Proof of The Realized Control as the Record-Frozen Control at the Observation Record, and Measurability of the Path-and-Record Closeness Set

Elementary facts used repeatedly. Throughout, $\#F$ denotes the number of elements of a finite set $F$, and a nonempty finite set of real numbers has a largest and a least element (by induction on the number of elements), so that maxima and minima over nonempty finite sets, such as $K_{E}$, are defined. (E1) For points $x,y$ of $\mathbb{R}^{\nu}$ ($\nu$ a natural number), claim 6 of the elementary properties of the Euclidean norm gives $|x+y|\le|x|+|y|$; since $y-x=(-1)(x-y)$, claim 5 there gives $|y-x|=|x-y|$; and applying the first inequality to $x=(x-y)+y$ and to $y=(y-x)+x$ yields the reverse form $\bigl||x|-|y|\bigr|\le|x-y|$. Consequently the Euclidean norm is sequentially continuous on $\mathbb{R}^{\nu}$: if the Euclidean distance from $x_{n}$ to $x$, which equals $|x_{n}-x|$ by claim 2 of the same lemma, converges to $0$, then $\bigl||x_{n}|-|x|\bigr|\le|x_{n}-x|$ and claim 3 of the order properties of limits give $|x_{n}|\to|x|$.

Proof of claim 1. Fix $\omega\in\Omega_{0}$. By condition 3 of the definition of a solution, the observation total $u\mapsto\tilde{c}_{u}(\omega)$ coincides on $[0,T]$ with the restriction of a counting path $c$, so by condition 1 of the definition of a counting path $\tilde{c}_{u}(\omega)$ is $0$ or a natural number for every $u\in[0,T]$. For $u\in[0,T]$ put $J_{u}=\{j\ge1:\tau_{j}(\omega)\le u\}$. By claim 1 of the realized-control lemma, $\tau_{j}(\omega)\le u$ if and only if $\tilde{c}_{u}(\omega)\ge j$; hence $J_{u}=\{j\ge1:j\le\tilde{c}_{u}(\omega)\}$, a set with exactly $\tilde{c}_{u}(\omega)$ elements (empty when $\tilde{c}_{u}(\omega)=0$), so that

Since $\tau_{j}(\omega)\le t$ implies $\tau_{j}(\omega)\le T$ for $t\in[0,T]$, we have $J_{t}\subseteq J_{T}$; and $J_{T}=\{1,\dots,K\}$ with $K=K_{T}(\omega)=\tilde{c}_{T}(\omega)$. In particular $\tilde{c}_{t}(\omega)\le K$.

The event times of condition 5 are $\tau_{1}(\omega),\dots,\tau_{K}(\omega)$. By condition 5 of the definition of a solution and the definition of the observation record, the event times of the solution at $\omega$ are the jump times of $c$ lying in $[0,T]$, listed in increasing order, and there are $K$ of them. By claim 1 of the realized-control lemma, for $1\le j\le K$ the number $\tau_{j}(\omega)$ is the $j$-th jump time of the observation total, that is, $\tau_{j}(\omega)=\inf\{u\ge0:c(u)\ge j\}$ in the sense of the definition of a counting path, and $\tau_{j}(\omega)\in[0,T]$ because $\tilde{c}_{T}(\omega)\ge j$. We check that these $K$ numbers are exactly the jump times of $c$ in $[0,T]$, in increasing order. First, $\tau_{j}(\omega)>0$ for every $j\ge1$, since $\tau_{j}(\omega)\le0$ would give $\tilde{c}_{0}(\omega)\ge j\ge1$, whereas $c(0)=0$. Next, the $\tau_{j}(\omega)$, $1\le j\le K$, are pairwise distinct: if $\tau_{j}(\omega)=\tau_{j+1}(\omega)=:u\in(0,T]$ for some $j<K$, then by the displayed counting identity $c(u)=\tilde{c}_{u}(\omega)\ge j+1$, while for every $s\in[0,u)$ the set $\{i\ge1:\tau_{i}(\omega)\le s\}$ omits $j$ and $j+1$ and is contained in $\{1,\dots,j-1\}$ (as $\tau_{i}(\omega)\le s<u=\tau_{j}(\omega)$ forces $i<j$ by the ordering $\tau_{1}(\omega)\le\tau_{2}(\omega)\le\cdots$ of claim 1 of the realized-control lemma), so $c(s)\le j-1$, whence $c(u-)\le j-1$ and $c(u)-c(u-)\ge2$, contradicting condition 4 of the definition of a counting path. Hence $\tau_{1}(\omega)<\dots<\tau_{K}(\omega)$. Each $\tau_{j}(\omega)$, $1\le j\le K$, is a jump time of $c$: with $u=\tau_{j}(\omega)$ we have $c(u)\ge j$ and, by the same argument, $c(s)\le j-1$ for $s<u$, so $c(u-)\le j-1<c(u)$. Conversely, let $u\in(0,T]$ be a jump time of $c$. If $u$ were none of the $\tau_{j}(\omega)$, $1\le j\le K$, then, the finitely many $\tau_{j}(\omega)$ ($j\le K$) that are smaller than $u$ having a largest one, say $u'<u$ (or $u'=0$ if there is none), we would have $\{j\ge1:\tau_{j}(\omega)\le s\}=\{j\ge1:\tau_{j}(\omega)\le u\}$ for every $s\in[u',u)$ — using $J_{u}\subseteq J_{T}=\{1,\dots,K\}$, so that only the $\tau_{j}(\omega)$ with $j\le K$ can be at most $u$, none of which equals $u$ — hence $c(s)=c(u)$ for $s\in[u',u)$ by the counting identity, so $c(u-)\ge c(u)$ and $u$ would not be a jump time. Therefore the jump times of $c$ in $[0,T]$ are exactly $\tau_{1}(\omega)<\dots<\tau_{K}(\omega)$, and listing them in increasing order returns $\tau_{j}(\omega)$ in the $j$-th place; the channel of condition 5 attached to the $j$-th event time is then $\upsilon_{j}(\omega)$ by claim 1 of the realized-control lemma. This justifies the identification recorded in the Data.

