Proof of The Realized Control of the Controlled N-Agent Dynamics as a Random Element of the Control Set

Write $\mathcal{G}=\mathcal{B}_{[0,T]}\otimes\mathcal{F}$ for the product $\sigma$-algebra on $[0,T]\times\Omega$. A real-valued map on a measurable space is called measurable when it is measurable with respect to the given $\sigma$-algebra and the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of the real line.

Three devices. Let $(\Xi,\mathcal{H})$ be a measurable space.

(D1) (Arithmetic.) If $f^{1},\dots,f^{d}:\Xi\to\mathbb{R}$ are measurable, $E\subseteq\mathbb{R}^{d}$ is nonempty with $(f^{1}(\xi),\dots,f^{d}(\xi))\in E$ for every $\xi$, and $g:E\to\mathbb{R}$ is sequentially continuous on $E$, then $g(f^{1},\dots,f^{d})$ is measurable, by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable. Since sums, differences, products, absolute values, maxima and minima of finitely many real arguments are given by sequentially continuous functions on $\mathbb{R}^{d}$, all of these operations preserve measurability of real-valued maps; the same device applied to the indicator of a set $A\in\mathcal{H}$, which is measurable because its preimages are $\emptyset$, $A$, $\Xi\setminus A$ and $\Xi$, shows that a product of a measurable map with such an indicator is measurable. In particular $\{f^{1}\le f^{2}\}=\{f^{2}-f^{1}\ge0\}\in\mathcal{H}$ for measurable $f^{1},f^{2}$.

(D2) (Piecewise measurability.) Let $(E_{n})_{n}$ be a countable family in $\mathcal{H}$ with $\bigcup_{n}E_{n}=\Xi$, and let $f:\Xi\to\mathbb{R}$ be such that $E_{n}\cap f^{-1}(A)\in\mathcal{H}$ for every $n$ and every $A\in\mathcal{B}(\mathbb{R})$. Then $f$ is measurable, since $f^{-1}(A)=\bigcup_{n}\bigl(E_{n}\cap f^{-1}(A)\bigr)$.

(D3) (Sublevel criterion.) Let $f:\Xi\to\mathbb{R}$ be bounded, say $0\le f\le K$, and suppose $\{f\le q\}\in\mathcal{H}$ for every real $q$. Then $f$ is measurable. Indeed, for each natural number $n$ let

a finite sum of constants times indicators of sets of $\mathcal{H}$, hence measurable by (D1); the sets involved are pairwise disjoint and cover $\Xi$, and $0\le f-f_{n}\le2^{-n}$ on each of them, so $(f_{n}(\xi))_{n}$ converges to $f(\xi)$ for every $\xi$ and $|f_{n}|\le K+1$. Claim 2 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions gives the measurability of $f$.

Step 1: the observation totals are counting paths. By condition 3 of Solution of the Controlled N-Agent Dynamics each $\tilde{N}^{i,\upsilon}_{t}$ is a random variable, so by (D1) so are the channel subtotals $\tilde{c}^{\upsilon}_{t}=\sum_{i=1}^{N}\tilde{N}^{i,\upsilon}_{t}$ and the observation total $\tilde{c}_{t}=\sum_{\upsilon=1}^{\tilde{l}}\tilde{c}^{\upsilon}_{t}$, for each fixed $t\in[0,T]$.

Fix $\omega\in\Omega$. For every $i$ and $\upsilon$ the map $t\mapsto\tilde{N}^{i,\upsilon}_{t}(\omega)$ on $[0,T]$ agrees with the restriction of a counting path: at points of $\Omega_{0}$ this is required by condition 3, and outside $\Omega_{0}$ the counter vanishes identically, which is the restriction of the counting path that is identically zero. In particular each such map is nondecreasing, takes values in the nonnegative integers, vanishes at $t=0$, and satisfies the right-continuity property of claim 3 of Counting Path and Its Jump Times. For a nondecreasing integer-valued map that property says exactly that the map is constant on some interval $[t,t+\eta]$ with $\eta>0$ to the right of each $t<T$; a finite sum of such maps is again nondecreasing, integer-valued, zero at $t=0$, and constant to the right of each $t<T$ on the intersection of finitely many such intervals. Hence for every $\omega$ the maps $t\mapsto\tilde{c}^{\upsilon}_{t}(\omega)$ and $t\mapsto\tilde{c}_{t}(\omega)$ on $[0,T]$ are nondecreasing, integer-valued, vanish at $t=0$, and are right-continuous in that sense.

