Proof of Time Shift of the Mean-Field Control Problem: Restriction of a Stationary Triple, Cost Splitting, and the Dynamic Programming Principle

Throughout, $\lambda_{[a,b]}$ and $\mathcal{B}_{[a,b]}$ are the restricted Lebesgue measure and $\sigma$-algebra on a compact interval, measurability of real-valued maps on $[a,b]$ is with respect to $\mathcal{B}_{[a,b]}$ and the Borel $\sigma$-algebra, and we use linearity and monotonicity of the integral freely. A bounded measurable real-valued function on $[a,b]$ is integrable, its absolute value being dominated by a constant, which is integrable on the finite measure space of claim 1 of the toolkit. For a mean-field trajectory pair, the mean-field cost $J^{MF}$, defined with Riemann integrals, agrees with the generalized mean-field cost of the same pair by claim 3 of the toolkit; both notions occur below and are identified where used.

Step 0: shift and splitting of interval integrals. Let $a<b$ and $c$ be real numbers and let $f:[a+c,b+c]\to\mathbb{R}$ be measurable and bounded. We claim that $g:[a,b]\to\mathbb{R}$ defined by $g(s)=f(s+c)$ is measurable and bounded, and

$$\int_{[a,b]}g\,d\lambda_{[a,b]}=\int_{[a+c,b+c]}f\,d\lambda_{[a+c,b+c]}.\tag{0.1}$$

Measurability: for a Borel set $E\subseteq\mathbb{R}$ write $f^{-1}(E)=Q\cap[a+c,b+c]$ with $Q$ Borel; then $s\in g^{-1}(E)$ if and only if $s\in[a,b]$ and $s+c\in Q\cap[a+c,b+c]$, so $g^{-1}(E)=\big(Q+(-c)\big)\cap[a,b]$, and $Q+(-c)$ is Borel by claim 1 of translation invariance; hence $g^{-1}(E)\in\mathcal{B}_{[a,b]}$. For (0.1), let $\tilde{g},\tilde{f}:\mathbb{R}\to\mathbb{R}$ be the zero extensions of claim 2 of the toolkit. For every real $x$: if $x\in[a,b]$ then $x+c\in[a+c,b+c]$ and $\tilde{g}(x)=f(x+c)=\tilde{f}(x+c)$, while if $x\notin[a,b]$ then $x+c\notin[a+c,b+c]$ and $\tilde{g}(x)=0=\tilde{f}(x+c)$; so $\tilde{g}(x)=\tilde{f}(x+c)$ for every $x$. By claim 2 of the toolkit $\tilde{f}$ and $\tilde{g}$ are measurable, hence so are the nonnegative parts $\tilde{f}^{+}=\max(\tilde{f},0)$, $\tilde{f}^{-}=\max(-\tilde{f},0)$ of the integral definition, and pointwise $\tilde{g}^{\pm}(x)=\tilde{f}^{\pm}(x+c)$. Claim 2 of translation invariance (applied with $t=c$ to the nonnegative measurable functions $\tilde{f}^{\pm}$) gives $\int_{\mathbb{R}}\tilde{g}^{\pm}\,d\lambda=\int_{\mathbb{R}}\tilde{f}^{\pm}\,d\lambda$. Since $f$ and $g$ are bounded and measurable they are integrable, and by claim 2 of the toolkit on each interval, both sides of (0.1) equal the common value $\int_{\mathbb{R}}\tilde{f}^{+}\,d\lambda-\int_{\mathbb{R}}\tilde{f}^{-}\,d\lambda$.

Next, let $a<c<b$ and let $f:[a,b]\to\mathbb{R}$ be measurable and bounded. The restrictions of $f$ to $[a,c]$ and $[c,b]$ are measurable (writing $f^{-1}(E)=Q\cap[a,b]$ with $Q$ Borel, the preimage of $E$ under the restriction to $[a,c]$ is $Q\cap[a,c]\in\mathcal{B}_{[a,c]}$, and likewise on $[c,b]$), and

$$\int_{[a,b]}f\,d\lambda_{[a,b]}=\int_{[a,c]}f\,d\lambda_{[a,c]}+\int_{[c,b]}f\,d\lambda_{[c,b]},\tag{0.2}$$

