Proof of Restricted First Moments and Tail Bounds for the Martingale Part of the Empirical State Measure

We use linearity and monotonicity of the integral throughout, and write $\|X\|_2=\mathbb{E}[X^2]^{1/2}$ for the mean-square norm of a square-integrable random variable.

(a) Since $X^2\le1+X^4$ pointwise, $X$ is square-integrable; $X^2$ is square-integrable because $\mathbb{E}[(X^2)^2]=\mathbb{E}[X^4]<\infty$; $\mathbf{1}_D$ is square-integrable, being bounded, with $\|\mathbf{1}_D\|_2=P(D)^{1/2}$; and $\mathbf{1}_DX$ is square-integrable since $(\mathbf{1}_DX)^2\le X^2$. Claim 1 of the Cauchy-Schwarz inequality applied to $\mathbf{1}_D$ and $\mathbf{1}_DX$ gives, using $\mathbf{1}_D^2=\mathbf{1}_D$,

$$\mathbb{E}[\mathbf{1}_DX]=\mathbb{E}\big[\mathbf{1}_D\cdot(\mathbf{1}_DX)\big]\le P(D)^{1/2}\,\mathbb{E}\big[\mathbf{1}_DX^2\big]^{1/2},$$

and applied to $\mathbf{1}_D$ and $X^2$ gives $\mathbb{E}[\mathbf{1}_DX^2]\le P(D)^{1/2}\,\mathbb{E}[X^4]^{1/2}$. Combining, $\mathbb{E}[\mathbf{1}_DX]\le P(D)^{1/2}\big(P(D)^{1/2}\mathbb{E}[X^4]^{1/2}\big)^{1/2}=P(D)^{3/4}\mathbb{E}[X^4]^{1/4}$. The nonnegative random variable $\mathbf{1}_DX$ has finite expectation, hence is integrable.

(b) Joint measurability. Write $\mathcal{B}_{[0,T]}\otimes\mathcal{F}$ for the product $\sigma$-algebra. The map $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)\Sigma^\gamma_t(\omega)$ is $\mathcal{B}_{[0,T]}\otimes\mathcal{F}$-measurable by part (b) of the joint measurability lemma, and so is $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)\Sigma^\gamma_0(\omega)$: it is the composition of the projection $(t,\omega)\mapsto\omega$, which is measurable because the preimage of $E\in\mathcal{F}$ is the rectangle $[0,T]\times E$, with the random variable $\mathbf{1}_{\Omega_0}\Sigma^\gamma_0$. Next, by the definition of the aggregate state drift and the definition of the empirical state measure,

$$\mathbf{1}_{\Omega_0}(\omega)\,b^\gamma(\Sigma_s(\omega),\alpha_s(\omega))=\frac{1}{N}\sum_{\sigma:\sigma\neq\gamma}\sum_{i=1}^{N}\Big(\mathbf{1}_{\Omega_0}\eta^{i,\sigma}_s\,\beta(\sigma,\gamma,\Sigma_s,\alpha_s)-\mathbf{1}_{\Omega_0}\eta^{i,\gamma}_s\,\beta(\gamma,\sigma,\Sigma_s,\alpha_s)\Big)(\omega),$$

a finite linear combination of the maps that condition 2 of the definition of a solution asserts to be $\mathcal{B}_{[0,T]}\otimes\mathcal{F}$-measurable; it is therefore measurable by measurability of sums and scalar multiples, and it is bounded in absolute value by $2(l-1)B$ by part (a) of the martingale decomposition theorem. Call this map $X_s(\omega)$. Consider the constant filtration $\mathcal{F}_t=\mathcal{F}$ ($t\in[0,T]$). The family $X=(X_s)_{s\in[0,T]}$ is progressively measurable with respect to it: for each $t$, the restriction of $X$ to $[0,t]\times\Omega$ is measurable with respect to $\mathcal{B}_{[0,t]}\otimes\mathcal{F}$, because it is the composition $X\circ\iota$ of $X$ with the inclusion $\iota:[0,t]\times\Omega\to[0,T]\times\Omega$, and $\iota$ is measurable from $\mathcal{B}_{[0,t]}\otimes\mathcal{F}$ to $\mathcal{B}_{[0,T]}\otimes\mathcal{F}$: the preimage $\iota^{-1}(B\times E)=(B\cap[0,t])\times E$ of a rectangle with $B\in\mathcal{B}_{[0,T]}$ and $E\in\mathcal{F}$ is a rectangle of $\mathcal{B}_{[0,t]}\otimes\mathcal{F}$, and such rectangles generate the product $\sigma$-algebra by its definition, so the generator criterion of the definition of a measurable function applies. Hence by claim 4 of the progressive measurability toolkit the family $Y_t(\omega)=\int_{[0,t]}X_s(\omega)\,ds$ is progressively measurable with continuous paths, and by claim 1 of the same toolkit $(t,\omega)\mapsto Y_t(\omega)$ is $\mathcal{B}_{[0,T]}\otimes\mathcal{F}$-measurable. Since $\mathbf{1}_{\Omega_0}(\omega)M^\gamma_t(\omega)=\mathbf{1}_{\Omega_0}(\omega)\Sigma^\gamma_t(\omega)-\mathbf{1}_{\Omega_0}(\omega)\Sigma^\gamma_0(\omega)-Y_t(\omega)$ (the factor $\mathbf{1}_{\Omega_0}$ being already contained in the integrand of $Y$), the first assertion of (b) follows from measurability of sums. The bound $|\mathbf{1}_{\Omega_0}M^\gamma_t|\le1+2(l-1)BT=K_M-1\le K_M$ holds since $\Sigma^\gamma_t,\Sigma^\gamma_0\in[0,1]$ and $|Y_t|\le2(l-1)BT$.

