^\upsilon_s(\omega)$; for $\omega\in\Omega_0$ this difference is the section $s\mapsto\tilde{e}^\upsilon_s(\omega)$, which is bounded by $4\tilde{\Lambda}\sqrt{N}$ by (a), hence Lebesgue integrable over $[0,t]$ for every $t\in[0,T]$ (preamble). Thus $\tilde{\mathcal{R}}_t$ is well defined. Moreover, by the Tonelli theorem the maps $\omega\mapsto\int_{[0,t]}p^\upsilon(s,\omega)\,ds$ and $\omega\mapsto\int_{[0,t]}q^\upsilon(s,\omega)\,ds$ are measurable $[0,\infty]$-valued functions of $\omega$; both are finite everywhere (integrands bounded by $4\tilde{\Lambda}\sqrt{N}$), and at every $\omega$

so $\tilde{\mathcal{R}}^\upsilon_t$ is a random variable. This proves (b).

Step 4 (Proof of (c)). Let $\Omega'$ be the intersection of $\Omega_0$ with the almost-sure event of clause (a) of the martingale decomposition; as an intersection of two events of probability $1$ it has probability $1$, so a statement holding at every $\omega\in\Omega'$ holds almost surely. Fix $\omega\in\Omega'$, $t\in[0,T]$, and $\upsilon$. By clause (a) of the decomposition, $s\mapsto\tilde{b}^\upsilon(\Sigma_s)$ is measurable on $[0,T]$ and bounded by $\tilde{B}$, hence Lebesgue integrable over $[0,t]$; and $s\mapsto\tilde{b}^\upsilon(S_s)$ is continuous (statement), hence also integrable. By the defining formula of $\tilde{M}^\upsilon$ in clause (b) of the decomposition,

so by linearity of the Lebesgue integral

Moreover $s\mapsto\tilde{e}^\upsilon_s(\omega)$ is measurable and bounded on $[0,T]$ by (b) and (a), so $s\mapsto(\tilde{\mathcal{E}}_s\mathfrak{s}_s(\omega))^\upsilon=\tilde{g}^\upsilon_s(\omega)-\tilde{e}^\upsilon_s(\omega)$ is measurable and bounded as a difference of such functions, and by linearity

all integrals existing. Combining the two displays proves (c).

Step 5 (Proof of (d)). Fix $t\in[0,T]$ and set, at each point of $[0,T]\times\Omega$,

so that $|\tilde{e}_s|\le m_s\le4\,\tilde{\Lambda}\,\sqrt{N}$ everywhere by (a). Let $\varphi,\psi:\mathbb{R}^l\to\mathbb{R}$ be the sequentially continuous maps $\varphi(x)=\min\big(\frac{3l\sqrt{\tilde{l}}\tilde{K}}{2\sqrt{N}}|x|^2,\,2\tilde{\Lambda}|x|\big)^2$ and $\psi(x)=|x|^4$, which vanish at $x=0$. The map $(s,\omega)\mapsto\mathbf{1}_{\Omega_0}\mathfrak{s}_s=\sqrt{N}\,(\mathbf{1}_{\Omega_0}\Sigma_s-\mathbf{1}_{\Omega_0}S_s)$ is product-measurable componentwise, by the joint measurability lemma for the first term and the continuous-composition lemma (product of $\mathbf{1}_{\Omega_0}$ with a continuous function of $s$) for the second. Since $\varphi(0)=\psi(0)=0$,

at every point, and both are product-measurable by the continuous-composition lemma. By the Tonelli theorem (the measures being $\sigma$-finite, preamble), $s\mapsto\mathbb{E}[\mathbf{1}_{\Omega_0}m_s^2]$ and $s\mapsto\mathbb{E}[\mathbf{1}_{\Omega_0}|\mathfrak{s}_s|^4]$ are measurable functions of $s$; and for fixed $s$ the functions $m_s^2$ and $|\mathfrak{s}_s|^4$ are random variables (each $\Sigma^\gamma_s$ is a random variable by the definition of a solution, and $\varphi$, $\psi$ are continuous — continuous-composition lemma) agreeing with $\mathbf{1}_{\Omega_0}m_s^2$ and $\mathbf{1}_{\Omega_0}|\mathfrak{s}_s|^4$ on the probability-$1$ event $\Omega_0$, so the corresponding expectations coincide. This justifies the measurability assertions of (d).

Fix $\omega\in\Omega_0$ and $\upsilon$. The section $s\mapsto\tilde{e}^\upsilon_s(\omega)$ is measurable and bounded on $[0,t]$ (Step 3 and (a)), so it and its square are Lebesgue integrable over $[0,t]$ (preamble), and the Cauchy-Schwarz claim of the toolkit, applied to the pair consisting of the constant function $1$ and this section, gives

Summing over $\upsilon\in\{1,\dots,\tilde{l}\}$ and using $\sum_\upsilon(\tilde{e}^\upsilon_s)^2=|\tilde{e}_s|^2\le m_s^2$ with monotonicity,

while $|\tilde{\mathcal{R}}_t|^2=0$ off $\Omega_0$. Hence at every $\omega$

Taking expectations and applying the Tonelli theorem to the nonnegative product-measurable map $\mathbf{1}_{\Omega_0}m^2$,

which is the min-form bound of (d). All quantities are finite because $m_s\le4\tilde{\Lambda}\sqrt{N}$ everywhere, so $\mathbb{E}[m_s^2]\le16\,\tilde{\Lambda}^2N$ and $\mathbb{E}[|\tilde{\mathcal{R}}_t|^2]\le16\,\tilde{\Lambda}^2N\,t^2<\infty$. Finally, from $m_s\le\frac{3l\sqrt{\tilde{l}}\tilde{K}}{2\sqrt{N}}|\mathfrak{s}_s|^2$ we get $m_s^2\le\frac{9\,l^2\,\tilde{l}\,\tilde{K}^2}{4\,N}\,|\mathfrak{s}_s|^4$ pointwise, and monotonicity of the integral and the expectation yields

completing the proof of (d). $\blacksquare$