the integrals on the right being those of the restrictions. Indeed, let $\phi_{1},\phi_{2},\phi_{3}:[a,b]\to\mathbb{R}$ equal $f$ on $[a,c]$, on $(c,b]$, and on $[c,b]$ respectively, and $0$ elsewhere on $[a,b]$; each $\phi_{i}$ is the product of $f$ with the indicator of a member of $\mathcal{B}_{[a,b]}$, hence measurable by arithmetic of measurable functions, and bounded; $f=\phi_{1}+\phi_{2}$ on $[a,b]$, and $\phi_{2}$ and $\phi_{3}$ agree off the single point $c$, a set of $\lambda_{[a,b]}$-measure zero, so $\int_{[a,b]}\phi_{2}=\int_{[a,b]}\phi_{3}$ by claim 2 of the null-set lemma. By claim 2 of the toolkit, the zero extensions of $\phi_{1}$ and of the restriction of $f$ to $[a,c]$ coincide, as do those of $\phi_{3}$ and of the restriction of $f$ to $[c,b]$; so $\int_{[a,b]}\phi_{1}=\int_{[a,c]}f$ and $\int_{[a,b]}\phi_{3}=\int_{[c,b]}f$, and (0.2) follows by linearity.

Step 1: proof of claim 1. The map $\tau:[0,T^{\sharp}]\to[0,T]$, $\tau(s)=t_{0}+s$, satisfies $|\tau(s)-\tau(s')|=|s-s'|$, hence is continuous. Every component of $S^{\sharp}$, $A^{\sharp}$ and $P^{\sharp}$ is the composition of the corresponding component of $S$, $A$ or $P$ — continuous on $[0,T]$ by clause 1 of the trajectory-pair definition and clause 1 of the co-state definition — with $\tau$, hence continuous on $[0,T^{\sharp}]$ by continuity of compositions, the metric and Euclidean notions agreeing by claim 1 of the agreement lemma.

Trajectory pair. The map $s\mapsto b^{\gamma}(S_{s},A_{s})$ is continuous on $[0,T]$ by clause 2 of the trajectory-pair definition, so its restriction to any compact subinterval is continuous by claim 1 of restriction stability, and over such subintervals its Riemann and Lebesgue integrals agree by claim 3 of the toolkit. Fix $\gamma$ and $t\in(0,T^{\sharp}]$. By clause 2 of the trajectory-pair definition at times $t_{0}+t$ and $t_{0}$, and by additivity of the Riemann integral on adjacent intervals (when $t_{0}>0$; for $t_{0}=0$ the subtraction is trivial),

$$S^{\sharp\gamma}_{t}-S^{\sharp\gamma}_{0}=S^{\gamma}_{t_{0}+t}-S^{\gamma}_{t_{0}}=\int_{t_{0}}^{t_{0}+t}b^{\gamma}(S_{s},A_{s})\,ds=\int_{[t_{0},t_{0}+t]}b^{\gamma}(S_{s},A_{s})\,ds=\int_{[0,t]}b^{\gamma}(S^{\sharp}_{s},A^{\sharp}_{s})\,ds,$$

the last equality by (0.1) with $c=t_{0}$ — its hypotheses holding because the integrand on $[t_{0},t_{0}+t]$ is continuous, hence measurable and bounded — since $b^{\gamma}(S_{t_{0}+s},A_{t_{0}+s})=b^{\gamma}(S^{\sharp}_{s},A^{\sharp}_{s})$ for $s\in[0,t]$. The map $s\mapsto b^{\gamma}(S^{\sharp}_{s},A^{\sharp}_{s})$ is continuous on $[0,T^{\sharp}]$ (the composition of the continuous $s\mapsto b^{\gamma}(S_{s},A_{s})$ with $\tau$, as above), so by claim 3 of the toolkit the last integral equals the Riemann integral required by clause 2 of the trajectory-pair definition; for $t=0$ the required identity holds trivially, both sides vanishing by the convention of that clause. With the continuity already shown and $S^{\sharp}_{t}\in\Delta^{l}$, $A^{\sharp}_{t}\in\mathcal{A}$ by definition, $(S^{\sharp},A^{\sharp})$ is a mean-field trajectory pair for $\beta$ with horizon $T^{\sharp}$.

