By Seppalainen T.

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4). Properties of random objects are often interpreted in such a way that they are not affected by events of probability zero. For example, let X = {Xt : t ∈ R+ } and Y = {Yt : t ∈ R+ } be two stochastic processes defined on the same probability space (Ω, F, P ). As functions on R+ × Ω, X and Y are equal if Xt (ω) = Yt (ω) for each ω ∈ Ω and t ∈ R+ . A useful relaxation of this strict notion of equality is called indistinguishability. 1. Filtrations and stopping times 37 Y are indistinguishable if there exists an event Ω0 ⊆ Ω such that P (Ω0 ) = 1 and for each ω ∈ Ω0 , Xt (ω) = Yt (ω) for all t ∈ R+ .

Tn ) and any B ∈ B t , P {x ∈ X : (xt1 , xt2 , . . , xtn ) ∈ B} = Qt (B). 35) We refer the reader to [1, Chapter 12] for a proof of Kolmogorov’s theorem in this generality. The appendix in [2] gives a proof for the case where I is countable and Xt = R for each t. The main idea of the proof is no different for the more abstract result. We will not discuss the proof. Let us observe that hypotheses (i) and (ii) are necessary for the existence of P , so nothing unnecessary is assumed in the theorem.

Proof. Part (i). Let A ∈ Fσ . For the first statement, we need to show that (A ∩ {σ ≤ τ }) ∩ {τ ≤ t} ∈ Ft . Write (A ∩ {σ ≤ τ }) ∩ {τ ≤ t} = (A ∩ {σ ≤ t}) ∩ {σ ∧ t ≤ τ ∧ t} ∩ {τ ≤ t}. All terms above lie in Ft . (i) The first by the definition of A ∈ Fσ . (ii) The second because both σ ∧ t and τ ∧ t are Ft -measurable random variables: for any u ∈ R, {σ ∧ t ≤ u} equals Ω if u ≥ t and {σ ≤ u} if u < t, a member of Ft in both cases. (iii) {τ ≤ t} ∈ Ft since τ is a stopping time. In particular, if σ ≤ τ , then Fσ ⊆ Fτ .