Proof of Uniform Mean-Square Bound for the Martingale Part of the Empirical State Measure

Step 1: an almost sure event of good paths. By clause (a) of the martingale decomposition, there is an event $\Omega_1\in\mathcal{F}$ with $P(\Omega_1)=1$ such that for every $\omega\in\Omega_1$ and every $\gamma$ the path $s\mapsto b^\gamma(\Sigma_s(\omega),\alpha_s(\omega))$ is measurable on $[0,T]$ and bounded in absolute value by $2(l-1)B$. Put $\Omega_*=\Omega_0\cap\Omega_1$. Its complement is the union of the two events $\Omega\setminus\Omega_0$ and $\Omega\setminus\Omega_1$, each of probability $0$; these two events need not be disjoint, but countable subadditivity of a measure gives $P(\Omega\setminus\Omega_*)=0$ and hence $P(\Omega_*)=1$. Also $\Omega_*\subseteq\Omega_0$.

Fix $\omega\in\Omega_*$ and $\gamma$. By condition 1 of the definition of a solution, each path $t\mapsto\sigma^i_t(\omega)$ is constant on each interval of a finite partition of $[0,T]$ into intervals that are closed on the left, so each occupation indicator $\eta^{i,\gamma}$ has this property too, and hence so does the empirical state measure $\Sigma^\gamma=\frac{1}{N}\sum_i\eta^{i,\gamma}$, the relevant partition for the latter being the common refinement generated by the finitely many times $t^{(i)}_k$ with $1\le i\le N$. In particular $t\mapsto\Sigma^\gamma_t(\omega)$ is right-continuous at every $t\in[0,T)$, being constant on a right-neighbourhood of each such $t$, and it is measurable in $t$. The map $t\mapsto\int_{[0,t]}\mathbf{1}_{\Omega_0}b^\gamma(\Sigma_s,\alpha_s)\,ds$ satisfies, for $0\le r\le t\le T$,

$$\Big|\int_{[0,t]}\mathbf{1}_{\Omega_0}b^\gamma(\Sigma_s,\alpha_s)ds-\int_{[0,r]}\mathbf{1}_{\Omega_0}b^\gamma(\Sigma_s,\alpha_s)ds\Big|=\Big|\int_{[r,t]}\mathbf{1}_{\Omega_0}b^\gamma(\Sigma_s,\alpha_s)ds\Big|\le2(l-1)B\,(t-r).$$

Indeed the indicator functions of the three intervals satisfy $\mathbf{1}_{[0,t]}=\mathbf{1}_{[0,r]}+\mathbf{1}_{[r,t]}$ off the Lebesgue null set $\{r\}$, so linearity of the integral gives the equality, and monotonicity of the integral together with the bound $2(l-1)B$ of part (a) of the martingale decomposition gives the inequality. Hence that map is continuous in $t$. Therefore $t\mapsto M^\gamma_t(\omega)$ is right-continuous at every $t\in[0,T)$, and since $\Sigma_t\in\Delta^l$ gives $|\Sigma^\gamma_t|\le1$ and $|\Sigma^\gamma_0|\le1$,

$$|M^\gamma_t(\omega)|\le1+1+2(l-1)BT=K_M .$$

Step 2: Doob's inequality. By clause (b) of the martingale decomposition, each $(M^\gamma_t)_{t\in[0,T]}$ is a square-integrable martingale with respect to the system filtration, with time index restricted to $[0,T]$, and $M^\gamma_0=0$. Step 1 verifies the path hypotheses of Doob's L2 maximal inequality for bounded right-continuous martingales on the event $\Omega_*$, with constant $K_M$. Only the properties $P(\Omega_*)=1$, $\Omega_*\subseteq\Omega_0$, and, for every $\omega\in\Omega_*$ and every $\gamma$, the right-continuity of $t\mapsto\Sigma^\gamma_t(\omega)$ and the bound $|M^\gamma_t(\omega)|\le K_M$ for every $t\in[0,T]$ and the right-continuity of $t\mapsto M^\gamma_t(\omega)$, are used in Steps 2--4; hence the conclusions hold for any event with those properties. Hence each $\overline{M^\gamma}$ is a random variable, with

$$\mathbb{E}\big[(\overline{M^\gamma})^2\big]\le4\,\mathbb{E}\big[(M^\gamma_T)^2\big] .$$

The map $\overline{M}$ is then a random variable, being the composition of the sequentially continuous map $z\mapsto|z|$ on $\mathbb{R}^l$ with the map whose components are the $\overline{M^\gamma}$, by measurability of sequentially continuous functions of measurable Euclidean maps.

Step 3: the terminal second moment. Apply the first covariation identity of clause (c) of the martingale decomposition with $r=0$, $t=T$, $\gamma=\delta$, and with the conditioning event of that clause taken to be $\Omega$, which lies in $\mathcal{F}^{\mathrm{sys}}_0$, and let $\Theta$ be the aggregate fluctuation covariance of $\beta$, as in that theorem. (That clause writes $D$ for its conditioning event; in the present statement $D$ denotes the set of dyadic partition points, a different object.) Since $M^\gamma_0=0$,

$$\mathbb{E}\big[(M^\gamma_T)^2\big]=\frac{1}{N}\,\mathbb{E}\Big[\int_{[0,T]}\mathbf{1}_{\Omega_0}\,\Theta^{\gamma\gamma}(\Sigma_s,\alpha_s)\,ds\Big]\le\frac{2(l-1)B\,T}{N},$$

using the bound $|\Theta^{\gamma\gamma}|\le2(l-1)B$ from clause (a) of that theorem and monotonicity of the integral.

Step 4: conclusion. Combining Steps 2 and 3 and summing over the $l$ values of $\gamma$,

$$\mathbb{E}\big[\overline{M}^{\,2}\big]=\sum_{\gamma=1}^l\mathbb{E}\big[(\overline{M^\gamma})^2\big]\le\sum_{\gamma=1}^l\frac{8(l-1)B\,T}{N}=\frac{8\,l\,(l-1)B\,T}{N} .$$

Finally, for $\omega\in\Omega_*$ and $t\in[0,T]$, the supremum clause of the supremum lemma gives $|M^\gamma_t(\omega)|\le\overline{M^\gamma}(\omega)$ for each $\gamma$, so

$$|M_t(\omega)|^2=\sum_{\gamma=1}^l\big(M^\gamma_t(\omega)\big)^2\le\sum_{\gamma=1}^l\big(\overline{M^\gamma}(\omega)\big)^2=\overline{M}(\omega)^2 . \qquad\blacksquare$$