Define f : [ 0 , T ] × R l → R l f:[0,T]\times\mathbb{R}^l\to\mathbb{R}^l f : [ 0 , T ] × R l → R l by f ( t , y ) = b ^ ( y , A t ) f(t,y)=\hat{b}(y,A_t) f ( t , y ) = b ^ ( y , A t ) .
Step 1: f f f satisfies the hypotheses of the differential equation theorem. Fix y ∈ R l y\in\mathbb{R}^l y ∈ R l and γ \gamma γ . By the lemma on affine-controlled data , b ^ ( y , α ) = b ( π Δ l ( y ) , α ) \hat{b}(y,\alpha)=b(\pi_{\Delta^l}(y),\alpha) b ^ ( y , α ) = b ( π Δ l ( y ) , α ) and ∣ b ( Σ , α ) − b ( Σ , α ′ ) ∣ ≤ 2 l ( l − 1 ) K 1 ∣ α − α ′ ∣ |b(\Sigma,\alpha)-b(\Sigma,\alpha')|\le2\sqrt{l}(l-1)K_1|\alpha-\alpha'| ∣ b ( Σ , α ) − b ( Σ , α ′ ) ∣ ≤ 2 l ( l − 1 ) K 1 ∣ α − α ′ ∣ for Σ ∈ Δ l \Sigma\in\Delta^l Σ ∈ Δ l , so α ↦ b ^ γ ( y , α ) \alpha\mapsto\hat{b}^\gamma(y,\alpha) α ↦ b ^ γ ( y , α ) is sequentially continuous on R m \mathbb{R}^m R m . Since the components of A A A are measurable, measurability of sequentially continuous functions of measurable Euclidean maps shows that t ↦ f γ ( t , y ) = b ^ γ ( y , A t ) t\mapsto f^\gamma(t,y)=\hat{b}^\gamma(y,A_t) t ↦ f γ ( t , y ) = b ^ γ ( 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 ) ∣ = ∣ b ^ ( y , A t ) ∣ ≤ K b |f(t,y)|=|\hat{b}(y,A_t)|\le K_b ∣ f ( t , y ) ∣ = ∣ b ^ ( y , A t ) ∣ ≤ K b for all ( t , y ) (t,y) ( t , y ) , and
∣ f ( t , y ) − f ( t , y ′ ) ∣ = ∣ b ^ ( y , A t ) − b ^ ( y ′ , A t ) ∣ ≤ Λ b ∣ y − y ′ ∣ , |f(t,y)-f(t,y')|=|\hat{b}(y,A_t)-\hat{b}(y',A_t)|\le\Lambda_b|y-y'| , ∣ f ( t , y ) − f ( t , y ′ ) ∣ = ∣ b ^ ( y , A t ) − b ^ ( y ′ , A t ) ∣ ≤ Λ b ∣ y − y ′ ∣ ,
with Λ b = 2 l ( l − 1 ) ( B + Λ β ) \Lambda_b=2\sqrt{l}(l-1)(B+\Lambda_\beta) Λ b = 2 l ( l − 1 ) ( B + Λ β ) as in that lemma. Thus f f 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 K_b K b and Λ b \Lambda_b Λ b .
Step 2: the solution and its values. By that theorem there is exactly one continuous x : [ 0 , T ] → R l x:[0,T]\to\mathbb{R}^l x : [ 0 , T ] → R l with
x t γ = S 0 γ + ∫ [ 0 , t ] f γ ( s , x s ) d s = S 0 γ + ∫ [ 0 , t ] b ^ γ ( x s , A s ) d s ( t ∈ [ 0 , T ] , γ ∈ { 1 , … , l } ) , 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\}), x t γ = S 0 γ + ∫ [ 0 , t ] f γ ( s , x s ) d s = S 0 γ + ∫ [ 0 , t ] b ^ γ ( x s , A s ) d s ( t ∈ [ 0 , T ] , γ ∈ { 1 , … , l }) ,
its value at t = 0 t=0 t = 0 is S 0 S_0 S 0 , and it satisfies ∣ x t − x r ∣ ≤ K b ∣ t − r ∣ |x_t-x_r|\le K_b|t-r| ∣ x t − x r ∣ ≤ K b ∣ t − r ∣ , which is claim 3.