Co-state. For $\gamma\in\{1,\dots,l\}$ let $f^{\gamma}(s)=\sum_{\delta=1}^{l}\partial_{\gamma}\bar{b}^{\delta}(S_{s},A_{s})\,P^{\delta}_{s}-\partial_{\gamma}\bar{L}(S_{s},A_{s})$, continuous on $[0,T]$ by clause 2 of the co-state definition. The shifted integrand $s\mapsto f^{\gamma}(t_{0}+s)$ is continuous on $[0,T^{\sharp}]$ (composition with $\tau$) and equals $s\mapsto\sum_{\delta}\partial_{\gamma}\bar{b}^{\delta}(S^{\sharp}_{s},A^{\sharp}_{s})P^{\sharp\delta}_{s}-\partial_{\gamma}\bar{L}(S^{\sharp}_{s},A^{\sharp}_{s})$, the integrand required by clause 2 of the co-state definition for the shifted data. For $t\in[0,T^{\sharp})$, using claim 3 of the toolkit on $[t,T^{\sharp}]$ and on $[t_{0}+t,T]$ (restrictions continuous by restriction stability) and (0.1) with $c=t_{0}$,

$$\int_{t}^{T^{\sharp}}f^{\gamma}(t_{0}+s)\,ds=\int_{[t,T^{\sharp}]}f^{\gamma}(t_{0}+s)\,ds=\int_{[t_{0}+t,T]}f^{\gamma}(s)\,ds=\int_{t_{0}+t}^{T}f^{\gamma}(s)\,ds .$$

Hence, by clause 2 of the co-state definition at time $t_{0}+t$, and since $S^{\sharp}_{T^{\sharp}}=S_{T}$,

$$P^{\sharp\gamma}_{t}=P^{\gamma}_{t_{0}+t}=-\partial_{\gamma}\bar{G}(S_{T})+\int_{t_{0}+t}^{T}f^{\gamma}(s)\,ds=-\partial_{\gamma}\bar{G}(S^{\sharp}_{T^{\sharp}})+\int_{t}^{T^{\sharp}}f^{\gamma}(t_{0}+s)\,ds ,$$

which is clause 2 for the shifted data; at $t=T^{\sharp}$ both prescriptions equal $-\partial_{\gamma}\bar{G}(S_{T})$ by the convention that the integral is $0$ there. Clause 1 (continuity) was shown above, and clause 3 (stationarity) at time $t$ for the shifted triple is clause 3 at time $t_{0}+t$ for the original. So $P^{\sharp}$ is a stationary co-state for the shifted data.

Constants and coefficients. The bound $\sum_{\delta}|P^{\sharp\delta}_{t}|=\sum_{\delta}|P^{\delta}_{t_{0}+t}|\le C_{P}$ is immediate since $t_{0}+t\in[0,T]$. By the definition of the mean-field Hamiltonian along a triple in the quadratic growth lemma, $\mathcal{H}^{\sharp}_{t}(a)=\bar{L}(S^{\sharp}_{t},a)-\sum_{\delta}P^{\sharp\delta}_{t}\bar{b}^{\delta}(S^{\sharp}_{t},a)=\bar{L}(S_{t_{0}+t},a)-\sum_{\delta}P^{\delta}_{t_{0}+t}\bar{b}^{\delta}(S_{t_{0}+t},a)=\mathcal{H}_{t_{0}+t}(a)$. By the definition of the fluctuation Hessian coefficients, $H^{\sharp}_{ij}(t)=\partial_{j}\partial_{i}\bar{L}(S^{\sharp}_{t},A^{\sharp}_{t})-\sum_{\delta}P^{\sharp\delta}_{t}\,\partial_{j}\partial_{i}\bar{b}^{\delta}(S^{\sharp}_{t},A^{\sharp}_{t})=H_{ij}(t_{0}+t)$.

