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 $\mathbb{R}^m$, 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

where the integral is the Riemann integral.

The pair is a generalized pair. By hypothesis $A_t\in\mathcal{A}$ for every $t$, so $A$ maps $[0,T]$ into $\mathcal{A}$. 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 its Riemann integral over $[0,t]$ coincides with its Lebesgue integral over the compact interval $[0,t]$, by the agreement of the two integrals for continuous integrands recorded in that toolkit. 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

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. $\blacksquare$