The hypotheses of the forward invariance lemma hold for this x x x : the components of A A A are measurable, x x x is continuous, x 0 = S 0 ∈ Δ l x_0=S_0\in\Delta^l x 0 = S 0 ∈ Δ l , and the displayed integral equation is exactly the one required. Hence x t ∈ Δ l x_t\in\Delta^l x t ∈ Δ l for every t ∈ [ 0 , T ] t\in[0,T] t ∈ [ 0 , T ] and b ^ ( x t , A t ) = b ( x t , A t ) \hat{b}(x_t,A_t)=b(x_t,A_t) b ^ ( x t , A t ) = b ( x t , A t ) for every t ∈ [ 0 , T ] t\in[0,T] t ∈ [ 0 , T ] .
Step 3: ( S , A ) (S,A) ( S , A ) is a generalized mean-field trajectory pair. Put S = x S=x S = x . Then S S S maps [ 0 , T ] [0,T] [ 0 , T ] into Δ l \Delta^l Δ l and A A A maps [ 0 , T ] [0,T] [ 0 , T ] into A \mathcal{A} A by hypothesis. Condition 1 of the definition holds: the components of S S S are continuous and those of A A A are measurable. For condition 2, the identity b ^ γ ( S s , A s ) = b γ ( S s , A s ) \hat{b}^\gamma(S_s,A_s)=b^\gamma(S_s,A_s) b ^ γ ( S s , A s ) = b γ ( S s , A s ) , valid for every s ∈ [ 0 , T ] s\in[0,T] s ∈ [ 0 , T ] by Step 2, turns the displayed integral equation into
S t γ = S 0 γ + ∫ [ 0 , t ] b γ ( S s , A s ) d s ( t ∈ [ 0 , T ] , γ ∈ { 1 , … , l } ) , 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\}), S t γ = S 0 γ + ∫ [ 0 , t ] b γ ( S s , A s ) d s ( t ∈ [ 0 , T ] , γ ∈ { 1 , … , l }) ,
as required. This proves claim 1.
Step 4: uniqueness. Let ( S ~ , A ) (\tilde{S},A) ( S ~ , A ) be a generalized mean-field trajectory pair with horizon T T T whose value at t = 0 t=0 t = 0 is S 0 S_0 S 0 . Then S ~ \tilde{S} S ~ is continuous with values in Δ l \Delta^l Δ l , so the fixed-point clause of the projection lemma gives π Δ l ( S ~ s ) = S ~ s \pi_{\Delta^l}(\tilde{S}_s)=\tilde{S}_s π Δ l ( S ~ s ) = S ~ s and hence b γ ( S ~ s , A s ) = b ^ γ ( S ~ s , A s ) b^\gamma(\tilde{S}_s,A_s)=\hat{b}^\gamma(\tilde{S}_s,A_s) b γ ( S ~ s , A s ) = b ^ γ ( S ~ s , A s ) for every s s s . Condition 2 of the definition therefore states that
S ~ t γ = S 0 γ + ∫ [ 0 , t ] b ^ γ ( S ~ s , A s ) d s = S 0 γ + ∫ [ 0 , t ] f γ ( s , S ~ s ) d s , \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 , S ~ t γ = S 0 γ + ∫ [ 0 , t ] b ^ γ ( S ~ s , A s ) d s = S 0 γ + ∫ [ 0 , t ] f γ ( s , S ~ s ) d s ,
so S ~ \tilde{S} S ~ is a continuous solution of the same integral equation as x x x with the same initial value. By the uniqueness clause of that theorem, S ~ = x = S \tilde{S}=x=S S ~ = x = S . ■ \blacksquare ■