Step 2: proof of claim 2. The matrix $R^{\sharp}_{t}$ of the quadratic growth lemma for the shifted triple has entries $R^{\sharp ij}_{t}=\tfrac14\big(H^{\sharp}_{l+i,l+j}(t)+H^{\sharp}_{l+j,l+i}(t)\big)=\tfrac14\big(H_{l+i,l+j}(t_{0}+t)+H_{l+j,l+i}(t_{0}+t)\big)=R^{ij}_{t_{0}+t}$ by claim 1, so if $a\cdot R_{t}a\ge r|a|^{2}$ for all $t\in[0,T]$ and $a\in\mathbb{R}^{m}$ then $a\cdot R^{\sharp}_{t}a=a\cdot R_{t_{0}+t}a\ge r|a|^{2}$ for all $t\in[0,T^{\sharp}]$: hypothesis (H1) transports with the same $r$. For (U): by claim 1, for each $t\in[0,T^{\sharp}]$ the functions $\mathcal{H}^{\sharp}_{t}$ and $\mathcal{H}_{t_{0}+t}$ agree on $V$ and $A^{\sharp}_{t}=A_{t_{0}+t}$, so if $A_{t_{0}+t}$ is the unique minimizer of $\mathcal{H}_{t_{0}+t}$ on $\mathcal{A}$ then $A^{\sharp}_{t}$ is the unique minimizer of $\mathcal{H}^{\sharp}_{t}$ on $\mathcal{A}$. The transport of $r_{0}$ is the same substitution: for $t\in[0,T^{\sharp}]$ and $a\in\mathcal{A}$, $\mathcal{H}^{\sharp}_{t}(a)-\mathcal{H}^{\sharp}_{t}(A^{\sharp}_{t})=\mathcal{H}_{t_{0}+t}(a)-\mathcal{H}_{t_{0}+t}(A_{t_{0}+t})\ge r_{0}|a-A_{t_{0}+t}|^{2}=r_{0}|a-A^{\sharp}_{t}|^{2}$.

Step 3: proof of claim 3. The components of $A$ are continuous, hence measurable by claim 3 of the toolkit, and bounded in absolute value by $R=\sup_{\alpha\in\mathcal{A}}|\alpha|$, finite by claim 1 of the affine rate-family lemma, so each satisfies $\int_{[0,T]}(A^{j})^{2}\,d\lambda_{[0,T]}\le R^{2}\,T$ by monotonicity and claim 1 of the toolkit; thus $A\in\mathcal{L}^{2}([0,T];\mathbb{R}^{m})$, and since $A_{t}\in\mathcal{A}$ for every $t\in[0,T]$, the definition of the control set gives $[A]\in\mathcal{U}_{\mathcal{A}}$ with admissible representative $A$. The same applies to $A^{\sharp}$ on $[0,T^{\sharp}]$ (its components are continuous by Step 1), giving $[A^{\sharp}]\in\mathcal{U}^{[T^{\sharp}]}_{\mathcal{A}}$.

The pair $(S,A)$ is a generalized mean-field trajectory pair for $(\beta_{0},\beta_{1})$ with horizon $T$: condition 1 holds by clause 1 of the trajectory-pair definition (continuity implying measurability of the control components as above), and condition 2 holds because the Riemann integrals of clause 2 of that definition equal the Lebesgue integrals by claim 3 of the toolkit. Its value at $t=0$ is $S_{0}$, so by claim 2 of the existence and uniqueness theorem (applied to the initial value $S_{0}$ and the control $A$) $S$ is the map furnished by claim 1 of that theorem, which by claim 2 of the flow stability lemma (with the admissible representative $A$) is the flow: $S=S(S_{0},[A])$. By the definition of the mean-field cost (evaluated with the admissible representative $A$), $F(S_{0},[A])$ is the generalized mean-field cost of $(S,A)$, namely $\int_{[0,T]}L(S_{t},A_{t})\,dt+G(S_{T})$; since the Riemann integral defining the mean-field cost $J^{MF}[(S),(A)]$ equals the Lebesgue integral by claim 3 of the toolkit, $F(S_{0},[A])=J^{MF}[(S),(A)]$. The same argument applied to the shifted pair $(S^{\sharp},A^{\sharp})$ (a trajectory pair by Step 1, with value $S_{t_{0}}$ at $t=0$) gives $S^{\sharp}=S^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])$ and $F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])=J^{MF}[(S^{\sharp}),(A^{\sharp})]=\int_{[0,T^{\sharp}]}L(S^{\sharp}_{t},A^{\sharp}_{t})\,dt+G(S^{\sharp}_{T^{\sharp}})$. By (0.1) with $c=t_{0}$ (the integrand $t\mapsto L(S_{t},A_{t})$ restricted to $[t_{0},T]$ being measurable and bounded, as a continuous function on a compact interval by the provisions of the mean-field cost definition and restriction stability) and $S^{\sharp}_{T^{\sharp}}=S_{T}$,

$$F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])=\int_{[t_{0},T]}L(S_{t},A_{t})\,dt+G(S_{T}).$$