Step 2: the times $\tau_{j}$ and $\tau^{\upsilon}_{n}$. Let $c$ be any map $[0,T]\to\mathbb{R}$ with the four properties just listed, let $j\ge1$ be a natural number, and put $\theta=\inf\{t\in[0,T]:c(t)\ge j\}$, with $\theta=T+1$ when this set is empty. We claim that for $q\in[0,T]$,

If $c(q)\ge j$ then $q$ belongs to the set $E=\{t\in[0,T]:c(t)\ge j\}$, so $\theta\le q$. Conversely suppose $\theta\le q$; then $E$ is nonempty, and by monotonicity of $c$ it is an up-set: if $t\in E$ and $t\le t'\le T$ then $t'\in E$.

We claim $\theta\in E$. Suppose not. Then $\theta<T$: for if $\theta=T$, then $E$ is a nonempty up-set contained in $[0,T]$ with greatest lower bound $T$, so $E=\{T\}$, whence $\theta=T\in E$, a contradiction. Also, since $\theta$ is the greatest lower bound of $E$ and $\theta\notin E$, every $s$ with $\theta<s\le T$ satisfies $s\in E$: there is a point of $E$ in $[\theta,s]$, and $E$ is an up-set. Hence $c(s)\ge j$ for all such $s$, while $c(\theta)<j$, so $c$ is not constant on any interval $[\theta,\theta+\eta]$ with $\eta>0$. This contradicts the right-continuity property, which for a nondecreasing integer-valued map on $[0,T]$ provides exactly such an interval at every point $\theta<T$. So $\theta\in E$ after all, and then $c(q)\ge c(\theta)\ge j$ by monotonicity, since $\theta\le q$.

Apply this to $c=\tilde{c}(\omega)$. The resulting $\theta$ is $\tau_{j}(\omega)$, so the displayed equivalence of claim 1 holds for every $\omega\in\Omega$, every $j\ge1$ and every $t\in[0,T]$. The sets $\{t\in[0,T]:\tilde{c}_{t}\ge j\}$ decrease as $j$ increases, so their infima increase and $\tau_{1}(\omega)\le\tau_{2}(\omega)\le\cdots$. Also $\tau_{j}$ takes values in $[0,T]\cup\{T+1\}$ and, for every real $q$,

all of which lie in $\mathcal{F}$ because $\tilde{c}_{q}$ is a random variable. Since $0\le\tau_{j}\le T+1$, device (D3) shows that $\tau_{j}$ is a random variable. For $\upsilon\in\{1,\dots,\tilde{l}\}$ and a natural number $n\ge1$ the same argument applied to $c=\tilde{c}^{\upsilon}(\omega)$ shows that

is a random variable with values in $[0,T]\cup\{T+1\}$, and that $\tilde{c}^{\upsilon}_{t}(\omega)\ge n$ if and only if $\tau^{\upsilon}_{n}(\omega)\le t$.

Step 3: the channels $\upsilon_{j}$. For $j\ge1$ put

when that set is nonempty, and $\upsilon_{j}=1$ otherwise. For each $\upsilon$ the set $A_{j,\upsilon}=\bigcup_{n\ge1}\{\tau^{\upsilon}_{n}=\tau_{j}\}$ lies in $\mathcal{F}$, since each $\{\tau^{\upsilon}_{n}=\tau_{j}\}=\{\tau^{\upsilon}_{n}\le\tau_{j}\}\cap\{\tau_{j}\le\tau^{\upsilon}_{n}\}$ lies in $\mathcal{F}$ by (D1) and the union is countable; and $\{\upsilon_{j}=\upsilon\}$ is a finite intersection of members of $\mathcal{F}$ and of their complements, formed from the $A_{j,\upsilon'}$. As $\upsilon_{j}$ takes finitely many values, it is a random variable.