The record-frozen control at the record. Fix $t\in[0,T]$. Suppose first $K\ge1$. By the definition of the observation record, $W(\omega)=\bigl(K,(\tau_{1}(\omega),\dots,\tau_{K}(\omega)),(\upsilon_{1}(\omega),\dots,\upsilon_{K}(\omega))\bigr)$. By the definition of the record-frozen control path, $k_{W(\omega)}(t)$ is the number of indices $j\in\{1,\dots,K\}$ with $\tau_{j}(\omega)\le t$, that is, $k_{W(\omega)}(t)=\#(J_{t}\cap J_{T})=\#J_{t}=\tilde{c}_{t}(\omega)$. Writing $k=\tilde{c}_{t}(\omega)$, that definition gives

read as $h_{0}(t)$ when $k=0$ (the first $k$ event times of the record are $\tau_{1}(\omega),\dots,\tau_{k}(\omega)$ because $J_{t}=\{1,\dots,k\}$ is an initial segment); and by claim 2 of the realized-control lemma $\hat{\alpha}(t,\omega)$ is the same expression, since $\omega\in\Omega_{0}$ and $k=\tilde{c}_{t}(\omega)$. Suppose next $K=0$. Then $W(\omega)=r_{\emptyset}$ by the definition of the observation record, the record $r_{\emptyset}$ has no event times, so $k_{r_{\emptyset}}(t)=0$ and $a^{r_{\emptyset}}(t)=h_{0}(t)$; on the other hand $\tilde{c}_{t}(\omega)\le K=0$, so $\tilde{c}_{t}(\omega)=0=k_{W(\omega)}(t)$ and $\hat{\alpha}(t,\omega)=h_{0}(t)$ by claim 2 of the realized-control lemma. In both cases $k_{W(\omega)}(t)=\tilde{c}_{t}(\omega)$ and $\hat{\alpha}(t,\omega)=a^{W(\omega)}(t)$. Finally $\hat{\alpha}(t,\omega)=\alpha_{t}(\omega)$ for $\omega\in\Omega_{0}$ by claim 2 of the realized-control lemma.

Proof of claim 2. Measurability and bounds for fixed $r$. Fix $r\in\mathbf{R}$. Each component of $a^{r}$ is $\mathcal{B}_{[0,T]}$-measurable by the definition of the record-frozen control path, and each component of $A$ is $\mathcal{B}_{[0,T]}$-measurable by assumption; so by claim 2 of the arithmetic lemma for measurable functions each component of $u\mapsto a^{r}(u)-A_{u}$ is $\mathcal{B}_{[0,T]}$-measurable, and by (E1) and the composition lemma for sequentially continuous functions the map $f_{r}(u)=|a^{r}(u)-A_{u}|$ is $\mathcal{B}_{[0,T]}$-measurable. Since $a^{r}(u)$ and $A_{u}$ lie in $\mathcal{A}$, (E1) with claim 5 of the elementary properties of the Euclidean norm gives $0\le f_{r}(u)\le|a^{r}(u)|+|A_{u}|\le2R$. A nonnegative real-valued map that is $\mathcal{B}_{[0,T]}$-measurable is measurable in the sense of the definition of the integral of a nonnegative function, as noted there, so $\mathsf{d}(r)=\int_{[0,T]}f_{r}\,du$ is defined in $[0,\infty]$. By the indicator lemma and claim 1 of the integral toolkit on a compact interval, $\int_{[0,T]}\mathbf{1}_{[0,T]}\,du=\lambda_{[0,T]}([0,T])=T$; hence, by the scalar-multiple and monotonicity assertions of claim 1 of the linearity and monotonicity theorem applied to $f_{r}\le2R\,\mathbf{1}_{[0,T]}$ (with $2R\ge0$), we get $0\le\mathsf{d}(r)\le2RT$.