Finally, for $t_{0}>0$, (0.2) with $c=t_{0}$ splits $F(S_{0},[A])=\int_{[0,T]}L(S_{t},A_{t})\,dt+G(S_{T})$ as $\int_{[0,t_{0}]}L(S_{t},A_{t})\,dt+\int_{[t_{0},T]}L(S_{t},A_{t})\,dt+G(S_{T})$, which is the asserted splitting; for $t_{0}=0$ the first term is $0$ by the stated convention and the splitting reads $F(S_{0},[A])=F^{[T]}(S_{0},[A])$, which holds because the two instances coincide.

Step 4: proof of claim 4. Suppose first $t_{0}=0$. Then $\zeta_{s}=v_{s}$ for every $s\in[0,T]$ (the interval $[0,t_{0})$ being empty and $T^{\sharp}=T$), so $\zeta=v$ is an admissible representative of $[\zeta]=\eta$; any two admissible representatives of $\eta$ are equal off a set of measure zero by the definition of the Lebesgue space (two representatives of one class agree off a null set), so $[\zeta]$ does not depend on the choice; and the asserted identity reads $F(S_{0},\eta)=0+F^{[T]}(S_{0},\eta)$, which holds as in Step 3. Assume now $t_{0}>0$.

Admissible representative. Fix $j\in\{1,\dots,m\}$ and a Borel set $E\subseteq\mathbb{R}$. Then

$$(\zeta^{j})^{-1}(E)=\Big((A^{j})^{-1}(E)\cap[0,t_{0})\Big)\cup\big\{s\in[t_{0},T]:v^{j}_{s-t_{0}}\in E\big\}.$$

The first set is the intersection of a member of $\mathcal{B}_{[0,T]}$ with the Borel set $[0,t_{0})$, hence a member of $\mathcal{B}_{[0,T]}$. The second set is a member of $\mathcal{B}_{[t_{0},T]}$ by the measurability part of Step 0 (applied with $[a,b]=[t_{0},T]$, $c=-t_{0}$ and $f=v^{j}$, measurable on $[0,T^{\sharp}]$), hence a Borel subset of $[0,T]$ (claim 1 of the toolkit) and so a member of $\mathcal{B}_{[0,T]}$. Thus $\zeta^{j}$ is measurable; moreover $\zeta_{s}\in\mathcal{A}$ for every $s$, so $|\zeta^{j}|\le R$ with $R=\sup_{\alpha\in\mathcal{A}}|\alpha|$, finite by claim 1 of the affine rate-family lemma; being measurable and bounded, each component is square-integrable, so $\zeta\in\mathcal{L}^{2}([0,T];\mathbb{R}^{m})$ and $\zeta$ is an admissible representative of $[\zeta]\in\mathcal{U}_{\mathcal{A}}$. If $v'$ is another admissible representative of $\eta$, then $v=v'$ off a set $N\in\mathcal{B}_{[0,T^{\sharp}]}$ with $\lambda_{[0,T^{\sharp}]}(N)=0$, and the two concatenations agree off $N+t_{0}$, which is Borel with $\lambda(N+t_{0})=\lambda(N)=0$ by claim 1 of translation invariance (members of $\mathcal{B}_{[0,T^{\sharp}]}$ being Borel by claim 1 of the toolkit), and is a subset of $[t_{0},T]$, hence a member of $\mathcal{B}_{[0,T]}$ of measure zero. Two elements of $\mathcal{L}^{2}$ agreeing off a null set have the same class by the definition of the Lebesgue space, so $[\zeta]$ is independent of the choice of $v$.

