Proof of Joint Measurability of the State and Control of the Controlled N-Agent Dynamics

By the level-set criterion of measurability, it suffices to show for each map $X$ in question and each real $c$ that $\{(t,\omega):X(t,\omega)\le c\}$ belongs to the product $\sigma$-algebra. Write $Q_T=(\mathbb{Q}\cap[0,T])\cup\{T\}$, a countable set.

Step 1 (a level-set identity for monotone right-continuous integer paths). Let $X:[0,T]\times\Omega\to\mathbb{R}$ be such that for every $\omega$ the path $t\mapsto X_t(\omega)$ is nondecreasing, right-continuous, and takes nonnegative integer values, and such that $X_q$ is $\mathcal{F}$-measurable for every $q\in Q_T$. Then for every real $c$

For the inclusion $\supseteq$: if $t\le q$ and $X_q(\omega)\le c$ then $X_t(\omega)\le X_q(\omega)\le c$ by monotonicity. For $\subseteq$: suppose $X_t(\omega)\le c$. If $t=T$ take $q=T$. If $t<T$, right-continuity at $t$ gives $X_s(\omega)\to X_t(\omega)$ as $s$ decreases to $t$; since the path is integer-valued and nondecreasing, there is $\varepsilon>0$ with $X_s(\omega)=X_t(\omega)$ for all $s\in[t,t+\varepsilon)\cap[0,T]$, and the nonempty open interval $(t,\min(t+\varepsilon,T))$ contains a rational $q$; then $t\le q$ and $X_q(\omega)=X_t(\omega)\le c$. Each set $[0,q]\times\{X_q\le c\}$ is a rectangle with $[0,q]$ in the trace Borel $\sigma$-algebra and $\{X_q\le c\}\in\mathcal{F}$, so the countable union lies in the product $\sigma$-algebra.

Step 2 (part (a)). Fix indices and set $X_t=\mathbf{1}_{\Omega_0}N^{i,\sigma\gamma}_t$. Off $\Omega_0$ the path is identically $0$; on $\Omega_0$ it agrees with $t\mapsto N^{i,\sigma\gamma}_t$, which by condition 3 of the solution definition coincides on $[0,T]$ with the restriction of a counting path, hence is nondecreasing, right-continuous, and nonnegative-integer valued. Each $X_q$ is $\mathcal{F}$-measurable because $N^{i,\sigma\gamma}_q$ is a random variable (condition 3) and $\Omega_0\in\mathcal{F}$. Step 1 applies. The observation counters $\mathbf{1}_{\Omega_0}\tilde{N}^{i,\upsilon}_t$ are handled identically.

Step 3 (part (b)). By condition 6 of the solution definition, at every $\omega\in\Omega_0$ and every $t$, $\eta^{i,\gamma}_t=\eta^{i,\gamma}_0+\sum_{\sigma\neq\gamma}N^{i,\sigma\gamma}_t-\sum_{\gamma'\neq\gamma}N^{i,\gamma\gamma'}_t$; multiplying by $\mathbf{1}_{\Omega_0}$ makes this an identity on all of $[0,T]\times\Omega$, both sides vanishing off $\Omega_0$. The map $(t,\omega)\mapsto\mathbf{1}_{\Omega_0}(\omega)\eta^{i,\gamma}_0(\omega)$ is the indicator of the rectangle $[0,T]\times(\Omega_0\cap\{\sigma^i_0=\gamma\})$, hence product-measurable. Finite sums and differences of product-measurable real maps are product-measurable, being compositions of the (componentwise measurable, hence jointly measurable into Euclidean space) tuple with the sequentially continuous arithmetic maps, by measurability of sequentially continuous functions of measurable Euclidean maps. Hence $\mathbf{1}_{\Omega_0}\eta^{i,\gamma}$ is product-measurable, and so is $\mathbf{1}_{\Omega_0}\Sigma^\gamma_t=\frac{1}{N}\sum_{i=1}^N\mathbf{1}_{\Omega_0}\eta^{i,\gamma}_t$.

Step 4 (part (c)). Let $K_t=\tilde{c}_t=\sum_{i,\upsilon}\tilde{N}^{i,\upsilon}_t$ be the observation total. By condition 3 it coincides pathwise with the restriction of a counting path and each $K_q$ is a random variable, so as in Steps 1-2 the map $\mathbf{1}_{\Omega_0}K_t$ is product-measurable, and consequently for every nonnegative integer $k$ the set

is product-measurable. By part (iv) of the existence theorem applied at time $T$, for each $j$ the observation event time $\tau_j$ and channel $\upsilon_j$ are measurable on the event $\{K_T\ge j\}$ (that is, sets of the form $\{\tau_j\le c\}\cap\{K_T\ge j\}$ and $\{\upsilon_j=v\}\cap\{K_T\ge j\}$ are events), and $\{K_T\ge k\}\in\mathcal{F}$.

Fix $k\ge1$, a component $j$, and a mark vector $v=(v_1,\dots,v_k)\in\{1,\dots,\tilde{l}\}^k$, and set $A_{k,v}=\Omega_0\cap\{K_T\ge k\}\cap\{\upsilon_1=v_1,\dots,\upsilon_k=v_k\}\in\mathcal{F}$. On $[0,T]\times A_{k,v}$ define $\Phi(t,\omega)=(t,\tau_1(\omega),\dots,\tau_k(\omega))$. By condition 5 of the solution definition, at every $\omega\in A_{k,v}$ the event times satisfy $0\le\tau_1<\dots<\tau_k\le T$, so $\Phi$ maps into $[0,T]\times R_k(T)$, the record space of the policy definition. $\Phi$ is measurable from the trace of the product $\sigma$-algebra on $[0,T]\times A_{k,v}$ to the $\sigma$-algebra generated by the relatively open subsets of $[0,T]\times R_k(T)$: each component of $\Phi$ is measurable ($(t,\omega)\mapsto t$ has rectangle preimages; $(t,\omega)\mapsto\tau_j(\omega)$ has rectangle preimages by the measurability above); every open subset of $\mathbb{R}^{1+k}$ is a countable union of open boxes with rational vertices, whose $\Phi$-preimages are finite intersections of component preimages, hence measurable; and the collection of subsets of $[0,T]\times R_k(T)$ whose $\Phi$-preimage is measurable is a $\sigma$-algebra containing the relatively open sets, hence containing the $\sigma$-algebra they generate. Since $h^j_k(\cdot,\cdot,v)$ is measurable with respect to that $\sigma$-algebra by the policy definition, the composition $G_{k,v}(t,\omega)=h^j_k(t,\tau_1(\omega),\dots,\tau_k(\omega),v)$ is measurable on $[0,T]\times A_{k,v}$. For $k=0$, $G_0(t,\omega)=h^j_0(t)$ is product-measurable on $[0,T]\times\Omega$, its level sets being rectangles.

By condition 5 of the solution definition, at every $\omega\in\Omega_0$ and every $t\in[0,T]$, $\alpha^j_t=h^j_{K_t}(t,\tau_1,\dots,\tau_{K_t},\upsilon_1,\dots,\upsilon_{K_t})$, equal to $h^j_0(t)$ when $K_t=0$; moreover $K_t=k$ implies $K_T\ge k$ by monotonicity, and the marks $(\upsilon_1,\dots,\upsilon_k)$ equal exactly one $v$. Hence for every real $c$

where $D_c=[0,T]\times(\Omega\setminus\Omega_0)$ if $c\ge0$ and $D_c=\varnothing$ otherwise (off $\Omega_0$ the map is $0$). Every set on the right is product-measurable and the unions are countable, so the left-hand side is product-measurable. This proves (c).