Joint measurability. Let $\mathrm{pr}:[0,T]\times\mathbf{R}\to[0,T]$ be the projection $(u,r)\mapsto u$. For $B\in\mathcal{B}_{[0,T]}$ the preimage $\mathrm{pr}^{-1}(B)=B\times\mathbf{R}$ is a measurable rectangle, hence belongs to $\mathcal{B}_{[0,T]}\otimes\mathcal{R}$ by the definition of the product $\sigma$-algebra. Consequently, for each component $A^{j}$ of $A$ and each Borel set $C$ of the real line, the preimage of $C$ under $(u,r)\mapsto A^{j}_{u}$ is $(A^{j})^{-1}(C)\times\mathbf{R}\in\mathcal{B}_{[0,T]}\otimes\mathcal{R}$, so $(u,r)\mapsto A^{j}_{u}$ is $\mathcal{B}_{[0,T]}\otimes\mathcal{R}$-measurable. Each component of $(u,r)\mapsto a^{r}(u)$ is $\mathcal{B}_{[0,T]}\otimes\mathcal{R}$-measurable by claim 1 of the record-frozen control lemma. By claim 2 of the arithmetic lemma, (E1) and the composition lemma, the map $F(u,r)=|a^{r}(u)-A_{u}|$ on $[0,T]\times\mathbf{R}$ is $\mathcal{B}_{[0,T]}\otimes\mathcal{R}$-measurable; it is real-valued and nonnegative.

Measurability of $\mathsf{d}$. Recall that a measure is $\sigma$-finite when its space is the union of a sequence of measurable sets of finite measure. The measure space $([0,T],\mathcal{B}_{[0,T]},\lambda_{[0,T]})$ has total mass $T<\infty$ by claim 1 of the integral toolkit, hence is $\sigma$-finite (take the constant sequence $[0,T]$). The measure space $(\mathbf{R},\mathcal{R},\varrho)$ is, by the definition of the observation record space, the countable disjoint union of the cell measure spaces: the cell $C_{\emptyset}$ carries the one-point measure space with unit mass, of total mass $1$, and for $k\ge1$ and a mark vector $v$ the cell $C_{k,v}$ carries the transport along a bijection of the restriction of $(\mathbb{R}^{k},\mathcal{B}_{k},\lambda_{k})$ to the ordered time simplex $D_{k}(T)$, whose total mass is $\lambda_{k}(D_{k}(T))=T^{k}/k!<\infty$ by claims 1 and 2 of the assembly lemma and the ordered time simplex lemma. Since every cell has finite mass, claim 4(a) of the assembly lemma shows that $\mathbf{R}$ is the union of countably many members of $\mathcal{R}$ of finite $\varrho$-measure, that is (listing them as a sequence, repeating a set if the family is finite), $\varrho$ is $\sigma$-finite. Both factors being $\sigma$-finite and $F$ being $\mathcal{B}_{[0,T]}\otimes\mathcal{R}$-measurable with values in $[0,\infty)$, Tonelli's theorem shows that the map $r\mapsto\int_{[0,T]}F(u,r)\,du=\mathsf{d}(r)$ is $\mathcal{R}$-measurable as a $[0,\infty]$-valued map, that is, $\{r:\mathsf{d}(r)>a\}\in\mathcal{R}$ for every real $a$; as $\mathsf{d}$ is real-valued, this is measurability with respect to $\mathcal{R}$ and the Borel $\sigma$-algebra of the real line, as noted in the definition of the integral of a nonnegative function.

The record of the solution. Let $\omega\in\Omega_{0}$. By claim 1, $a^{W(\omega)}(u)=\hat{\alpha}(u,\omega)$ for every $u\in[0,T]$, so the integrands $u\mapsto|a^{W(\omega)}(u)-A_{u}|$ and $u\mapsto|\hat{\alpha}(u,\omega)-A_{u}|$ coincide at every point of $[0,T]$, and their integrals agree: $\mathsf{d}(W(\omega))=\int_{[0,T]}|\hat{\alpha}(u,\omega)-A_{u}|\,du$.