The flow follows $S$, then the shifted flow. Write $x=S(S_{0},[\zeta])$, by claim 2 of the flow stability lemma (with the admissible representative $\zeta$) the map furnished by claim 1 of the existence and uniqueness theorem for the initial value $S_{0}$ and the control $\zeta$: $x$ is continuous, $x_{t}\in\Delta^{l}$, and

$$x^{\gamma}_{t}=S^{\gamma}_{0}+\int_{[0,t]}\hat{b}^{\gamma}(x_{s},\zeta_{s})\,ds\qquad(t\in[0,T],\ \gamma\in\{1,\dots,l\}),$$

the integrand being measurable and bounded by $2\sqrt{l}\,(l-1)B$: by claim 1 of the existence theorem the pair $(x,\zeta)$ is a generalized mean-field trajectory pair whose drift integrand agrees with the displayed one — $\hat{b}$ agreeing with $b$ on $\Delta^{l}\times\mathcal{A}$ by claim 6 of the affine rate-family lemma and $x_{s}\in\Delta^{l}$ — and the closing provisions of the generalized-pair definition record that measurability and bound. Restricting to $t\in[0,t_{0}]$, the pair of restrictions satisfies the same equation for the instance with horizon $t_{0}$ and the control $\zeta|_{[0,t_{0}]}$ (restrictions of measurable maps being measurable as in Step 0). On the other hand, define $q^{\gamma}:[0,t_{0}]\to\mathbb{R}$ by $q^{\gamma}(s)=\hat{b}^{\gamma}(S_{s},\zeta_{s})$. Off the single point $t_{0}$ we have $\zeta_{s}=A_{s}$ and hence, by claim 6 of the affine rate-family lemma ($\hat{b}=b$ on $\Delta^{l}\times\mathcal{A}$), $q^{\gamma}(s)=b^{\gamma}(S_{s},A_{s})$, a continuous function of $s$; at $s=t_{0}$, $q^{\gamma}(t_{0})$ is some real number. On $[0,t_{0}]$ we may write $q^{\gamma}=b^{\gamma}(S_{\cdot},A_{\cdot})\,\mathbf{1}_{[0,t_{0})}+q^{\gamma}(t_{0})\,\mathbf{1}_{\{t_{0}\}}$, where $\mathbf{1}_{D}$ denotes the indicator of a set $D$; the first factor is continuous, hence measurable by measurability of continuous functions, the indicators are measurable as indicators of the members $[0,t_{0})$ and $\{t_{0}\}$ of $\mathcal{B}_{[0,t_{0}]}$, and $q^{\gamma}$ is measurable by arithmetic of measurable functions; moreover $|q^{\gamma}|\le2\sqrt{l}\,(l-1)B$ by claim 6 of the affine rate-family lemma; so, by claim 2 of the null-set lemma and clause 2 of the trajectory-pair definition together with claim 3 of the toolkit, for every $t\in[0,t_{0}]$,

$$\int_{[0,t]}\hat{b}^{\gamma}(S_{s},\zeta_{s})\,ds=\int_{[0,t]}b^{\gamma}(S_{s},A_{s})\,ds=S^{\gamma}_{t}-S^{\gamma}_{0}.$$

Thus the restriction of $S$ to $[0,t_{0}]$ also satisfies the horizon-$t_{0}$ equation with the control $\zeta|_{[0,t_{0}]}$. By the uniqueness in claim 1 of the existence theorem (horizon $t_{0}$, initial value $S_{0}$, control $\zeta|_{[0,t_{0}]}$), $x_{t}=S_{t}$ for all $t\in[0,t_{0}]$; in particular $x_{t_{0}}=S_{t_{0}}$.

Now define $x^{\natural}:[0,T^{\sharp}]\to\Delta^{l}$ by $x^{\natural}_{t}=x_{t_{0}+t}$, continuous as in Step 1. For $t\in(0,T^{\sharp}]$, by the displayed equation for $x$ at $t_{0}+t$ and at $t_{0}$, (0.2) (split at $t_{0}$) and (0.1) (shift by $c=t_{0}$; $\zeta_{t_{0}+s}=v_{s}$ and $x_{t_{0}+s}=x^{\natural}_{s}$ for $s\in[0,T^{\sharp}]$),

$$x^{\natural\gamma}_{t}=x^{\gamma}_{t_{0}}+\int_{[t_{0},t_{0}+t]}\hat{b}^{\gamma}(x_{s},\zeta_{s})\,ds=S^{\gamma}_{t_{0}}+\int_{[0,t]}\hat{b}^{\gamma}(x^{\natural}_{s},v_{s})\,ds ,$$

and this also holds trivially at $t=0$. By the uniqueness in claim 1 of the existence theorem (horizon $T^{\sharp}$, initial value $S_{t_{0}}$, control $v$) and claim 2 of the flow stability lemma ($v$ being an admissible representative of $\eta$), $x^{\natural}=S^{[T^{\sharp}]}(S_{t_{0}},\eta)$.