Now fix $\omega\in\Omega_{0}$ and $j$ with $1\le j\le\tilde{c}_{T}(\omega)$. By Step 1 the map $t\mapsto\tilde{c}_{t}(\omega)$ agrees on $[0,T]$ with the restriction of a counting path $c$, and condition 3 requires this; since $c(t)\ge j$ for some $t\le T$, the $j$-th jump time of $c$ in the sense of Counting Path and Its Jump Times is the infimum of $\{t\ge0:c(t)\ge j\}$, which equals $\tau_{j}(\omega)$ because the qualifying times already occur in $[0,T]$. As $\tilde{c}_{0}(\omega)=0<j$ we have $\tau_{j}(\omega)>0$. Condition 5 lists $\tau_{1}<\dots<\tau_{K_{T}}$ as the jump times of the observation total, so the number $\tau_{j}(\omega)$ is the $j$-th of them, as asserted.

It remains to identify $\upsilon_{j}(\omega)$ with the channel of condition 5. Write $\theta=\tau_{j}(\omega)$. For a channel $\upsilon$, say that $\tilde{c}^{\upsilon}$ jumps at $\theta$ if $\tilde{c}^{\upsilon}_{\theta}(\omega)$ exceeds $\tilde{c}^{\upsilon}_{s}(\omega)$ for every $s\in[0,\theta)$. If $\tilde{c}^{\upsilon}$ jumps at $\theta$, then with $n=\tilde{c}^{\upsilon}_{\theta}(\omega)$ we have $\tilde{c}^{\upsilon}_{s}(\omega)<n$ for $s<\theta$ and $\tilde{c}^{\upsilon}_{\theta}(\omega)\ge n$, so $\tau^{\upsilon}_{n}(\omega)=\theta$. Conversely if $\tau^{\upsilon}_{n}(\omega)=\theta$ for some $n\ge1$ then, by the equivalence of Step 2, $\tilde{c}^{\upsilon}_{s}(\omega)<n$ for every $s<\theta$ while $\tilde{c}^{\upsilon}_{\theta}(\omega)\ge n$, so $\tilde{c}^{\upsilon}$ jumps at $\theta$. Thus $A_{j,\upsilon}$ is, at $\omega$, exactly the condition that $\tilde{c}^{\upsilon}$ jumps at $\theta$. Next, $\tilde{c}^{\upsilon}$ jumps at $\theta$ precisely when some observation counter with channel $\upsilon$ jumps at $\theta$. Indeed, $\tilde{c}^{\upsilon}$ is the sum over $i$ of the counters $\tilde{N}^{i,\upsilon}$, each nondecreasing. If some $\tilde{N}^{i,\upsilon}$ jumps at $\theta$, so that $\tilde{N}^{i,\upsilon}_{\theta}(\omega)>\tilde{N}^{i,\upsilon}_{s}(\omega)$ for every $s\in[0,\theta)$, then adding to this strict inequality the inequalities $\tilde{N}^{i',\upsilon}_{\theta}(\omega)\ge\tilde{N}^{i',\upsilon}_{s}(\omega)$ for the remaining indices $i'$, using claim 3 of Elementary Order Arithmetic in an Ordered Field, gives $\tilde{c}^{\upsilon}_{\theta}(\omega)>\tilde{c}^{\upsilon}_{s}(\omega)$ for every $s\in[0,\theta)$. Conversely, if no counter with channel $\upsilon$ jumps at $\theta$, then for each $i$ there is $s_{i}\in[0,\theta)$ with $\tilde{N}^{i,\upsilon}_{\theta}(\omega)=\tilde{N}^{i,\upsilon}_{s_{i}}(\omega)$, and taking $s$ to be the largest of these finitely many numbers, which lies in $[0,\theta)$ since $\theta>0$, monotonicity gives $\tilde{N}^{i,\upsilon}_{\theta}(\omega)=\tilde{N}^{i,\upsilon}_{s}(\omega)$ for every $i$ and hence $\tilde{c}^{\upsilon}_{\theta}(\omega)=\tilde{c}^{\upsilon}_{s}(\omega)$, so $\tilde{c}^{\upsilon}$ does not jump at $\theta$. Condition 5 states that exactly one channel has this property and calls it $\upsilon_{j}$. Hence the set in the definition of $\upsilon_{j}(\omega)$ is a singleton, whose unique element is the channel of condition 5, and the minimum picks it out. This proves claim 1.

