Reason: First published version of the cost convergence proof, splitting on whether the uniform tracking error exceeds the modulus threshold and estimating the exceptional probability by Markov's inequality.
whose expectation is the N-agent costJN[hA], satisfies ∣V∣≤(T+1)C at every point of Ω∗. At every point of Ω it satisfies V≥−(TCL+CG), where CL and CG are the lower bounds belonging to the population cost data. Since P(Ω∗)=1, the one-sided version of the passage of almost sure inequalities to expectations, applied with the constant random variable U=(T+1)C, shows that JN[hA] is a real number with ∣JN[hA]∣≤(T+1)C.
By the tracking proposition, ∣Σt(ω)−St∣≤Ψ(ω) for every t∈[0,T]. If Ψ(ω)≤δ then the uniform continuity clause gives
for every t; and in all cases these differences are at most 2C. Writing 1{Ψ>δ} for the function equal to 1 where Ψ>δ and 0 elsewhere, we therefore have, at every ω∈Ω∗,
Adding the two displays and using the triangle inequality, the random variable V satisfies V−JMF[(S),(A)]≤(T+1)(ε+2C1{Ψ>δ}) at every ω∈Ω∗, the number JMF[(S),(A)] being exactly the sum of the two mean-field terms by the definition of the generalized mean-field cost. Since P(Ω∗)=1 and JN[hA]=E[V], the one-sided version of the passage of almost sure inequalities to expectations, applied to the random variable V−JMF[(S),(A)], which is bounded below at every point of Ω, and to the bounded random variable U=(T+1)(ε+2C1{Ψ>δ}), gives
the last equality by linearity of the integral, the expectation of the indicator of an event being its probability.
Finally, Ψ2 is a nonnegative random variable, so Markov's inequality gives
P(Ψ>δ)≤P(Ψ2≥δ2)≤δ2E[Ψ2],
which yields the stated bound.
Claim 2. Let η>0. Choose ε=η/(2(T+1)) and let δ>0 be as in the boundedness and uniform continuity lemma for this ε; note that C and δ do not depend on N. By the tracking proposition,
E[ΨN2]≤2e2ΛbT(E[∣Σ0N−S0∣2]+N8l(l−1)BT),
where ΨN is the random variable of that proposition for the N-th system. Both terms in the bracket converge to 0 as N increases, the first by hypothesis, so E[ΨN2] converges to 0. Choose N0 such that (T+1)2CE[ΨN2]/δ2<η/2 for every N≥N0. Then Claim 1 gives
JN[hA]−JMF[(S),(A)]<2η+2η=ηfor every N≥N0.
As η>0 was arbitrary, JN[hA]converges to JMF[(S),(A)]. ■