Cost splitting. By the definition of the mean-field cost with the admissible representative $\zeta$ and by the generalized cost, whose provisions record that the integrand $t\mapsto L(x_{t},\zeta_{t})$ is measurable and bounded,

$$F(S_{0},[\zeta])=\int_{[0,T]}L(x_{t},\zeta_{t})\,dt+G(x_{T})=\int_{[0,t_{0}]}L(x_{t},\zeta_{t})\,dt+\int_{[t_{0},T]}L(x_{t},\zeta_{t})\,dt+G(x_{T})$$

by (0.2). On $[0,t_{0}]$ the integrand agrees off the single point $t_{0}$ with $t\mapsto L(S_{t},A_{t})$ (using $x=S$ on $[0,t_{0}]$ and $\zeta=A$ on $[0,t_{0})$), both being integrable (measurable and bounded, the latter continuous), so the first integral equals $\int_{[0,t_{0}]}L(S_{t},A_{t})\,dt$ by claim 2 of the null-set lemma. By (0.1) — its hypotheses holding because the restriction to $[t_{0},T]$ of the measurable bounded integrand $t\mapsto L(x_{t},\zeta_{t})$ is measurable (as in Step 0) and bounded — the second integral equals $\int_{[0,T^{\sharp}]}L(x^{\natural}_{t},v_{t})\,dt$, and $G(x_{T})=G(x^{\natural}_{T^{\sharp}})$; by the definition of $F^{[T^{\sharp}]}$ with the admissible representative $v$ and the identification $x^{\natural}=S^{[T^{\sharp}]}(S_{t_{0}},\eta)$, their sum is $F^{[T^{\sharp}]}(S_{t_{0}},\eta)$. This proves the asserted identity.

Step 5: proof of claim 5. Let $\eta\in\mathcal{U}^{[T^{\sharp}]}_{\mathcal{A}}$ be arbitrary and let $v$ be an admissible representative of $\eta$ (existing by claim 2 of the flow stability lemma for the horizon-$T^{\sharp}$ instance), with concatenation $\zeta$ as in claim 4. Since $J^{*}_{S_{0}}$ is the infimum of the value set $\mathcal{V}_{S_{0}}$ (notation of the statement), it is a lower bound of it, so $F(S_{0},[A])=J^{*}_{S_{0}}\le F(S_{0},[\zeta])$ (the equality by the assumption $[A]\in\mathcal{M}^{*}_{S_{0}}$ and the definition of the optimal control set). By claims 3 and 4,

$$\int_{[0,t_{0}]}L(S_{t},A_{t})\,dt+F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])=F(S_{0},[A])\le F(S_{0},[\zeta])=\int_{[0,t_{0}]}L(S_{t},A_{t})\,dt+F^{[T^{\sharp}]}(S_{t_{0}},\eta),$$

so $F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])\le F^{[T^{\sharp}]}(S_{t_{0}},\eta)$ for every $\eta\in\mathcal{U}^{[T^{\sharp}]}_{\mathcal{A}}$. Hence $F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])$ is a lower bound of the value set $\mathcal{V}^{[T^{\sharp}]}_{S_{t_{0}}}$ and a member of it; since $J^{*[T^{\sharp}]}_{S_{t_{0}}}$ is the greatest lower bound of that set, $J^{*[T^{\sharp}]}_{S_{t_{0}}}\ge F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])$, while as a lower bound it also satisfies $J^{*[T^{\sharp}]}_{S_{t_{0}}}\le F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])$; so the two are equal, $[A^{\sharp}]\in\mathcal{M}^{*[T^{\sharp}]}_{S_{t_{0}}}$ by the definition of the optimal control set, and the first asserted display follows from claim 3. Subtracting the splitting of claim 3 from $J^{*}_{S_{0}}=F(S_{0},[A])$ gives the second: $J^{*[T^{\sharp}]}_{S_{t_{0}}}=F^{[T^{\sharp}]}(S_{t_{0}},[A^{\sharp}])=F(S_{0},[A])-\int_{[0,t_{0}]}L(S_{t},A_{t})\,dt=J^{*}_{S_{0}}-\int_{[0,t_{0}]}L(S_{t},A_{t})\,dt$. $\blacksquare$