Proof of claim 3. Fix $p\in\mathsf{Path}(E,T)$. For every $u\in[0,T]$, $p(u)\in E$ gives $|p(u)|\le K_{E}$, so by (E1) $0\le|p(u)-S^{*}_{u}|\le|p(u)|+|S^{*}_{u}|\le K_{E}+K^{*}$. The set $\{|p(u)-S^{*}_{u}|:u\in[0,T]\}$ is therefore nonempty and bounded above by $K_{E}+K^{*}$, so its least upper bound $\mathsf{s}(p)$ exists by the least upper bound property of the real numbers and satisfies $0\le\mathsf{s}(p)\le K_{E}+K^{*}$; the same property furnishes the supremum over the nonempty subset $\mathcal{D}$ of $[0,T]$ used below.

Reduction to dyadic times. The set $\mathcal{D}$ is a nonempty subset of $[0,T]$ (it contains $0$ and $T$), so $\sup_{q\in\mathcal{D}}|p(q)-S^{*}_{q}|\le\mathsf{s}(p)$. For the reverse inequality let $u\in[0,T]$; we show $|p(u)-S^{*}_{u}|\le\sup_{q\in\mathcal{D}}|p(q)-S^{*}_{q}|$. If $u=T$ this is clear since $T\in\mathcal{D}$. Let $u<T$. By claim 2 of the path-space lemma there is a real $\delta>0$ with $p(s)=p(u)$ for all $s\in[u,u+\delta)\cap[0,T]$. For every natural number $n$ let $i_{n}$ be the least integer $i\ge0$ with $iT2^{-n}\ge u$ (such integers exist by clause 1 of the Archimedean property, and a least one exists by the well ordering of the natural numbers, applied to the nonempty set of natural numbers $i$ with $iT2^{-n}\ge u$ when $u>0$, while $i_{n}=0$ when $u=0$), and put $q_{n}=i_{n}T2^{-n}$. Then $q_{n}\ge u$; if $i_{n}\ge1$ the minimality of $i_{n}$ gives $(i_{n}-1)T2^{-n}<u$, so $q_{n}<u+T2^{-n}$, and if $i_{n}=0$ then $u\le0$, so $u=0=q_{n}$; in both cases $u\le q_{n}<u+T2^{-n}$. Moreover $q_{n}\le T$: if $i_{n}\ge1$ then $(i_{n}-1)T2^{-n}<u<T$ gives $i_{n}-1<2^{n}$, so $i_{n}\le2^{n}$; and $i_{n}=0$ is trivial. Hence $q_{n}\in\mathcal{D}\cap[u,u+T2^{-n})$ for every $n$. Now $2^{n}\ge n+1$ for every natural number $n$ (by induction: $2^{1}=2$, and $2^{n+1}=2\cdot2^{n}\ge2n+2\ge n+2$), so $0\le q_{n}-u<T2^{-n}\le T/(n+1)<T/n$. The sequence $(T/n)_{n\ge1}$ converges to $0$: given $\varepsilon>0$, clause 3 of the Archimedean property gives a natural number $n_{1}$ with $1/n_{1}<\varepsilon/T$, and then $0<T/n\le T/n_{1}<\varepsilon$ for all $n\ge n_{1}$. Hence $(q_{n})_{n\ge1}$ converges to $u$ by claim 3 of the order properties of limits. In particular there is a natural number $n_{0}$ with $T2^{-n}<\delta$ for all $n\ge n_{0}$ (take $n_{0}=n_{1}$ for $\varepsilon=\delta$), and for such $n$ we have $q_{n}\in[u,u+\delta)\cap[0,T]$, hence $p(q_{n})=p(u)$.

Convergence of $S^{*}_{q_{n}}$. We show that $|S^{*}_{q_{n}}-S^{*}_{u}|\to0$. Let $\varepsilon>0$. For each $\gamma\in\{1,\dots,l\}$ the component $S^{*,\gamma}$ is continuous at $u$ relative to $[0,T]$, so there is $\delta_{\gamma}>0$ with $|S^{*,\gamma}_{s}-S^{*,\gamma}_{u}|<\varepsilon/(2l)$ for all $s\in[0,T]$ with $|s-u|<\delta_{\gamma}$. Let $\delta'$ be the least of $\delta_{1},\dots,\delta_{l}$; since $q_{n}\to u$ there is $n_{2}$ with $|q_{n}-u|<\delta'$ for all $n\ge n_{2}$, and for such $n$ every coordinate of $S^{*}_{q_{n}}-S^{*}_{u}$ has absolute value at most $\varepsilon/(2l)$, so $|S^{*}_{q_{n}}-S^{*}_{u}|\le l\cdot\varepsilon/(2l)=\varepsilon/2<\varepsilon$ by the coordinate bound for the Euclidean norm. This being true for every $\varepsilon>0$, the limit of $|S^{*}_{q_{n}}-S^{*}_{u}|$ is $0$.

