Proof of Existence and Uniqueness of the Generalized Mean-Field Trajectory for a Measurable Control

Define $f:[0,T]\times\mathbb{R}^l\to\mathbb{R}^l$ by $f(t,y)=\hat{b}(y,A_t)$.

Step 1: $f$ satisfies the hypotheses of the differential equation theorem. Fix $y\in\mathbb{R}^l$ and $\gamma$. By the lemma on affine-controlled data, $\hat{b}(y,\alpha)=b(\pi_{\Delta^l}(y),\alpha)$ and $|b(\Sigma,\alpha)-b(\Sigma,\alpha')|\le2\sqrt{l}(l-1)K_1|\alpha-\alpha'|$ for $\Sigma\in\Delta^l$, so $\alpha\mapsto\hat{b}^\gamma(y,\alpha)$ is sequentially continuous on $\mathcal{A}$. Since the components of $A$ are measurable and $A$ takes values in $\mathcal{A}$, measurability of sequentially continuous functions of measurable Euclidean maps, applied with $E=\mathcal{A}$, shows that $t\mapsto f^\gamma(t,y)=\hat{b}^\gamma(y,A_t)$ is measurable. The bound and the state-Lipschitz property of the projected drift in the lemma on affine-controlled data give $|f(t,y)|=|\hat{b}(y,A_t)|\le K_b$ for all $(t,y)$, and

$$|f(t,y)-f(t,y')|=|\hat{b}(y,A_t)-\hat{b}(y',A_t)|\le\Lambda_b|y-y'| ,$$

with $\Lambda_b=2\sqrt{l}(l-1)(B+\Lambda_\beta)$ as in that lemma. Thus $f$ satisfies hypotheses 1, 2 and 3 of the existence and uniqueness theorem for ordinary differential equations with measurable time dependence, with constants $K_b$ and $\Lambda_b$.

Step 2: the solution and its values. By that theorem there is exactly one continuous $x:[0,T]\to\mathbb{R}^l$ with

$$x^\gamma_t=S^\gamma_0+\int_{[0,t]}f^\gamma(s,x_s)\,ds=S^\gamma_0+\int_{[0,t]}\hat{b}^\gamma(x_s,A_s)\,ds\qquad(t\in[0,T],\ \gamma\in\{1,\dots,l\}),$$

its value at $t=0$ is $S_0$, and it satisfies $|x_t-x_r|\le K_b|t-r|$, which is claim 3.

The hypotheses of the forward invariance lemma hold for this $x$: the components of $A$ are measurable, $x$ is continuous, $x_0=S_0\in\Delta^l$, and the displayed integral equation is exactly the one required. Hence $x_t\in\Delta^l$ for every $t\in[0,T]$ and $\hat{b}(x_t,A_t)=b(x_t,A_t)$ for every $t\in[0,T]$.

Step 3: $(S,A)$ is a generalized mean-field trajectory pair. Put $S=x$. Then $S$ maps $[0,T]$ into $\Delta^l$ and $A$ maps $[0,T]$ into $\mathcal{A}$ by hypothesis. Condition 1 of the definition holds: the components of $S$ are continuous and those of $A$ are measurable. For condition 2, the identity $\hat{b}^\gamma(S_s,A_s)=b^\gamma(S_s,A_s)$, valid for every $s\in[0,T]$ by Step 2, turns the displayed integral equation into

$$S^\gamma_t=S^\gamma_0+\int_{[0,t]}b^\gamma(S_s,A_s)\,ds\qquad(t\in[0,T],\ \gamma\in\{1,\dots,l\}),$$

as required. This proves claim 1.

Step 4: uniqueness. Let $(\tilde{S},A)$ be a generalized mean-field trajectory pair with horizon $T$ whose value at $t=0$ is $S_0$. Then $\tilde{S}$ is continuous with values in $\Delta^l$, so the fixed-point clause of the projection lemma gives $\pi_{\Delta^l}(\tilde{S}_s)=\tilde{S}_s$ and hence $b^\gamma(\tilde{S}_s,A_s)=\hat{b}^\gamma(\tilde{S}_s,A_s)$ for every $s$. Condition 2 of the definition therefore states that

$$\tilde{S}^\gamma_t=S^\gamma_0+\int_{[0,t]}\hat{b}^\gamma(\tilde{S}_s,A_s)\,ds=S^\gamma_0+\int_{[0,t]}f^\gamma(s,\tilde{S}_s)\,ds ,$$

so $\tilde{S}$ is a continuous solution of the same integral equation as $x$ with the same initial value. By the uniqueness clause of that theorem, $\tilde{S}=x=S$. $\blacksquare$