Proof of Continuous Mean-Field Trajectory Pairs are Generalized Mean-Field Trajectory Pairs

By the definition of a mean-field trajectory pair, $S$ maps $[0,T]$ into $\Delta^l$, $A$ maps $[0,T]$ into $\mathcal{A}$, all components of $S$ and of $A$ are continuous on $[0,T]$, the map $s\mapsto b^\gamma(S_s,A_s)$ is continuous for every $\gamma$, and

$$S^\gamma_t=S^\gamma_0+\int_0^tb^\gamma(S_s,A_s)\,ds\qquad(t\in[0,T]),$$

where the integral is the Riemann integral.

The pair is a generalized pair. Condition 1 of the definition of a generalized mean-field trajectory pair holds: the components of $S$ are continuous, and the components of $A$, being continuous, are measurable with respect to the trace Borel $\sigma$-algebra on $[0,T]$ by measurability of continuous functions.

For condition 2, the integrand $s\mapsto b^\gamma(S_s,A_s)$ is continuous on $[0,T]$, hence for $t>0$ its Riemann integral over $[0,t]$ coincides with its Lebesgue integral over the compact interval $[0,t]$, by claim 3 of that toolkit; for $t=0$ both definitions set the integral to $0$, so the two dynamics conditions agree there as well. Since $S_s\in\Delta^l$ for every $s$, claim 6 of the affine-rate lemma gives $\hat b(S_s,A_s)=b(S_s,A_s)$, so the two definitions use the same integrand here. The displayed identity is therefore exactly condition 2 of the generalized definition. Hence $(S,A)$ is a generalized mean-field trajectory pair for $(\beta_0,\beta_1)$ with horizon $T$.

The two costs agree. By the definition of the mean-field cost, the map $t\mapsto L(S_t,A_t)$ is continuous on $[0,T]$ and

$$J^{MF}[(S),(A)]=\int_0^TL(S_t,A_t)\,dt+G(S_T)$$

with a Riemann integral, while by the definition of the generalized mean-field cost the same expression is formed with the Lebesgue integral of the same function over $[0,T]$ and the same terminal term $G(S_T)$. The two integrals agree because the integrand is continuous, again by the compact-interval toolkit. Hence the two costs are equal. $\blacksquare$