Step 4: $\hat{\alpha}$ is well defined, $\mathcal{A}$-valued, and agrees with $\alpha$ on $\Omega_{0}$. Let $\omega\in\Omega_{0}$, $t\in[0,T]$ and $k=\tilde{c}_{t}(\omega)$. If $k\ge1$ then $k\le\tilde{c}_{T}(\omega)$ by monotonicity, so by claim 1 the numbers $\tau_{1}(\omega)\le\dots\le\tau_{k}(\omega)$ are jump times in $[0,T]$, and $\tau_{k}(\omega)\le t$ by the equivalence of claim 1 applied with $j=k$. Hence $(\tau_{1}(\omega),\dots,\tau_{k}(\omega))\in R_{k}(T)$ and $(t,\tau_{1}(\omega),\dots,\tau_{k}(\omega))\in[0,T]\times R_{k}(T)$, so the value of $h_{k}$ is defined; for $k=0$ the value $h_{0}(t)$ is defined. Thus $\hat{\alpha}$ is well defined. All values of $h$ lie in $\mathcal{A}$ because $h$ is $\mathcal{A}$-valued, and $a_{0}\in\mathcal{A}$, so $\hat{\alpha}(t,\omega)\in\mathcal{A}$ for every $t$ and $\omega$. Finally, condition 5 of Solution of the Controlled N-Agent Dynamics states that at every point of $\Omega_{0}$,

with $\tau_{1},\dots,\tau_{K_{t}}$ and $\upsilon_{1},\dots,\upsilon_{K_{t}}$ the jump times of the observation total and their channels; by claim 1 these are exactly the values of our random variables $\tau_{j}$ and $\upsilon_{j}$ for $j\le K_{t}\le\tilde{c}_{T}(\omega)$. Hence $\hat{\alpha}(t,\omega)=\alpha_{t}(\omega)$ for every $t\in[0,T]$ and every $\omega\in\Omega_{0}$.

Step 5: joint measurability of $\hat{\alpha}$. The maps $(t,\omega)\mapsto t$ and, for each $j$, $(t,\omega)\mapsto\tau_{j}(\omega)$ are $\mathcal{G}$-measurable: preimages of Borel sets are $A\times\Omega$ and $[0,T]\times\tau_{j}^{-1}(A)$ respectively, which lie in $\mathcal{G}$ by Product Sigma-Algebra. Hence for each $j$ the set $\{(t,\omega):\tau_{j}(\omega)\le t\}$ lies in $\mathcal{G}$ by (D1), and therefore so does

with $Q_{0}$ the complement of $\{(t,\omega):\tau_{1}(\omega)\le t\}$; the displayed identity is the equivalence of claim 1.

Set $E_{\ast}=[0,T]\times(\Omega\setminus\Omega_{0})$ and, for each $k\ge0$ and each $v\in\{1,\dots,\tilde{l}\}^{k}$ (the empty tuple when $k=0$),

These sets, together with $E_{\ast}$, form a countable family covering $[0,T]\times\Omega$. On $E_{\ast}$ every component of $\hat{\alpha}$ is constant, so its preimages meet $E_{\ast}$ in $\emptyset$ or $E_{\ast}$.

Fix $k\ge0$ and $v$, and put $P_{k}=[0,T]\times R_{k}(T)\subseteq\mathbb{R}^{1+k}$ (so $P_{0}=[0,T]$), and let $\mathcal{P}_{k}$ be the $\sigma$-algebra on $P_{k}$ generated by the relatively open subsets of $P_{k}$, that is by the sets $U\cap P_{k}$ with $U$ an open subset of $\mathbb{R}^{1+k}$; by Observation-Driven Control Policy each component of $h_{k}(\cdot,\cdot,v)$ is measurable with respect to $\mathcal{P}_{k}$. Let $Z_{k}(t,\omega)=(t,\tau_{1}(\omega),\dots,\tau_{k}(\omega))$. Its components are $\mathcal{G}$-measurable, so $Z_{k}$ is measurable with respect to $\mathcal{G}$ and the $\sigma$-algebra $\mathcal{B}_{1+k}$ of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets by claim 2 there, and $\mathcal{B}_{1+k}$ is the Borel $\sigma$-algebra of $\mathbb{R}^{1+k}$ by claim 5 there. By Step 4, $Z_{k}(t,\omega)\in P_{k}$ for every $(t,\omega)\in E_{k,v}$.

