Reason: First published version of the proof of the martingale bound, combining the path regularity of the empirical state measure, the right-continuous Doob inequality, and the published covariation identity.
Proof
Step 1: an almost sure event of good paths. By clause (a) of the martingale decomposition, there is an event Ω1∈F with P(Ω1)=1 such that for every ω∈Ω1 and every γ the path s↦bγ(Σs(ω),αs(ω)) is measurable on [0,T] and bounded in absolute value by 2(l−1)B. Put Ω∗=Ω0∩Ω1. Its complement is the union of the two events Ω∖Ω0 and Ω∖Ω1, each of probability 0, so P(Ω∗)=1 by additivity and monotonicity of the measure, as recorded in the assembly of a measure space; and Ω∗⊆Ω0.
Fix ω∈Ω∗ and γ. By condition 1 of the definition of a solution, each path t↦σti(ω) is constant on each interval of a finite partition of [0,T] into intervals that are closed on the left, so each occupation indicator ηi,γ and hence the empirical state measure Σγ=N1∑iηi,γ has this property too. In particular t↦Σtγ(ω) is right-continuous at every t∈[0,T), being constant on a right-neighbourhood of each such t, and it is measurable in t. The map t↦∫[0,t]bγ(Σs,αs)ds satisfies, for 0≤r≤t≤T,
by linearity and monotonicity of the integral, the single point r being a Lebesgue null set; hence it is continuous in t. Therefore t↦Mtγ(ω) is right-continuous at every t∈[0,T), and since Σt∈Δl gives ∣Σtγ∣≤1 and ∣Σ0γ∣≤1,
∣Mtγ(ω)∣≤1+1+2(l−1)BT=KM.
Step 2: Doob's inequality. By clause (b) of the martingale decomposition, each (Mtγ)t∈[0,T] is a square-integrable martingale with respect to the system filtration, with time index restricted to [0,T], and M0γ=0. Step 1 verifies the path hypotheses of Doob's L2 maximal inequality for bounded right-continuous martingales on the event Ω∗, with constant KM. Hence each Mγ is a random variable, with
Step 3: the terminal second moment. Apply the first covariation identity of clause (c) of the martingale decomposition with r=0, t=T, γ=δ and D=Ω, which lies in F0sys. Since M0γ=0,