Paths and the random variable $I$. The map $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)|M_t(\omega)|=\big(\sum_\gamma(\mathbf{1}_{\Omega_0}M^\gamma_t)^2\big)^{1/2}$ is a continuous function of jointly measurable maps, hence jointly measurable by measurability of continuous functions of measurable maps, nonnegative, and bounded by $\sqrt{l}K_M$; likewise $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}|M_t|^4$. By the Tonelli theorem (the trace Lebesgue measure on $[0,T]$ and $P$ being finite, hence $\sigma$-finite, measures), for every $\omega$ the section $t\mapsto\mathbf{1}_{\Omega_0}(\omega)|M_t(\omega)|$ is measurable, and $\omega\mapsto\int_{[0,T]}\mathbf{1}_{\Omega_0}(\omega)|M_t(\omega)|\,dt$ is a measurable $[0,\infty]$-valued map, bounded by $\sqrt{l}K_MT$ by monotonicity; this map is $I$, which is therefore a random variable with $0\le I\le\sqrt{l}K_MT$, and for $\omega\in\Omega_0$ its integrand is $|M_t(\omega)|$ while for $\omega\notin\Omega_0$ it vanishes identically, which gives the two descriptions of $I$ in the statement. (The map $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)|M_t(\omega)|$, being nonnegative and real-valued and measurable into the real line, is also measurable into $[0,\infty]$ in the sense of the integral of a nonnegative function, as the Tonelli theorem requires; the same remark applies to $\mathbf{1}_{\Omega_0}|M_t|^4$ below.)

Fourth moments. By part (c) of the moment bounds for the aggregate compensated counters, for every $t\in[0,T]$, $\mathbb{E}[|\sqrt{N}M_t|^4]\le c_M\big(Bt/N+(Bt)^2\big)\le c_M\kappa_T$, since $N\ge1$ and $t\le T$; as $|\sqrt{N}M_t|^4=N^2|M_t|^4$, this gives $\mathbb{E}[|M_t|^4]\le c_M\kappa_TN^{-2}$. For $I$: at $\omega\in\Omega_0$, claim 4 of the integral toolkit with $f=|M_\cdot(\omega)|$ and $g=1$ gives $\big(\int_{[0,T]}|M_t|\,dt\big)^2\le T\int_{[0,T]}|M_t|^2\,dt$, and with $f=|M_\cdot(\omega)|^2$ and $g=1$ gives $\big(\int_{[0,T]}|M_t|^2\,dt\big)^2\le T\int_{[0,T]}|M_t|^4\,dt$ (all these path functions being bounded and measurable by the above); hence $I(\omega)^4\le T^3\int_{[0,T]}|M_t(\omega)|^4\,dt$ on $\Omega_0$, while $I=0$ off $\Omega_0$. Therefore, by monotonicity and the Tonelli theorem applied to $\mathbf{1}_{\Omega_0}|M_t|^4$,

$$\mathbb{E}[I^4]\le T^3\,\mathbb{E}\Big[\int_{[0,T]}\mathbf{1}_{\Omega_0}|M_t|^4\,dt\Big]=T^3\int_{[0,T]}\mathbb{E}\big[\mathbf{1}_{\Omega_0}|M_t|^4\big]\,dt\le T^3\cdot T\cdot c_M\kappa_TN^{-2},$$

using $\mathbb{E}[\mathbf{1}_{\Omega_0}|M_t|^4]\le\mathbb{E}[|M_t|^4]$.

(c) Each $M^\gamma_t$ is a random variable by part (b) of the martingale decomposition theorem, so $|M_t|$ is a nonnegative random variable by the composition lemma, with $\mathbb{E}[|M_t|^4]<\infty$ by (b). Part (a) with $X=|M_t|$ gives $\mathbb{E}[\mathbf{1}_D|M_t|]\le P(D)^{3/4}\big(c_M\kappa_TN^{-2}\big)^{1/4}=(c_M\kappa_T)^{1/4}N^{-1/2}P(D)^{3/4}$, and part (a) with $X=I$ gives $\mathbb{E}[\mathbf{1}_DI]\le P(D)^{3/4}\big(T^4c_M\kappa_TN^{-2}\big)^{1/4}=T(c_M\kappa_T)^{1/4}N^{-1/2}P(D)^{3/4}$.

(d) Since $|M_t|\ge\epsilon$ if and only if $|M_t|^4\ge\epsilon^4$, Markov's inequality applied to the nonnegative random variable $|M_t|^4$ with $a=\epsilon^4$ gives $P(|M_t|\ge\epsilon)\le\mathbb{E}[|M_t|^4]\epsilon^{-4}\le c_M\kappa_T\epsilon^{-4}N^{-2}$; the bound for $I$ follows in the same way from $\mathbb{E}[I^4]\le T^4c_M\kappa_TN^{-2}$. $\blacksquare$