Let $\mathcal{V}$ be the family of subsets $V\subseteq P_{k}$ with $E_{k,v}\cap Z_{k}^{-1}(V)\in\mathcal{G}$. Because $Z_{k}$ maps $E_{k,v}$ into $P_{k}$, preimages of complements and of countable unions taken within $P_{k}$ intersect $E_{k,v}$ in complements within $E_{k,v}$ and in countable unions, so $\mathcal{V}$ is a $\sigma$-algebra on $P_{k}$. It contains every relatively open $V=U\cap P_{k}$, since $E_{k,v}\cap Z_{k}^{-1}(U\cap P_{k})=E_{k,v}\cap Z_{k}^{-1}(U)$ and $U$ lies in $\mathcal{B}_{1+k}$ by claim 4 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets. Hence $\mathcal{P}_{k}\subseteq\mathcal{V}$.

Now let $A\in\mathcal{B}(\mathbb{R})$ and let $\hat{\alpha}^{r}$ be a component of $\hat{\alpha}$. For $(t,\omega)\in E_{k,v}$ we have $\hat{\alpha}^{r}(t,\omega)=h^{r}_{k}(Z_{k}(t,\omega),v)$, so

which lies in $\mathcal{G}$. With the corresponding statement for $E_{\ast}$, device (D2) shows that $\hat{\alpha}^{r}$ is $\mathcal{G}$-measurable. This completes claim 2.

Step 6: claim 3. Fix $\omega\in\Omega$ and let $A\in\mathcal{B}(\mathbb{R})$. The set $\{t\in[0,T]:\hat{\alpha}^{r}(t,\omega)\in A\}$ is the section at $\omega$ of the set $(\hat{\alpha}^{r})^{-1}(A)\in\mathcal{G}$, hence lies in $\mathcal{B}_{[0,T]}$ by the statement on sections in Tonelli and Fubini Theorems, applied to the indicator of that set on the product of the finite, hence $\sigma$-finite, measure spaces $([0,T],\mathcal{B}_{[0,T]},\lambda_{[0,T]})$ of claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval and $(\Omega,\mathcal{F},P)$. So every component of the path $t\mapsto\hat{\alpha}(t,\omega)$ is measurable. All values of the path lie in $\mathcal{A}$ by Step 4, so $|\hat{\alpha}(t,\omega)|\le R$ for every $t$, and by monotonicity of the integral, claim 1 of Linearity and Monotonicity of the Lebesgue Integral, together with claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval,

Thus the path is square-integrable in the sense of claim 1 of The Lebesgue Space of Square-Integrable Vector-Valued Functions on a Compact Interval, and since every one of its values lies in $\mathcal{A}$ it exhibits its own class as an element of $\mathcal{U}_{\mathcal{A}}$, the exceptional null set of The Set of Controls with Values in a Prescribed Subset of Euclidean Space being taken empty. Being everywhere $\mathcal{A}$-valued, the path is an admissible representative of $\hat{\alpha}(\omega)$ in the sense of claim 2 of Stability of the Mean-Field Flow under Perturbation of the Initial State and Weak Convergence of Controls.