Conclusion of the reduction. For $n\ge n_{0}$, by $p(q_{n})=p(u)$ and the reverse form in (E1),

Applying claim 3 of the order properties of limits to the sequences indexed by $n\ge n_{0}$ (a shift of the index does not affect convergence, the definition of the limit involving only tails), $|p(u)-S^{*}_{u}|$ is the limit of the sequence $|p(q_{n})-S^{*}_{q_{n}}|$, $n\ge n_{0}$, each term of which is at most $M=\sup_{q\in\mathcal{D}}|p(q)-S^{*}_{q}|$ because $q_{n}\in\mathcal{D}$; by claim 1 of the order properties of limits (comparison with the constant sequence $M$, which converges to $M$), $|p(u)-S^{*}_{u}|\le M$. This holds for every $u\in[0,T]$, hence $\mathsf{s}(p)\le\sup_{q\in\mathcal{D}}|p(q)-S^{*}_{q}|$, and the two suprema agree.

Measurability of $\mathsf{s}$. For each natural number $n$ put $\mathcal{D}_{n}=\{iT2^{-n}:i\in\{0,1,\dots,2^{n}\}\}$, a finite set, and define $g_{n}:\mathsf{Path}(E,T)\to\mathbb{R}$ by $g_{n}(p)=\max_{q\in\mathcal{D}_{n}}|p(q)-S^{*}_{q}|$. For fixed $q\in[0,T]$ the map $p\mapsto|p(q)-S^{*}_{q}|$ is $\mathcal{C}_{T}$-measurable by claim 1 of the path-space lemma, applied to the function $\phi:E\to\mathbb{R}$, $\phi(y)=|y-S^{*}_{q}|$. The maximum of finitely many measurable real-valued maps is measurable by repeated application of claim 4 of the arithmetic lemma, so each $g_{n}$ is $\mathcal{C}_{T}$-measurable, and $|g_{n}(p)|\le K_{E}+K^{*}$ for every $p$ by the bound above. Since $iT2^{-n}=(2i)T2^{-(n+1)}$ with $2i\le2^{n+1}$, we have $\mathcal{D}_{n}\subseteq\mathcal{D}_{n+1}$, and $\mathcal{D}=\bigcup_{n\ge1}\mathcal{D}_{n}$ (the points $iT2^{-n}$ of $\mathcal{D}$ with $n=0$, namely $0$ and $T$, also lie in $\mathcal{D}_{1}$). Hence, for every $p$, $\sup_{n\ge1}g_{n}(p)=\sup_{q\in\mathcal{D}}|p(q)-S^{*}_{q}|$: each $g_{n}(p)$ is a maximum over a subset of $\mathcal{D}$, so it is at most the right-hand side, and every $q\in\mathcal{D}$ lies in some $\mathcal{D}_{n}$, so $|p(q)-S^{*}_{q}|\le g_{n}(p)\le\sup_{n}g_{n}(p)$. By the reduction to dyadic times, $\sup_{n\ge1}g_{n}(p)=\mathsf{s}(p)$. Claim 1 of the toolkit for countable suprema applied to the uniformly bounded measurable maps $g_{n}$ ($n\ge1$) now shows that $\mathsf{s}$ is $\mathcal{C}_{T}$-measurable.

Proof of claim 4. The set $(-\infty,\varepsilon]$ is closed in the real line, its complement $(\varepsilon,\infty)$ being open, so it belongs to the Borel $\sigma$-algebra, which contains the open sets and is closed under complements. Hence $\{p:\mathsf{s}(p)\le\varepsilon\}=\mathsf{s}^{-1}((-\infty,\varepsilon])\in\mathcal{C}_{T}$ by claim 3, and likewise $\{r:\mathsf{d}(r)\le\varepsilon'\}=\mathsf{d}^{-1}((-\infty,\varepsilon'])\in\mathcal{R}$ by claim 2. The closeness set $\mathsf{E}(\varepsilon,\varepsilon')$ is the Cartesian product of these two sets, a measurable rectangle, and therefore belongs to $\mathcal{C}_{T}\otimes\mathcal{R}$ by the definition of the product $\sigma$-algebra. This completes the proof.