Step 7: claim 4. Let $\zeta\in\mathcal{U}_{\mathcal{A}}$ and let its representative, again written $\zeta$, be admissible, so that $|\zeta(t)|\le R$ for every $t$. Let $(w_{r})_{r\in\mathbb{N}}$ be the sequence used in the definition of $\rho$ in the weak metrizability and compactness theorem. Fix $r$ and put

Each component of $(t,\omega)\mapsto\zeta(t)$ and of $(t,\omega)\mapsto w_{r}(t)$ is $\mathcal{G}$-measurable, being the composition of a measurable map on $[0,T]$ with the projection, so $f$ is $\mathcal{G}$-measurable by (D1). Next, $|x\cdot y|\le|x|\,|y|$ for all $x,y\in\mathbb{R}^{m}$: claim 2 of Cauchy-Schwarz Inequality for a Positive Semidefinite Quadratic Form on $\mathbb{R}^n$, applied to the identity matrix, which is symmetric and positive semidefinite, gives $(x\cdot y)^{2}\le(x\cdot x)(y\cdot y)$, and $x\cdot x=|x|^{2}$ by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so taking nonnegative square roots gives the bound. Since $|\hat{\alpha}(t,\omega)-\zeta(t)|\le2R$ by claims 5 and 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, it follows that $|f(t,\omega)|\le2R|w_{r}(t)|$. Moreover $\int_{[0,T]}|w_{r}|\,d\lambda_{[0,T]}$ is finite by claim 4 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval. Writing $f^{+}=\max(f,0)$ and $f^{-}=\max(-f,0)$, both $\mathcal{G}$-measurable by (D1), the Tonelli statement of Tonelli and Fubini Theorems shows that

is $\mathcal{F}$-measurable with values in $[0,\infty]$; both are bounded by $2R\int_{[0,T]}|w_{r}|\,d\lambda_{[0,T]}$, hence real. By claim 4 of The Lebesgue Space of Square-Integrable Vector-Valued Functions on a Compact Interval their difference is $\langle\hat{\alpha}(\omega)-\zeta,w_{r}\rangle_{L^{2}}$, so this is a random variable by (D1), and so is

with values in $[0,1]$. By claim 1 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions the supremum of these over $r$ is a random variable; by claim 1 of that theorem it is $\rho(\hat{\alpha}(\omega),\zeta)$.

Step 8: claim 5. By claim 3 of the compactness lemma the metric space $(\mathcal{U}_{\mathcal{A}},\rho)$ is separable; let $D_{\mathcal{U}}$ be a countable dense subset, so that $\bigl((\mathcal{U}_{\mathcal{A}},\rho),D_{\mathcal{U}}\bigr)$ is a separable metric datum in the sense of Borel Sets and Measurable Maps in a Separable Metric Space. By claim 4 the map $\omega\mapsto\rho(\hat{\alpha}(\omega),q)$ is a random variable for every $q\in D_{\mathcal{U}}$, so claim 3 of Borel Sets and Measurable Maps in a Separable Metric Space makes $\omega\mapsto\hat{\alpha}(\omega)$ measurable with respect to $\mathcal{F}$ and the Borel $\sigma$-algebra of $(\mathcal{U}_{\mathcal{A}},\rho)$, that is, a random element of that space.

By condition 1 of Solution of the Controlled N-Agent Dynamics we have $\sigma^{i}_{0}=\varsigma^{i}_{0}$ on $\Omega_{0}$, and in general each occupation indicator $\eta^{i,\gamma}_{0}$ is the indicator of an event, so each component $\Sigma^{\gamma}_{0}=\frac{1}{N}\sum_{i=1}^{N}\eta^{i,\gamma}_{0}$ is a random variable by (D1). All values of $\Sigma_{0}$ lie in $\Delta^{l}$. By claim 2 of the compactness lemma the space $(\Delta^{l},d_{\Delta})$ is separable; let $D_{\Delta}$ be a countable dense subset. For $q\in D_{\Delta}$ the map $\omega\mapsto d_{\Delta}(\Sigma_{0}(\omega),q)=|\Sigma_{0}(\omega)-q|$ is a random variable by (D1), since $x\mapsto|x-q|$ is sequentially continuous on $\mathbb{R}^{l}$. So claim 3 of Borel Sets and Measurable Maps in a Separable Metric Space makes $\Sigma_{0}$ a random element of $(\Delta^{l},d_{\Delta})$.

Finally, claim 4 of Borel Sets and Measurable Maps in a Separable Metric Space, applied to the two separable metric data just produced, shows that $\omega\mapsto(\Sigma_{0}(\omega),\hat{\alpha}(\omega))$ is a random element of $(X,d_{X})$. $\blacksquare$